Sequentiality of String-to-Context Transducers

TL;DR

This paper extends the characterization of sequential transducers to string-to-context transducers, providing multiple criteria for their sequentiality and analyzing the computational complexity of the equivalence problem.

Contribution

It introduces the class of string-to-context transducers, extends Choffrut's results, and provides new characterizations and complexity results for their sequentiality.

Findings

Characterization based on Lipschitz property

Introduction of a new twinning property

Deciding sequentiality is in coNP

Abstract

Transducers extend finite state automata with outputs, and describe transformations from strings to strings. Sequential transducers, which have a deterministic behaviour regarding their input, are of particular interest. However, unlike finite-state automata, not every transducer can be made sequential. The seminal work of Choffrut allows to characterise, amongst the functional one-way transducers, the ones that admit an equivalent sequential transducer. In this work, we extend the results of Choffrut to the class of transducers that produce their output string by adding simultaneously, at each transition, a string on the left and a string on the right of the string produced so far. We call them the string-to-context transducers. We obtain a multiple characterisation of the functional string-to-context transducers admitting an equivalent sequential one, based on a Lipschitz property of the function realised by the transducer, and on a pattern (a new twinning property). Last, we prove that given a string-to-context transducer, determining whether there exists an equivalent sequential one is in coNP.

Equations

$$\text{textsl}{dist}_f(f(u),f(v))⩽|c_3{c}_2|+|d_3{d}_2|⩽M_{{\mathcal{T}}^{\prime }}.(2+|u^{\prime }|+|v^{\prime }|)⩽3M_{{\mathcal{T}}^{\prime }}\text{textsl}{dist}_p(u,v)$$

$$\begin{array}[]{lll}\textsl{dist}_{f}(e_{1}^{j}d_{1}c_{1}[\varepsilon],e_{2}^{j}d_{2}c_{2}[\varepsilon])&\leqslant&\textsl{dist}_{f}(g_{1}f_{1}e_{1}^{j}d_{1}c_{1}[\varepsilon],g_{2}f_{2}e_{2}^{j}d_{2}c_{2}[\varepsilon])+2(n+1)M_{\mathcal{T}}\\ &\leqslant&\textsl{dist}_{f}(f(\alpha_{j}),f(\beta_{j}))+2(n+1)M_{\mathcal{T}}\\ &\leqslant&K\textsl{dist}_{p}(\alpha_{j},\beta_{j})+2(n+1)M_{\mathcal{T}}\leqslant 2Kn+2(n+1)M_{\mathcal{T}}\end{array}$$

$$\text{textsl}{dist}_f(x,z)$$

$$⩽|x|+|z|-2|\gamma |$$

$$⩽|x_1|+|x_2|+|y_3|+|z_1|+|z_2|+|y_2|$$

$$⩽\text{textsl}{dist}_f(x,y)+\text{textsl}{dist}_f(y,z)$$

$$\text{textsl}{dist}_f(x,z)$$

$$⩽|x|+|z|-2|\beta |$$

$$⩽|x_1|+|x_2|+|y_3|+|y_4|+|z_1|+|z_2|$$

$$⩽\text{textsl}{dist}_f(x,y)+\text{textsl}{dist}_f(y,z)$$

$$\text{textsl}{dist}_f(x,z)$$

$$⩽|x|+|z|$$

$$⩽|x_1|+|x_2|+|y_3|+|z_1|+|z_2|+|y_2|$$

$$⩽\text{textsl}{dist}_f(x,y)+\text{textsl}{dist}_f(y,z)$$

$$\begin{array}[]{lll}\textsl{dist}_{f}(w,w^{\prime})&\leqslant&|w|+|w^{\prime}|-2|y|\\ &\leqslant&|w|+|w^{\prime}|-2|x|+2|c|+2|c^{\prime}|\\ &\leqslant&\textsl{dist}_{f}(c[w],c^{\prime}[w^{\prime}])+|c|+|c^{\prime}|\end{array}$$

$${\Delta }_{\overline{p}}[\epsilon ]$$

$$=(t_{\text{textsl}init}\bullet H_1{H}_2)\bullet H_3[\epsilon ]$$

$$=({\Delta }_{\overline{r}}c)\bullet H_3[\epsilon ]$$

$$=({\Delta }_{\overline{r}}\bullet H_3)c[\epsilon ]$$

$$={\Delta }_{\overline{s}}dc[\epsilon ]$$

$${\Delta }_{\overline{p}}[\epsilon ]$$

$$=(t_{\text{textsl}init}\bullet H_1{H}_2)\bullet H_3[\epsilon ]$$

$$=({\Delta }_{\overline{r}}c)\bullet H_3[\epsilon ]$$

$$={\Delta }_{\overline{r}}\bullet H_3[\epsilon ]$$

$$={\Delta }_{\overline{s}}d[\epsilon ]$$

$$={\Delta }_{\overline{s}}dc[\epsilon ]$$

$${\Delta }_{\overline{p}}[\epsilon ]$$

$$=(t_{\text{textsl}init}\bullet H_1{H}_2)\bullet H_3[\epsilon ]$$

$$=({\Delta }_{\overline{r}}c)\bullet H_3[\epsilon ]$$

$$={\Delta }_{\overline{r}}\bullet H_3[x^{\ell }]$$

$$={\Delta }_{\overline{s}}d[x^{\ell }]$$

$$={\Delta }_{\overline{s}}dc[\epsilon ]$$

$${\Delta }_{\overline{p}}[x^j]$$

$$=(t_{\text{textsl}init}\bullet H_1{H}_2)\bullet H_3[x^j]$$

$$=({\Delta }_{\overline{r}}c)\bullet H_3[x^j]$$

$$=({\Delta }_{\overline{r}}\bullet H_3)c[x^j]$$

$$={\Delta }_{\overline{s}}dc[x^j]$$

$${\Delta }_{\overline{p}}[x^j]$$

$$=(t_{\text{textsl}init}\bullet H_1{H}_2)\bullet H_3[x^j]$$

$$=({\Delta }_{\overline{r}}c)\bullet H_3[x^j]$$

$$={\Delta }_{\overline{r}}\bullet H_3[x^{j+\ell }]$$

$$={\Delta }_{\overline{s}}d[x^{j+\ell }]$$

$$={\Delta }_{\overline{s}}dc[x^j]$$

$$|d|⩽2(|Q{|}^2).(2M_{\mathcal{T}}|Q{|}^{|Q|})=4M_{\mathcal{T}}|Q{|}^{|Q|+2}$$

$$\begin{array}[]{ccl}u\in\textsf{dom}([\![\mathcal{T}]\!])&\iff&\textsf{dom}(t_{\textsl{init}}\bullet u)\cap\textsf{dom}(t_{\textsl{final}})\neq\varnothing\\ &\iff&\textsf{dom}(\Delta_{\overline{q}})\cap\textsf{dom}(t_{\textsl{final}})\neq\varnothing\\ &\iff&\overline{q}\in\textsf{dom}(\overline{t_{\textsl{final}}})\\ &\iff&u\in\textsf{dom}([\![\mathcal{D}]\!])\end{array}$$

$$\text{textsl}{dist}_f([[\mathcal{T}]](u_1),[[\mathcal{T}]](u_2))$$

Full text

In this work, we extend the results of Choffrut to the class of transducers that produce their output string by adding simultaneously, at each transition, a string on the left and a string on the right of the string produced so far. We call them the string-to-context transducers. We obtain a multiple characterisation of the functional string-to-context transducers admitting an equivalent sequential one, based on a Lipschitz property of the function realised by the transducer, and on a pattern (a new twinning property). Last, we prove that given a string-to-context transducer, determining whether there exists an equivalent sequential one is in coNP.

1 Introduction

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Transducers are a fundamental model to describe programs manipulating strings. They date back to the very first works in theoretical computer science, and are already present in the pioneering works on finite state automata [23, 1]. While finite state automata are very robust w.r.t. modifications of the model such as non-determinism and two-wayness, this is not the case for transducers. These two extensions do affect the expressive power of the model. Non-determinism is a feature very useful for modelisation and specification purposes. However, when one turns to implementation, deriving a sequential, i.e. input-deterministic, transducer is a major issue. A natural and fundamental problem thus consists, given a (non-deterministic) transducer, in deciding whether there exists an equivalent sequential transducer. This problem is called the sequentiality problem.

In [10], Choffrut addressed this problem for the class of functional (one-way) finite state transducers, which corresponds to so-called rational functions. He proved a multiple characterisation of the transducers admitting an equivalent sequential transducer. This characterisation includes a machine-independent property, namely a Lipschitz property of the function realised by the transducer. It also involves a pattern property, namely the twinning property, that allows to prove that the sequentiality problem is decidable in polynomial time for the class of functional finite state transducers [25]. This seminal work has led to developments on the sequentiality of finite state transducers [8, 7]. These results have also been extended to weighted automata [9, 19, 15] and to tree transducers [24]. See also [20] for a survey on sequentiality problems.

While the model of one-way transducers is now rather well-understood, a current challenge is to address the so-called class of regular functions, which corresponds to functions realised by two-way transducers. This class has attracted a lot of interest during the last years. It is closed under composition [11] and enjoys alternative presentations using logic [14], a deterministic one-way model equipped with registers, named streaming string transducers [2] (SST for short), as well as a set of regular combinators [4, 6, 12]. This class of functions is much more expressive, as it captures for instance the mirror image and the copy. Yet, it has good decidability properties: equivalence and type-checking are decidable in PSpace [18, 3]. We refer the interested reader to [16] for a recent survey. Intuitively, two-way finite state transducers (resp. SST) extend one-way finite state transducers with two important features: firstly, they can go through the input word both ways (resp. they can prepend and append words to registers), and secondly, they can perform multiple passes (resp. they can perform register concatenation).

In this paper, we lift the results of Choffrut [10] to a class of transducers that can perform the first of the two features mentioned above, thus generalising the class of rational functions. More precisely, we consider transducers which, at each transition, extend the output word produced so far by prepending and appending two words to it. This operation can be defined as the extension of a word with a context, and we call these transducers the string-to-context transducers. However, it is important to notice that that they still describe functions from strings to strings. We characterise the functional string-to-context transducers that admit an equivalent sequential string-to-context transducer through $i)$ a machine independent property: the function realised by the transducer satisfies a Lipschitz property that involves an original factor distance and $ii)$ a pattern property of the transducer which we call contextual twinning property, and that generalises the twinning property to contexts. We also prove that the sequentiality problem for these transducers is in the class coNP.

A key technical tool of the result of [10] was a combinatorial analysis of the loops, showing that the output words of synchronised loops have conjugate primitive roots. For string-to-context transducers, the situation is more complex, as the combinatorics may involve the words of the two sides of the context. Intuitively, when these words do commute with the output word produced so far, it is possible for instance to move to the right a part of the word produced on the left. In order to prove our results, we thus dig into the combinatorics of contexts associated with loops, identifying different possible situations, and we then use this analysis to describe an original determinisation construction.

Our results also have a strong connection with the register minimisation problem for SST. This problem consists in determining, given an SST and a natural number $k$, whether there exists an equivalent SST with $k$ registers. It has been proven in [13] that the problem is decidable for SST that can only append words to registers, and the proof crucially relies on the fact that the $k=1$ case exactly corresponds to the sequentiality problem of one-way finite state transducers. Hence, our results constitute a first step towards register minimisation for SST without register concatenation. The register minimisation problem for non-deterministic SST has also been studied in [5] for the case of concatenation-free SST. The targeted model being non-deterministic, the two problems are independent.

Due to lack of space, omitted proofs can be found in the Appendix.

2 Models

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Words, contexts and partial functions

Let $A$ be a finite alphabet. The set of finite words (or strings) over $A$ is denoted by $A^{*}$. The empty word is denoted by $\epsilon$. The length of a word $u$ is denoted by $|u|$. We say that a word $u$ is a prefix (resp. suffix) of a word $v$ if there exists a word $y$ such that $uy=v$ (resp. $yu=v$). We say that two words $u,v\in A^{*}$ are conjugates if there exist two words $t_{1},t_{2}\in A^{*}$ such that $u=t_{1}t_{2}$ and $v=t_{2}t_{1}$. If this holds, we write $u\sim v$. The primitive root of a word $u\in A^{*}$, denoted $\rho(u)$, is the shortest word $x$ such that $u=x^{p}$ for some $p\geqslant 1$.

Lemma 2.1** ([17]).**

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page\cref@result Let $u,v\in A^{*}$. There exists $n\in\mathbb{N}$ such that if there is a common factor of $u$ and $v$ of length at least $n$, then $\rho(u)\sim\rho(v)$.

Given two words $u,v\in A^{*}$, the longest common prefix (resp. suffix) of $u$ and $v$ is denoted by $\textsf{lcp}(u,v)$ (resp. $\textsf{lcs}(u,v)$). We define the prefix distance between $u$ and $v$, denoted by $\textsl{dist}_{p}(u,v)$, as $|u|+|v|-2|\textsf{lcp}(u,v)|$.

Given a word $u\in B^{*}$, we say that $v$ is a factor of $u$ if there exist words $x,y$ such that $u=xvy$. Given two words $u,v\in B^{*}$, a longest common factor of $u$ and $v$ is a word $w$ of maximal length that is a factor of both $u$ and $v$. Note that this word is not necessarily unique. We denote such a word by $\textsf{lcf}(u,v)$. The factor distance between $u$ and $v$, denoted by $\textsl{dist}_{f}(u,v)$, is defined as $\textsl{dist}_{f}(u,v)=|u|+|v|-2|\textsf{lcf}(u,v)|$. This definition is correct as $|\textsf{lcf}(u,v)|$ is independent of the choice of the common factor of maximal length.

Using a careful case analysis, we can prove that $\textsl{dist}_{f}$ is indeed a distance, the only difficulty lying in the subadditivity:

Lemma 2.2**.**

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page\cref@result $\textsl{dist}_{f}$ is a distance.

Given a finite alphabet $B$, a context on $B$ is a pair of words $(u,v)\in B^{*}\times B^{*}$. The set of contexts on $B$ is denoted $\mathcal{C}(B)$. The empty context is denoted by $c_{\varepsilon}$. For a context $c=(u,v)$, we denote by \overleftarrow{\makebox{c\vphantom{x}}} (resp. \overrightarrow{\makebox{c\vphantom{x}}}) its left (resp. right) component: \overleftarrow{\makebox{c\vphantom{x}}}=u (resp. \overrightarrow{\makebox{c\vphantom{x}}}=v). The length of a context $c$ is defined by |c|=|\overleftarrow{\makebox{c\vphantom{x}}}|+|\overrightarrow{\makebox{c\vphantom{x}}}|. The lateralized length of a context $c$ is defined by \|c\|=(|\overleftarrow{\makebox{c\vphantom{x}}}|,|\overrightarrow{\makebox{c\vphantom{x}}}|). For a context $c\in\mathcal{C}(B)$ and a word $w\in B^{*}$, we write $c[w]$ for the word \overleftarrow{\makebox{c\vphantom{x}}}w\overrightarrow{\makebox{c\vphantom{x}}}. We define the concatenation of two contexts $c_{1},c_{2}\in\mathcal{C}(B)$ as the context c_{1}c_{2}=(\overleftarrow{\makebox{c_{1}\vphantom{x}}}\overleftarrow{\makebox{c_{2}\vphantom{x}}},\overrightarrow{\makebox{c_{2}\vphantom{x}}}\overrightarrow{\makebox{c_{1}\vphantom{x}}}). Last, given a context $c$ and a word $u$, we denote by $c^{-1}[u]$ the unique word $v$ such that $c[v]=u$, when such a word exists.

Given a set of contexts $C\subseteq\mathcal{C}(B)$, we denote by $\textsf{lcc}(C)$ the longest common context of elements in $C$, defined as \textsf{lcc}(C)=(\textsf{lcs}(\{\overleftarrow{\makebox{c\vphantom{x}}}\mid c\in C\}),\textsf{lcp}(\{\overrightarrow{\makebox{c\vphantom{x}}}\mid c\in C\})). We also write $C.\textsf{lcc}(C)^{-1}=\{c^{\prime}\mid c^{\prime}.\textsf{lcc}(C)\in C\}$.

We consider two sets $X,Y$. Given $\Delta\subseteq X\times Y$, we let $\textsf{dom}(\Delta)=\{x\in X\mid\exists y,(x,y)\in\Delta\}$. We denote the set of partial functions from $X$ to $Y$ as ${\cal F}(X,Y)$. Given $f\in{\cal F}(X,Y)$, we write $f:X\hookrightarrow Y$, and we denote by $\textsf{dom}(f)$ its domain. When more convenient, we may also see elements of ${\cal F}(X,Y)$ as subsets of $X\times Y$. Last, given $\Delta\subseteq X\times Y$, we let $\mathsf{choose}(\Delta)$ denote some $\Delta^{\prime}\in{\cal F}(X,Y)$ such that $\Delta^{\prime}\subseteq\Delta$ and $\textsf{dom}(\Delta)=\textsf{dom}(\Delta^{\prime})$.

String-to-Context and String-to-String Transducers

Definition 2.3**.**

Let $A,B$ be two finite alphabets. A string-to-context transducer (S2C for short) $\mathcal{T}$ from $A^{*}$ to $B^{*}$ is a tuple $(Q,t_{\textsl{init}},t_{\textsl{final}},T)$ where $Q$ is a finite set of states, $t_{\textsl{init}}:Q\hookrightarrow\mathcal{C}(B)$ (resp. $t_{\textsl{final}}:Q\hookrightarrow\mathcal{C}(B)$) is the finite initial (resp. final) function, $T\subseteq Q\times A\times\mathcal{C}(B)\times Q$ is the finite set of transitions.

A state $q$ is said to be initial (resp. final) if $q\in\textsf{dom}(t_{\textsl{init}})$ (resp. $q\in\textsf{dom}(t_{\textsl{final}})$). We depict as as \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}q (resp. q\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}) the fact that $t_{\textsl{init}}(q)=c$ (resp. $t_{\textsl{final}}(q)=c$). A run $\rho$ from a state $q_{1}$ to a state $q_{k}$ on a word $w=w_{1}\dotsm w_{k}\in A^{*}$ where for all $i$, $w_{i}\in A$, is a sequence of transitions: $(q_{1},w_{1},c_{1},q_{2}),(q_{2},w_{2},c_{2},q_{3}),\ldots,(q_{k},w_{k},c_{k},q_{k+1})$. The output of such a run is the context $c=c_{k}\dots c_{2}c_{1}\in\mathcal{C}(B)$, and is denoted by $\textsf{out}(\rho)$. We depict this situation as q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}w}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}q_{k+1}. The set of runs of $\mathcal{T}$ is denoted $\mathcal{R}(\mathcal{T})$. The run $\rho$ is said to be accepting if $q_{1}$ is initial and $q_{k+1}$ final. This string-to-context transducer $\mathcal{T}$ computes a relation $[\![\mathcal{T}]\!]\subseteq A^{*}\times B^{*}$ defined by the set of pairs $(w,edc[\varepsilon])$ such that there are $p,q\in Q$ with \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}w}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}. Thus, even if its definition involves contexts on $B$, the semantics of $\mathcal{T}$ is a relation between words on $A$ and words on $B$. Given an S2C $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$, we define the constant $M_{\mathcal{T}}$ as $M_{\mathcal{T}}=\max\{|c|\mid(p,a,c,q)\in T\text{ or }(q,c)\in t_{\textsl{init}}\cup t_{\textsl{final}}\}$. Given $\Delta:Q\hookrightarrow\mathcal{C}(B)$, we denote by $\mathcal{T}_{\Delta}$ the S2C obtained by replacing $t_{\textsl{init}}$ with $\Delta$. An S2C is trimmed if each of its states appears in some accepting run. W.l.o.g., we assume that the string-to-context transducers we consider are trimmed. An S2C $\mathcal{T}$ from $A^{*}$ to $B^{*}$ is functional if the relation $[\![\mathcal{T}]\!]$ is a function from $A^{*}$ to $B^{*}$. An S2C $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$ is sequential if $\textsf{dom}(t_{\textsl{init}})$ is a singleton and if for every transitions $(p,a,c,q),(p,a,c^{\prime},q^{\prime})\in T$, we have $q=q^{\prime}$ and $c=c^{\prime}$.

The classical model of finite-state transducers is recovered in the following definition:

Definition 2.4**.**

Let $A,B$ be two finite alphabets. A string-to-context transducer $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$ is a string-to-string transducer (S2S for short) from $A^{*}$ to $B^{*}$ if, for all $(q,c)\in t_{\textsl{init}}\cup t_{\textsl{final}}$, \overleftarrow{\makebox{c\vphantom{x}}}=\varepsilon, and for all $(q,a,c,q^{\prime})\in T$, \overleftarrow{\makebox{c\vphantom{x}}}=\varepsilon.

Notations defined for S2C hold for classical transducers as is. For an S2S, we write \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w}}q (resp. q\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w}}, and q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w}}q^{\prime}) instead of \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(\varepsilon,w)}}q (resp. q\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(\varepsilon,w)}}, and q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(\varepsilon,w)}}q^{\prime}).

Given an S2C $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$, we define its right S2S, denoted \overrightarrow{\makebox{\mathcal{T}\vphantom{x}}}, as the tuple (Q,\overrightarrow{t_{\textsl{init}}},\overrightarrow{t_{\textsl{final}}},\overrightarrow{\makebox{T\vphantom{x}}}) where, for all $q\in Q$, $\overrightarrow{t_{\textsl{init}}}(q)=\overrightarrow{t_{\textsl{init}}(q)}$ and $\overrightarrow{t_{\textsl{final}}}(q)=\overrightarrow{t_{\textsl{final}}(q)}$, and, for all $(p,a,c,q)\in T$, (p,a,\overrightarrow{\makebox{c\vphantom{x}}},q)\in\overrightarrow{\makebox{T\vphantom{x}}}. Its left S2S \overleftarrow{\makebox{\mathcal{T}\vphantom{x}}} is defined similarly, and by applying the mirror image on its output labels.

Example 2.5**.**

Two examples of S2C (not realisable by S2S) are depicted on \Crefe:StoC.

3 Lipschitz and Twinning Properties

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We recall the properties considered in [10], and the associated results.

Definition 3.1**.**

*We say that a function $f:A^{*}\hookrightarrow B^{*}$ satisfies the Lipschitz property if there exists $K\in\mathbb{N}$ such that $\forall u,v\in\textsf{dom}(f),\textsl{dist}_{p}(f(u),f(v))\leqslant K.\textsl{dist}_{p}(u,v)$. *

Definition 3.2**.**

We consider an S2S and $L\in\mathbb{N}$. Two states $q_{1}$ and $q_{2}$ are said to be $L$-twinned if for any two runs \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}y_{1}}}q_{1} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}y_{2}}}q_{2}, where $p_{1}$ and $p_{2}$ are initial, we have for all $j\geqslant 0$, $\textsl{dist}_{p}(w_{1}x_{1}y_{1}^{j},w_{2}x_{2}y_{2}^{j})\leqslant L$. An S2S satisfies the twinning property (TP) if there exists $L\in\mathbb{N}$ such that any two of its states are $L$-twinned.

Theorem 3.3** ([10]).**

Let $\mathcal{T}$ be a functional S2S. The following assertions are equivalent:

there exists an equivalent sequential S2S, 2. 2.

$[\![\mathcal{T}]\!]$* satisfies the Lipschitz property,* 3. 3.

$\mathcal{T}$* satisfies the twinning property.*

We present the adaptation of these properties to string-to-context transducers.

Definition 3.4**.**

*We say that $f:A^{*}\hookrightarrow B^{*}$ satisfies the contextual Lipschitz property (CLip) if there exists $K\in\mathbb{N}$ such that $\forall u,v\in\textsf{dom}(f),\textsl{dist}_{f}(f(u),f(v))\leqslant K.\textsl{dist}_{p}(u,v)$. *

Definition 3.5**.**

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page\cref@result We consider an S2C and $L\in\mathbb{N}$. Two states $q_{1}$ and $q_{2}$ are said to be $L$-contextually twinned if for any two runs \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2}, where $p_{1}$ and $p_{2}$ are initial, we have for all $j\geqslant 0$, $\textsl{dist}_{f}(e_{1}^{j}d_{1}c_{1}[\varepsilon],e_{2}^{j}d_{2}c_{2}[\varepsilon])\leqslant L$. An S2C satisfies the contextual twinning property (CTP) if there exists $L\in\mathbb{N}$ such that any two of its states are $L$-contextually twinned.

4 Main Result

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page\cref@result

The main result of the paper is the following theorem, which extends to string-to-context transducers the characterisation of sequential transducers amongst functional ones.

Theorem 4.1**.**

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page\cref@result Let $\mathcal{T}$ be a functional S2C. The following assertions are equivalent:

there exists an equivalent sequential string-to-context transducer, 2. 2.

$[\![\mathcal{T}]\!]$* satisfies the contextual Lipschitz property,* 3. 3.

$\mathcal{T}$* satisfies the contextual twinning property.*

Proof 4.2**.**

The implications $1\Rightarrow 2$ and $2\Rightarrow 3$ are proved in \Crefr:det-implies-lip and \Crefr:lip-implies-ctp respectively. The implication $3\Rightarrow 1$ is more involved, and is based on a careful analysis of word combinatorics of loops of string-to-context transducers satisfying the CTP. This analysis is summarised in \Crefr:ctp-implies-2-loop and used in \Crefsec:construction to describe the construction of an equivalent sequential S2C.

Proposition 4.3**.**

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page\cref@result Let $\mathcal{T}$ be a functional S2C realizing the function $f$. If there exists an equivalent sequential S2C, then $f$ satisfies the contextual Lipschitz property.

Proof 4.4**.**

Let us consider $\mathcal{T}^{\prime}$ the equivalent sequential S2C. We claim that $f$ is context-Lipschitzian with coefficient $3M_{\mathcal{T}^{\prime}}$. Consider two input words $u,v$ in the domain of $f$. If $u=v$, then the result is trivial. Otherwise, let $w=\textsf{lcp}(u,v)$ and let $u=w.u^{\prime}$, with $0\leqslant|u^{\prime}|$. Then we have $[\![\mathcal{T}^{\prime}]\!](u)=c_{3}c_{2}c_{1}[\epsilon]$ where $c_{1}$ is the context produced along $w$, $c_{2}$ the one produced along $u^{\prime}$, and $c_{3}$ is the final output context. Similarly, we can write (with $v=w.v^{\prime}$, and $0\leqslant|v^{\prime}|$) $[\![\mathcal{T}^{\prime}]\!](v)=d_{3}d_{2}d_{1}[\epsilon]$. As $\mathcal{T}^{\prime}$ is sequential, we have $d_{1}=c_{1}$. We also have $|c_{3}|\leqslant M_{\mathcal{T}^{\prime}}$, $|d_{3}|\leqslant M_{\mathcal{T}^{\prime}}$, $|c_{2}|\leqslant M_{\mathcal{T}^{\prime}}.|u^{\prime}|$ and $|d_{2}|\leqslant M_{\mathcal{T}^{\prime}}.|v^{\prime}|$. Finally, as $u\neq v$, we have $\textsl{dist}_{p}(u,v)=|u^{\prime}|+|v^{\prime}|\geqslant 1$ and we obtain:

[TABLE]

Proposition 4.5**.**

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page\cref@result Let $\mathcal{T}$ be a functional S2C realizing the function $f$. If $f$ satisfies the contextual Lipschitz property, then $\mathcal{T}$ satisfies the contextual twinning property.

Proof 4.6**.**

We consider an instance of the CTP and stick to the notations of \Crefd:ctp. We denote by $n$ the number of states of $\mathcal{T}$. As $\mathcal{T}$ is trimmed, there exist runs q_{i}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}w_{i}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}f_{i}}}r_{i}\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}g_{i}}}, with $|w_{i}|\leqslant n$, for $i\in\{1,2\}$. We consider the input words $\alpha_{j}=uv^{j}w_{1}$ and $\beta_{j}=uv^{j}w_{2}$, for all $j\geqslant 0$. We have, for every $j$, $\textsl{dist}_{p}(\alpha_{j},\beta_{j})\leqslant|w_{1}|+|w_{2}|\leqslant 2n$.

*The following property of $\textsl{dist}_{f}$ can be proven using a case analysis:

Fact. For every $w,w^{\prime}\in B^{*}$, $c,c^{\prime}\in\mathcal{C}(B)$, we have $\textsl{dist}_{f}(w,w^{\prime})\leqslant\textsl{dist}_{f}(c[w],c^{\prime}[w^{\prime}])+|c|+|c^{\prime}|$.*

As $f$ is $K$ context-Lipschitzian, for some fixed $K$, we obtain, for all $j$:

[TABLE]

5 Analysis of Loop Combinatorics

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page\cref@result

The classical twinning property forces the outputs of two runs reading the same input to only diverge by a finite amount. This constraint in turn makes for strong combinatorial bindings between runs involving loops: for two runs \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}y_{1}}}q_{1} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}w_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}y_{2}}}q_{2}, we have $|y_{1}|=|y_{2}|$, and $\rho(y_{1})\sim\rho(y_{2})$. Similar behaviours are expected with string-to-context transducers and lead us to study the combinatorial properties of synchronised runs involving loops in those machines. Throughout this section, we consider a string-to-context transducer $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$ that satisfies the contextual twinning property.

5.1 Behaviours of Loops

We start with two examples illustrating how output contexts of synchronised loops can be modified to obtain an equivalent sequential S2C.

Example 5.1**.**

\cref@constructprefix

page\cref@result \Crefe:s2c-com-non-det shows an example of a non-sequential functional S2C transducer $\mathcal{T}_{1}$. The contexts produced on loops around states $q_{1}$ and $q_{2}$ both commute with word $a$. This observation can be used to build an equivalent sequential S2C $\mathcal{D}_{1}$, depicted on \Crefe:s2c-com-det. \Crefe:s2c-align-non-det shows an example of a non-sequential functional S2C transducer $\mathcal{T}_{2}$ where output contexts are non-commuting, but can be slightly shifted so as to be aligned. This observation can be used to build an equivalent sequential S2C $\mathcal{D}_{2}$, depicted on \Crefe:s2c-align-det.

The following definition follows from the intuition drawn by the previous example.

Definition 5.2** (Lasso, Aligned/Commuting/Non-commuting lasso).**

\cref@constructprefix

page\cref@result\cref@constructprefixpage\cref@result A lasso around a state $q$ is a run $\rho$ of the form \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q with $p$ an initial state. $\rho$ is said to be productive, if $|e|\neq 0$. We say that $\rho$ is:

• •

aligned w.r.t. $f$ and $w$, for some $f\in\mathcal{C}(B)$ and $w\in B^{*}$, denoted as $(f,w)\mathrm{-aligned}$, if there exists a context $g\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}$, $e^{i}dc[\varepsilon]=gf^{i}[w]$.

• •

commuting w.r.t. $x$, for some $x\in B^{+}$, denoted as $x\mathrm{-commuting}$, if there exists a context $f\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}_{>0}$, there exists $k\in\mathbb{N}$ such that $e^{i}dc[\varepsilon]=f[x^{k}]$.

• •

non-commuting if there exists no word $x\in B^{+}$ such that $\rho$ is commuting w.r.t $x$.

Two lassos \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2} are said to be synchronised if $u_{1}=u_{2}$ and $v_{1}=v_{2}$. They are said to be strongly balanced if $\|e_{1}\|=\|e_{2}\|$.

Given an integer $k\geqslant 1$, we consider the $k$-th power of $\mathcal{T}$, that we denote by $\mathcal{T}^{k}$. A run in $\mathcal{T}^{k}$ naturally corresponds to $k$ synchronised runs in $\mathcal{T}$, i.e. on the same input word. We lift the notion of lasso to $\mathcal{T}^{k}$, and we denote them by $H_{1}H_{2}$, where $H_{1}$ starts in initial states and ends in some state $v=(q_{i})_{i\in\{1,\ldots,k\}}\in Q^{k}$, and $H_{2}$ is a loop around state $v$. In the sequel, we will only consider lassos such that $v$ contains pairwise distinct states ($q_{i}\neq q_{j}$ for all $i\neq j$). Those lassos are included in the lassos in $\mathcal{T}^{\leqslant|Q|}=\cup_{1\leqslant k\leqslant|Q|}\mathcal{T}^{k}$.

The intuition given by \Crefe:commuting-aligned is formalised in the following Lemma:

Lemma 5.3**.**

\cref@constructprefix

page\cref@result \cref@constructprefixpage\cref@result\cref@constructprefixpage\cref@result Let $H_{1}H_{2}=(\rho_{j})_{j\in\{1,\dots,k\}}$ a lasso in $\mathcal{T}^{k}$, for some $1\leqslant k\leqslant|Q|$. We write \rho_{j}:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{j}}}p_{j}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{j}}}q_{j}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{j}}}q_{j} for each $j$. Then there exists an integer $m\in\mathbb{N}$ such that $|e_{j}|=m$ for all $j\in\{1,\ldots,k\}$. If $m>0$, we say that the lasso $H_{1}H_{2}$ is productive, and:

• •

either there exists $x\in B^{+}$ primitive such that $\rho_{j}$ is $x\mathrm{-commuting}$ for all $j\in\{1,\dots,k\}$. In this case, we say that the lasso $H_{1}H_{2}$ is $x\mathrm{-commuting}$, and we let $\mathsf{pow}_{c}(x,H_{1},H_{2})=m/{|x|}$ and $\mathsf{split}_{c}(x,H_{1},H_{2})=\{(q_{j},f_{j})\mid j\in\{1,\dots,k\}\}$ where $f_{j}\in\mathcal{C}(B)$ is such that $\forall\alpha\in\mathbb{N},e_{j}^{\alpha}d_{j}c_{j}[\varepsilon]=f_{j}[x^{\alpha\ \mathsf{pow}_{c}(x,H_{1},H_{2})}]$.

• •

or there exist $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that $\rho_{j}$ is non-commuting and $(f,w)\mathrm{-aligned}$ for all $j\in\{1,\dots,k\}$. In this case, we say that the lasso $H_{1}H_{2}$ is $(f,w)\mathrm{-aligned}$, and we let $\mathsf{split}_{nc}(f,w,H_{1},H_{2})=\{(q_{j},g_{j})\mid j\in\{1,\dots,k\}\}$ where $g_{j}\in\mathcal{C}(B)$ is such that $\forall\alpha\in\mathbb{N},e_{j}^{\alpha}d_{j}c_{j}[\varepsilon]=g_{j}f^{\alpha}[w]$.

Proof 5.4** (Proof Sketch).**

*As $\mathcal{T}$ satisfies the CTP, the outputs must grow at the same pace when the loops are pumped. This entails that the lengths of the $e_{j}$ must be equal. Next, the result is proved by considering two productive synchronised lassos, with loops producing respectively $e_{1}$ and $e_{2}$. If they are not strongly balanced or one of them is $x\mathrm{-commuting}$, for some $x\in B^{+}$, then, using the result of Fine and Wilf (\Crefl:Fine) between \overleftarrow{\makebox{e_{1}\vphantom{x}}},\overleftarrow{\makebox{e_{2}\vphantom{x}}},\overrightarrow{\makebox{e_{1}\vphantom{x}}} and \overrightarrow{\makebox{e_{2}\vphantom{x}}}, we can prove that the other one is also $x\mathrm{-commuting}$. Otherwise, they are both non-commuting and strongly balanced. Using again \Crefl:Fine but first between \overleftarrow{\makebox{e_{1}\vphantom{x}}} and \overleftarrow{\makebox{e_{2}\vphantom{x}}}, and then between \overrightarrow{\makebox{e_{1}\vphantom{x}}} and \overrightarrow{\makebox{e_{2}\vphantom{x}}}, we prove that there exist $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that $\rho_{1}$ and $\rho_{2}$ are $(f,w)\mathrm{-aligned}$. Finally, the result is lifted to $k$ productive synchronised lassos. *

Example 5.5**.**

We consider the example S2C in \Crefe:s2c-loops. The lasso in $\mathcal{T}_{1}^{2}$ around $(q_{1},q_{2})$ is $\mathsf{a}\mathrm{-commuting}$. We can compute a $\mathsf{pow}_{c}$ of $2$ and $\{(q_{1},(\mathsf{a},\mathsf{a})),(q_{2},(\mathsf{b},\mathsf{a}))\}$ as a possible $\mathsf{split}_{c}$. The lasso in $\mathcal{T}_{2}^{2}$ around $(q_{1},q_{2})$ is $((\mathsf{ab},\mathsf{de}),\mathsf{c})\mathrm{-aligned}$. We can compute $\{(q_{1},c_{\varepsilon}),(q_{2},(\mathsf{b},\mathsf{d}))\}$ as a possible $\mathsf{split}_{nc}$.

5.2 Analysis of Loops Consecutive to a Productive Loop

Consider a run that contains two consecutive productive loops. We can observe that the type (commuting or non-commuting) of the lasso involving the first loop impacts the possible types of the lasso involving the second loop. For instance, it is intuitive that a non-commuting lasso cannot be followed by a commuting lasso. Similarly, an $x\mathrm{-commuting}$ lasso cannot be followed by an $y\mathrm{-commuting}$ lasso, if $x$ and $y$ are not conjugates. We will see that loops following a first productive loop indeed satisfy stronger combinatorial properties. The following definition characterises their properties.

Definition 5.6** (Strongly commuting/Strongly aligned lasso).**

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page\cref@result\cref@constructprefixpage\cref@result Let $\rho$ be a productive lasso \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q and $x\in B^{+}$. We say that $\rho$ is:

• •

strongly commuting w.r.t. $x$, denoted as $\mathrm{strongly-}x\mathrm{-commuting}$, if there exists a context $f\in\mathcal{C}(B)$ such that for all $i,j\in\mathbb{N}_{>0}$, there exists $k\in\mathbb{N}$ such that $e^{i}dc[x^{j}]=f[x^{k}]$.

• •

*strongly aligned w.r.t. $g$, $f$ and $x$, denoted as $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$, if there exists a context $h\in\mathcal{C}(B)$ such that for all $i,j\in\mathbb{N}$, $e^{j}dc[x^{i}]=hg^{j}f[x^{i}]$. *

The following Lemma states the properties of a lasso consecutive to a commuting lasso. To prove it, we proceed as for \Crefr:all-commuting-or-aligned by proving the result first for two runs and then lifting it to $k$ runs. The case of two runs is obtained by distinguishing whether they are strongly balanced or not, and using \Crefl:Fine.

Lemma 5.7**.**

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page\cref@result Let $H_{1}H_{2}$ a productive $x\mathrm{-commuting}$ lasso in $\mathcal{T}^{\leqslant|Q|}$, for some $x\in B^{+}$. Let $\Delta=\mathsf{split}_{c}(x,H_{1},H_{2})$ and $H_{3}H_{4}=(\rho_{j})_{j\in\{1,\dots,k\}}$ a productive lasso in $\mathcal{T}^{k}_{\Delta}$, for some $1\leqslant k\leqslant|Q|$. We write \rho_{j}:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{j}}}p_{j}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{j}}}q_{j}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{j}}}q_{j} for each $j$. Then:

• •

*either every $\rho_{j}$ is $\mathrm{strongly-}x\mathrm{-commuting}$: we say that $H_{3}H_{4}$ is $\mathrm{strongly-}x\mathrm{-commuting}$, *

• •

or there exist $g,h\in\mathcal{C}(B)$ such that every $\rho_{j}$ is $\mathrm{strongly-}(h,g,x)\mathrm{-aligned}$. In this case, we say that $H_{3}H_{4}$ is $\mathrm{strongly-}(h,g,x)\mathrm{-aligned}$ and we let $\mathsf{extract}_{nc}(h,g,x,\Delta,H_{3},H_{4})=\{(q_{j},h_{j})\mid j\in\{1,\dots,k\}\}$ where $h_{j}\in\mathcal{C}(B)$ is s.t. $\forall\alpha,\beta\in\mathbb{N},e_{j}^{\alpha}d_{j}c_{j}[x^{\beta}]=h_{j}h^{\alpha}g[x^{\beta}]$.

The following Lemma states that once a non-commuting loop is encountered, then the alignment of production is fixed, i.e. no transfer between left and right productions is possible anymore. Hence, the left and right S2S derived from the S2C both satisfy the twinning property:

Lemma 5.8**.**

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page\cref@result Let $H_{1}H_{2}$ be a productive non-commuting lasso that is either

• •

$(f,w)\mathrm{-aligned}$* in $\mathcal{T}^{\leqslant|Q|}$, for some $f\in\mathcal{C}(B)$ and $w\in B^{*}$, and $\Delta^{\prime}=\mathsf{split}_{nc}(f,w,H_{1},H_{2})$,*

• •

or $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$ in $\mathcal{T}^{\leqslant|Q|}_{\Delta}$, for some $g,f\in\mathcal{C}(B)$ and $\Delta\in{\cal F}(Q,\mathcal{C}(B))$, and $\Delta^{\prime}=\mathsf{extract}_{nc}(g,f,x,\Delta,H_{1},H_{2})$.

Then \overleftarrow{\makebox{\mathcal{T}{\Delta^{\prime}}\vphantom{x}}} and \overrightarrow{\makebox{\mathcal{T}{\Delta^{\prime}}\vphantom{x}}} both satisfy the twinning property.

5.3 A Two-loop Pattern Property

The following 2-loop property summarises the combinatorial properties of the synchronised runs involving loops in string-to-context transducers that satisfy the CTP.

Definition 5.9** (2-loop property).**

\cref@constructprefix

page\cref@result Given four runs $H_{1},H_{2},H_{3},H_{4}$ in $\mathcal{T}^{\leqslant|Q|}$, such that $H_{1}H_{2}$ and $(H_{1}H_{3})H_{4}$ are lassos in $\mathcal{T}^{\leqslant|Q|}$, we say that they satisfy the 2-loop property if:

$H_{1}H_{2}$* is either non productive, or productive and $x\mathrm{-commuting}$, for some $x\in B^{+}$, or productive, non-commuting and $(f,w)\mathrm{-aligned}$, for some $f\in\mathcal{C}(B)$ and $w\in B^{*}$.* 2. 2.

if $H_{1}H_{2}$ is productive and $x\mathrm{-commuting}$, we let $\Delta=\mathsf{split}_{c}(x,H_{1},H_{2})$, then $H_{3}H_{4}$ is a lasso in $\mathcal{T}_{\Delta}^{\leqslant|Q|}$. If productive then it is:

1. (a)

either $\mathrm{strongly-}x\mathrm{-commuting}$, 2. (b)

or non-commuting and $\mathrm{strongly-}(h,g,x)\mathrm{-aligned}$, for some $g,h\in\mathcal{C}(B)$. We let $\Delta^{\prime}=\mathsf{extract}_{nc}(h,g,x,\Delta,H_{3},H_{4})$, then \overleftarrow{\makebox{\mathcal{T}_{\Delta^{\prime}}\vphantom{x}}} and \overrightarrow{\makebox{\mathcal{T}_{\Delta^{\prime}}\vphantom{x}}} both satisfy the twinning property. 3. 3.

if $H_{1}H_{2}$ is productive, non-commuting and $(f,w)\mathrm{-aligned}$, we let $\Delta=\mathsf{split}_{nc}(f,w,H_{1},H_{2})$, then \overleftarrow{\makebox{\mathcal{T}_{\Delta}\vphantom{x}}} and \overrightarrow{\makebox{\mathcal{T}_{\Delta}\vphantom{x}}} both satisfy the twinning property.

*A string-to-context transducer $\mathcal{T}$ is said to satisfy the 2-loop property if for all runs $H_{1},H_{2},H_{3},H_{4}$ as above, they satisfy the 2-loop property. *

As a consequence of \Crefr:all-commuting-or-aligned,r:all-strongly-commuting-or-aligned,r:all-fully-aligned, we have:

Lemma 5.10**.**

\cref@constructprefix

page\cref@result If an S2C $\mathcal{T}$ satisfies the CTP then it satisfies the 2-loop property.

6 Determinisation

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page\cref@result

Throughout this section, we consider a string-to-context transducer $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$ from $A^{*}$ to $B^{*}$ that satisfies the 2-loop property. Intuitively, our construction stores the set of possible runs of $\mathcal{T}$, starting in an initial state, on the input word read so far. These runs are incrementally simplified by erasing synchronised loops, and by replacing a prefix by a partial function $\Delta:Q\hookrightarrow\mathcal{C}(B)$. These simplifications are based on the 2-loop property.

Observation

It is worth noticing that, as $\mathcal{T}$ is functional, if two runs reach the same state, it is safe to keep only one of them. This allows us to maintain a set of at most $|Q|$ runs.

Notations

Given $\Delta\in{\cal F}(Q,\mathcal{C}(B))$, $c\in\mathcal{C}(B)$, $w\in B^{*}$, $a\in A$ and $H\in\mathcal{R}(\mathcal{T}^{\leqslant|Q|})$, we define the following notations and operations:

• •

$\Delta c=\{(q,dc)\mid(q,d)\in\Delta\}$,

• •

$\Delta[w]=\{(q,d[w])\mid(q,d)\in\Delta\}$,

• •

\Delta\bullet a=\mathsf{choose}(\{(q^{\prime},dc)\mid(q,c)\in\Delta\text{ and }q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}a}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q^{\prime}\}),

• •

$H\bullet a\in\mathcal{R}(\mathcal{T}^{\leqslant|Q|})$ is the run obtained by extending runs of $H$ with consecutive transitions of $\mathcal{T}$ associated with input symbol $a$, and by eliminating runs so as to ensure that runs reach pairwise distinct states of $\mathcal{T}$,

• •

\Delta\bullet H=\mathsf{choose}(\{(q^{\prime},dc)\mid(q,c)\in\Delta\text{ and there is a run }\rho:q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}x}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q^{\prime}\in H\}),

• •

$id_{\Delta}=(q_{i})_{1\leqslant i\leqslant k}\in\mathcal{R}(\mathcal{T}^{k})$, for some enumeration $\{q_{1},\ldots,q_{k}\}$ of $\textsf{dom}(\Delta)$.

Construction

We define an equivalent deterministic string-to-context transducer $\overline{\mathcal{D}}=(\overline{Q},\overline{t_{\textsl{init}}},\overline{t_{\textsl{final}}},\overline{T})$, and we denote by $\mathcal{D}$ its trim part. While $\overline{\mathcal{D}}$ may have infinitely many states, we will prove that $\mathcal{D}$ is finite. Formally, we define $\overline{Q}=\overline{Q}_{\textsf{start}}\uplus\overline{Q}_{\textsf{com}}\uplus\overline{Q}_{\neg\textsf{com}}$ where:

• •

$\overline{Q}_{\textsf{start}}=\{(\varepsilon,t_{\textsl{init}},H)\mid H\in\mathcal{R}(\mathcal{T}^{\leqslant|Q|})\}$

• •

$\overline{Q}_{\textsf{com}}=\{(x,\Delta,H)\mid x\in B^{+},\Delta\in{\cal F}(Q,\mathcal{C}(B)),H\in\mathcal{R}(\mathcal{T}^{\leqslant|Q|})\}$

• •

$\overline{Q}_{\neg\textsf{com}}=\{(\bot,\Delta,id_{\Delta})\mid\Delta\in{\cal F}(Q,\mathcal{C}(B))\}$.

By definition, we have $\overline{Q}\subseteq(B^{*}\cup\{\bot\})\times{\cal F}(Q,\mathcal{C}(B))\times\mathcal{R}(\mathcal{T}^{\leqslant|Q|})=\overline{Q}_{\infty}$. Given $\overline{q}=(x,\Delta,H)\in\overline{Q}_{\infty}$, we let $\Delta_{\overline{q}}=\Delta\bullet H\in{\cal F}(Q,{\cal C}(B))$. An invariant of our construction is that every starting state of a run in $H$ belongs to $\textsf{dom}(\Delta)$.

Intuitively, the semantics of a state $\overline{q}=(x,\Delta,H)\in\overline{Q}$ can be understood as follows: $x$ is used to code the type of state ($\overline{Q}_{\textsf{start}}$, $\overline{Q}_{\textsf{com}}$ or $\overline{Q}_{\neg\textsf{com}}$), and $\Delta$ and $H$ are used to represent the runs that remain to be executed to faithfully simulate the runs of $\mathcal{T}$ on the input word $u$ read so far. As we have seen in the previous section, loops may either be commuting, allowing to shift some parts of the output from one side of the context to the other side, or they are non-commuting, and then should be aligned, forbidding such modifications. Intuitively, states in $\overline{Q}_{\textsf{start}}$ correspond to situations in which no productive loop has been encountered yet. States in $\overline{Q}_{\textsf{com}}$ (with $x\in B^{+}$) correspond to situations in which only $x$-commuting loops have been encountered. States in $\overline{Q}_{\neg\textsf{com}}$ correspond to situations in which a non-commuting loop has been encountered. A representation of $\mathcal{D}$ is given in \Creff:states.

Initial and final states

They are defined as follows:

• •

$\overline{t_{\textsl{init}}}=\{(\overline{i},c_{\varepsilon})\}$ where $\overline{i}=(\varepsilon,t_{\textsl{init}},id_{t_{\textsl{init}}})\in\overline{Q}_{\textsf{start}}$

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$\overline{t_{\textsl{final}}}=\mathsf{choose}(\{(\bar{q},dc)\mid\overline{q}\in\overline{Q},(p,c)\in\Delta_{\bar{q}},(p,d)\in t_{\textsl{final}}\})$

Transitions

Intuitively, a transition of $\overline{\mathcal{D}}$ leaving some state $\overline{p}=(x,\Delta,H)\in\overline{Q}$ with letter $a\in A$ aims at first extending $H$ with $a$, obtaining the new set of runs $H\bullet a$, and then simplifying this set of runs by removing loops. Formally, we let $(\overline{q},c)=\textsc{simplify}((x,\Delta,H\bullet a))$ and define the transition \overline{p}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}a}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q}. The function simplify is performed by Algorithm 2, which calls Algorithm 1 to remove all loops of $H\bullet a$ one by one. Depending on the type of the loop encountered, the type of the state is updated. These two algorithms are described below.

We first define $\textsc{extend\_with\_loop}(\overline{p},H_{2})$ in Algorithm 1 that takes as input a state $\overline{p}=(x,\Delta,H_{1})\in\overline{Q}_{\textsf{start}}\cup\overline{Q}_{\textsf{com}}$ and a run $H_{2}$ in $\mathcal{T}^{\leqslant|Q|}$ such that $H_{1}H_{2}$ is a lasso in $\mathcal{T}^{\leqslant|Q|}_{\Delta}$. The algorithm enumerates the possible cases for the type of this lasso, depending on the type of $\overline{p}$. This enumeration strongly relies on the 2-loop property. Depending on the case, the loop is processed, and a pair composed of a new state and a context is returned. This context will be part of the output associated with the transition. By a case analysis, we prove:

Lemma 6.1**.**

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page\cref@result Let $\overline{p}=(x,\Delta,H_{1})\in\overline{Q}_{\textsf{start}}\cup\overline{Q}_{\textsf{com}}$ and $H_{2}\in\mathcal{R}(\mathcal{T}^{\leqslant|Q|})$ such that $H_{1}H_{2}$ is a lasso in $\mathcal{T}^{\leqslant|Q|}_{\Delta}$. We let $(\overline{q},c)=\textsc{extend\_with\_loop}(\overline{p},H_{2})$.

• •

If $x=\varepsilon$ then $(\Delta_{\overline{p}}\bullet H_{2})[\varepsilon]=\Delta_{\overline{q}}c[\varepsilon]$.

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If $x\in B^{+}$ then for all $k\in\mathbb{N}$, $(\Delta_{\overline{p}}\bullet H_{2})[x^{k}]=\Delta_{\overline{q}}c[x^{k}]$.

We then define $\textsc{simplify}(\overline{p})$ in Algorithm 2 that takes as input a state $\overline{p}\in\overline{Q}_{\infty}$ (we need to consider $\overline{Q}_{\infty}$ as input and not only $\overline{Q}$ because of the recursive calls) and returns a pair composed of a new state and a context. Intuitively, it recursively processes the lassos present in the runs stored by the state $\overline{p}$, by using calls to the previous algorithm. The following result is proved by induction, using \Crefr:corr-extend:

Lemma 6.2**.**

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page\cref@result Let $\overline{p}=(x,\Delta,H)\in\overline{Q}_{\infty}$ and $(\overline{q},c)=\textsc{simplify}(\overline{p})$. Then $\overline{q}\in\overline{Q}$ and we have:

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If $x=\varepsilon$ then $\Delta_{\overline{p}}[\varepsilon]=\Delta_{\overline{q}}c[\varepsilon]$.

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If $x\in B^{+}$ then for all $k\in\mathbb{N}$, $\Delta_{\overline{p}}[x^{k}]=\Delta_{\overline{q}}c[x^{k}]$.

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If $x=\bot$ then $\Delta_{\overline{p}}=\Delta_{\overline{q}}c$.

Theorem 6.3**.**

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page\cref@result $\mathcal{D}$ is a finite sequential string-to-context transducer equivalent to $\mathcal{T}$.

Proof 6.4** (Proof Sketch).**

First observe that $\mathcal{D}$ is sequential. The correctness of $\mathcal{D}$ is a consequence of the following property, that we prove using \Crefr:corr-simplify and an induction on $|u|$: for all $u\in A^{*}$, if we have \overline{i}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q} in $\mathcal{D}$, then $\Delta_{\overline{q}}c[\varepsilon]=(t_{\textsl{init}}\bullet u)[\varepsilon]$. Last, we prove that $\mathcal{D}$ is finite. By construction, for every state $\overline{q}=(x,\Delta,H)$ of $\mathcal{D}$, $H$ contains no loop, hence its length is bounded by $|Q|^{|Q|}$. This can be used to bound the size of $x$, as well as the size of $\Delta$, for states in $\overline{Q}_{\textsf{start}}\cup\overline{Q}_{\textsf{com}}$. The case of states in $\overline{Q}_{\neg\textsf{com}}$ is different: when such a state $(\bot,\Delta,id_{\Delta})$ is reached, then by the 2-loop property, the transducers \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} and \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} both satisfy the (classical) twinning property. It remains to observe that the operations performed on Line 24 precisely correspond to two determinisations of [10], on both sides of the S2C.

7 Decision

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page\cref@result

In this section, we prove the following result:

Theorem 7.1**.**

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page\cref@result Given a string-to-context transducer, determining whether there exists an equivalent sequential string-to-context transducer is in coNP.

In order to show this result, we introduce a restriction of the 2-loop property:

Definition 7.2** (small-2-loop property).**

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*page\cref@result A string-to-context transducer $\mathcal{T}$ is said to satisfy the small-2-loop property if, for all runs $H_{1},H_{2},H_{3},H_{4}\in\mathcal{T}^{2}$ with $|H_{i}|\leqslant|Q|^{2}$ for each $i$, $H_{1}H_{2}$, $H_{1}H_{3}H_{4}$ are lassos and they satisfy the 2-loop property (in the sense of \Cref2-loop). *

By definition, if a string-to-context transducer satisfies the 2-loop property then it also satisfies the small-2-loop property. We will show that the two properties are equivalent.

Lemma 7.3**.**

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page\cref@result If a string-to-context transducer $\mathcal{T}$ satisfies the small-2-loop property then $[\![\mathcal{T}]\!]$ satisfies the contextual Lipschitz property.

Proof 7.4** (Proof Sketch).**

We claim there exists $K\in\mathbb{N}$ such that for every pair of synchronised runs H:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{0},d_{0})}}(p_{0},q_{0})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{1},d_{1})}}(p_{1},q_{1}) in $\mathcal{T}^{2}$, we have $\textsl{dist}_{f}(c_{1}c_{0}[\varepsilon],d_{1}d_{0}[\varepsilon])\leqslant K$. The result then easily follows. To prove this claim, we apply the main procedure simplify (see \Crefsec:construction) to the state $\overline{p}=(\varepsilon,t_{\textsl{init}},H)$. This procedure can indeed be applied: as it always processes the first loop (see Line 29), the lassos considered satisfy the premises of the small-2-loop property. The claim follows from the proof of finiteness of $\mathcal{D}$.

Proof 7.5** (Proof Sketch of \Crefr:decision-det).**

By \Creft:main and \Crefr:small-2-loop-implies-lip, $\mathcal{T}$ admits an equivalent sequential S2C transducer iff $\mathcal{T}$ satisfies the small-2-loop property (see also the figure below). Thus, we describe a procedure to decide whether $\mathcal{T}$ satisfies the small-2-loop property.

The procedure first non-deterministically guesses a counter-example to the small-2-loop property and then verifies that it is indeed a counter-example. By definition of the small-2-loop property, the counter-example can have finitely many shapes. Those shapes require the verification of the properties of the involved lassos: being productive or not, being commuting or not, being aligned or not, satisfying the (classical) twinning property, etc.

Verifying that a lasso in $\mathcal{T}^{2}$ is not commuting (resp. not aligned) boils down to checking whether there exists no $x\in B^{+}$ such that the lasso is $x\mathrm{-commuting}$ (resp. no $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that the lasso is $(f,w)\mathrm{-aligned}$). In both cases, the search space for the words $x,w$ and context $f$ can be narrowed down to factors of the output contexts of the given lasso. Thus these verifications can be done in polynomial time. The classical twinning property can also be checked in polynomial time. As a summary, we can show that the verifications for all the shapes can be done in polynomial time. Furthermore, all the shapes are of polynomial size, by definition of the small-2-loop property, yielding the result.

8 Conclusion

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page\cref@result

We have proposed a multiple characterisation of string-to-context transducers that admit an equivalent sequential S2C, including a machine independent property, a pattern property, as well as a "small" pattern property allowing to derive a decision procedure running in non-deterministic polynomial time. All these equivalences are summarised on the diagram on the right. Future work includes a lower bound for the complexity of the problem, the extension of this work to the register minimisation problem for streaming string transducers without register concatenation, and the extension of our results to infinite words.

Appendix A Proofs of \Crefsec:preliminaries: Models

Proof A.1** (Proof of \Crefr:distf).**

Let $x,y,z$ words.

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Symmetry:* It is trivial to prove that $\textsl{dist}_{f}(x,y)=\textsl{dist}_{f}(y,x)$.*

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Identity:* It is trivial to prove that $\textsl{dist}_{f}(x,y)=0\iff x=y$.*

• •

Triangle Inequality:* We want to prove that $\textsl{dist}_{f}(x,z)\leqslant\textsl{dist}_{f}(x,y)+\textsl{dist}_{f}(y,z)$. By definition, $\textsl{dist}_{f}(x,y)=|x|+|y|-2|\textsf{lcf}(x,y)|$, $\textsl{dist}_{f}(y,z)=|y|+|z|-2|\textsf{lcf}(y,z)|$ and $\textsl{dist}_{f}(x,z)=|x|+|z|-2|\textsf{lcf}(x,z)|$. Let $\alpha$ a longest common factor of $x$ and $y$, and $\beta$ a longest common factor of $y$ and $z$. Let $x_{1},x_{2},y_{1},y_{2},y_{3},y_{4},z_{1},z_{2}$ words such that $x=x_{1}\alpha x_{2}$, $y=y_{1}\alpha y_{2}$, and $y=y_{3}\beta y_{4}$, $z=z_{1}\beta z_{2}$. Observe that $\textsl{dist}_{f}(x,y)=|x_{1}|+|x_{2}|+|y_{1}|+|y_{2}|$ and $\textsl{dist}_{f}(y,z)=|y_{3}|+|y_{4}|+|z_{1}|+|z_{2}|$. We observe six cases:*

1. (i)

$|y_{1}|\leqslant|y_{3}|<|y_{1}|+|\alpha|$* and $|y_{4}|\leqslant|y_{2}|<|y_{4}|+|\beta|$.*

There exists $\gamma$ such that $y_{1}\alpha=y_{3}\gamma$ and $\gamma y_{2}=\beta y_{4}$. Then we have $|\alpha|\leqslant|y_{3}|+|\gamma|$ and $|\beta|\leqslant|y_{2}|+|\gamma|$. Yet, $|x|=|x_{1}|+|x_{2}|+|\alpha|$ and we obtain that $|x|-|\gamma|=|x_{1}|+|x_{2}|+|\alpha|-|\gamma|\leqslant|x_{1}|+|x_{2}|+|y_{3}|$. Also, $|z|=|z_{1}|+|z_{2}|+|\beta|$ and we obtain that $|z|-|\gamma|=|z_{1}|+|z_{2}|+|\beta|-|\gamma|\leqslant|z_{1}|+|z_{2}|+|y_{2}|$. Finally, we have that $|\textsf{lcf}(x,z)|\geqslant|\gamma|$ as $\gamma$ indeed is a common factor of $x$ and $z$. Then,

[TABLE] 2. (ii)

$|y_{3}|\leqslant|y_{1}|<|y_{3}|+|\alpha|$* and $|y_{2}|\leqslant|y_{4}|<|y_{2}|+|\beta|$.*

This case is symmetrical to the previous case. 3. (iii)

$|y_{1}|\leqslant|y_{3}|<|y_{1}|+|\alpha|$* and $|y_{2}|\leqslant|y_{4}|<|y_{2}|+|\beta|$.*

Then we have $|\alpha|\leqslant|y_{3}|+|\beta|+|y_{4}|$. Yet, $|x|=|x_{1}|+|x_{2}|+|\alpha|$ and we obtain that $|x|-|\beta|=|x_{1}|+|x_{2}|+|\alpha|-|\beta|\leqslant|x_{1}|+|x_{2}|+|y_{3}|+|y_{4}|$. Also, $|z|=|z_{1}|+|z_{2}|+|\beta|$ and thus $|z|-|\beta|=|z_{1}|+|z_{2}|$. Finally, we have that $|\textsf{lcf}(x,z)|\geqslant|\beta|$ as $\beta$ indeed is a common factor of $x$ and $z$. Then,

[TABLE] 4. (iv)

$|y_{3}|\leqslant|y_{1}|<|y_{3}|+|\alpha|$* and $|y_{4}|\leqslant|y_{2}|<|y_{4}|+|\beta|$.*

This case is symmetrical to the previous case. 5. (v)

$|y_{1}|+|\alpha|\leqslant|y_{3}|$* and $|y_{4}|+|\beta|\leqslant|y_{2}|$.*

Then, $|\alpha|\leqslant|y_{3}|$ and $|\beta|\leqslant|y_{2}|$. Yet, $|x|=|x_{1}|+|x_{2}|+|\alpha|$ and we obtain that $|x|\leqslant|x_{1}|+|x_{2}|+|y_{3}|$. Also, $|z|=|z_{1}|+|z_{2}|+|\beta|$ and we obtain that $|z|\leqslant|z_{1}|+|z_{2}|+|y_{2}|$. Then, as $|\textsf{lcf}(x,z)|\geqslant 0$,

[TABLE] 6. (vi)

$|y_{3}|+|\alpha|\leqslant|y_{1}|$* and $|y_{2}|+|\beta|\leqslant|y_{4}|$.*

This case is symmetrical to the previous case.

Lemma A.2**.**

Let $c,c^{\prime}\in\mathcal{C}(B)$ and $w,w^{\prime}\in B^{*}$. $\textsl{dist}_{f}(c[w],c^{\prime}[w^{\prime}])\leqslant\textsl{dist}_{f}(w,w^{\prime})+|c|+|c^{\prime}|$.

Proof A.3**.**

It is easy to see that $|\textsf{lcf}(c[w],c^{\prime}[w^{\prime}])|\geqslant|\textsf{lcf}(w,w^{\prime})|$. Then $|c|+|c^{\prime}|+|w|+|w^{\prime}|-2|\textsf{lcf}(c[w],c^{\prime}[w^{\prime}])|\leqslant|c|+|c^{\prime}|+|w|+|w^{\prime}|-2|\textsf{lcf}(w,w^{\prime})|$. And we obtain the result.

Corollary A.4**.**

Let $c,c^{\prime}\in\mathcal{C}(B)$ and $w\in B^{*}$. $\textsl{dist}_{f}(c[w],c^{\prime}[w])\leqslant|c|+|c^{\prime}|$.

Lemma A.5**.**

Let $c,c^{\prime}\in\mathcal{C}(B)$ and $w,w^{\prime}\in B^{*}$. $\textsl{dist}_{f}(w,w^{\prime})\leqslant\textsl{dist}_{f}(c[w],c^{\prime}[w^{\prime}])+|c|+|c^{\prime}|$.

Proof A.6**.**

First observe that for any common factor $y$ of $w$ and $w^{\prime}$, we have $\textsl{dist}_{f}(w,w^{\prime})\leqslant|w|+|w^{\prime}|-2|y|$. To prove the result, consider a longest common factor $x$ of $c[w]$ and $c^{\prime}[w^{\prime}]$. We claim that there exists a common factor $y$ of $w$ and $w^{\prime}$ (which may be the empty word) such that $|y|\geqslant|x|-|c|-|c^{\prime}|$. Indeed, one obtains $y$ from $x$ by removing symbols of $x$ that correspond to positions in $c$ or $c^{\prime}$. As a consequence, we obtain:

[TABLE]

This concludes the proof.

Appendix B Proofs of \Crefsec:combinatorics: Analysis of Loop Combinatorics

Throughout this section, we consider a string-to-context transducer $\mathcal{T}=(Q,t_{\textsl{init}},t_{\textsl{final}},T)$ that satisfies the contextual twinning property (CTP). Let $L\in\mathbb{N}$ such that any two states of $\mathcal{T}$ are $L$-contextually twinned.

B.1 Additional Word Combinatorics Notations

The size of a word $x$ is denoted by $|x|$. Given two words $x,y\in A^{*}$, we write $x\preceq_{p}y$, resp. $x\preceq_{s}y$, if $x$ is a prefix, resp. a suffix, of $y$. If we have $x\preceq_{p}y$, resp. $x\preceq_{s}y$, then we note $x^{-1}y$, resp. $yx^{-1}$, the unique word $z$ such that $y=xz$, resp. $y=zx$. A word $x\in A^{*}$ is primitive if there is no word $y$ such that $|y|<|x|$ and $x\in y^{*}$. The primitive root of a word $x\in A^{*}$, denoted by $\rho(x)$, is the (unique) primitive word $y$ such that $x\in y^{*}$. In particular, if $x$ is primitive, then its primitive root is $x$. The primitive period of a word $x\in A^{*}$, denoted by $\bar{\rho}(x)$, is the (unique) primitive word $y$ such that $x\in y^{+}z$ for some $z\preceq_{p}y$. Two words $x$ and $y$ are conjugates, written $x\sim y$, if there exists $z\in A^{*}$ such that $xz=zy$. It is well-known that two words are conjugates iff there exist $t_{1},t_{2}\in A^{*}$ such that $x=t_{1}t_{2}$ and $y=t_{2}t_{1}$.

Example B.1**.**

The primitive root and primitive period act differently. For instance, $\rho(abcab)=abcab$ but $\bar{\rho}(abcab)=abc$.

For $n,m\in\mathbb{N}_{>0}$, we note by $\gcd(n,m)$ the greatest common divisor of $n$ and $m$.

Lemma B.2** **(Fine and Wilf,

[17], Chapter 9 of [21]).

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page\cref@result Let $x,y\in A^{*}$ and $m,n\in\mathbb{N}$. If $x^{m}$ and $y^{n}$ have a common subword of length at least $|x|+|y|-\gcd(|x|,|y|)$, then their primitive roots are conjugates.

Lemma B.3** (Saarela, Theorem 4.3 of [22]).**

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page\cref@result Let $m,n\geqslant 1$, $s_{j},t_{j}\in A^{*}$ and $u_{j},v_{j}\in A^{+}$. If $s_{0}u_{1}^{i}s_{1}\dots u_{m}^{i}s_{m}=t_{0}v_{1}^{i}t_{1}\dots v_{n}^{i}t_{n}$ holds for $m+n$ values of $i$, then it holds for all $i$.

Lemma B.4**.**

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page\cref@result Let $x,y\in A^{*}$ such that $|y|\geqslant|x|$. If $y$ is a factor of $x^{*}$ then $\bar{\rho}(y)\sim\rho(x)$.

Proof B.5**.**

Without loss of generality, consider $x$ to be primitive. Let $u,v\in A^{*}$ and $i\in\mathbb{N}_{>0}$ such that $uyv=x^{i}$. There exist $t_{1},t_{1}^{\prime}$ such that $x=t_{1}t_{1}^{\prime}$ and $u\in(t_{1}t_{1}^{\prime})^{*}t_{1}$. Then $yv\in t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{+}$, as $|y|\geqslant|t_{1}^{\prime}t_{1}|$. There exist $t_{2},t_{2}^{\prime}$ such that $t_{1}^{\prime}t_{1}=t_{2}t_{2}^{\prime}$ and $y\in(t_{1}^{\prime}t_{1})^{+}t_{2}$. Thus, by definition of $\bar{\rho}(y)$, we have $\bar{\rho}(y)\sim x$.

B.2 Lassos

Two lassos \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2} are said to be weakly balanced if $|e_{1}|=|e_{2}|$.

The following Lemma states a consequence of the definition of aligned lassos.

Lemma B.6**.**

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page\cref@result Let $\rho$ be a productive lasso \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q and $f\in\mathcal{C}(B)$ and $w\in B^{*}$. $\rho$ is $(f,w)\mathrm{-aligned}$ if and only if there exist $t_{1},t_{2},t_{3},t_{4}$ and $\alpha,\beta\geqslant 0$ such that \rho(\overleftarrow{\makebox{e\vphantom{x}}})=t_{1}t_{2}, \rho(\overleftarrow{\makebox{f\vphantom{x}}})=t_{2}t_{1}, \rho(\overrightarrow{\makebox{e\vphantom{x}}})=t_{3}t_{4}, \rho(\overrightarrow{\makebox{f\vphantom{x}}})=t_{4}t_{3}, and $dc[\varepsilon]=(t_{1}t_{2})^{\alpha}t_{1}wt_{3}(t_{4}t_{3})^{\beta}$.

As a corollary, any productive lasso \rho:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q is $(e,dc[\varepsilon])\mathrm{-aligned}$. Also, note that a lasso can be commuting and aligned at the same time.

The following lemma states the combinatorial properties of two synchronised lassos.

Lemma B.7**.**

\cref@constructprefix

page\cref@result For any two synchronised lassos $\rho_{1}$ and $\rho_{2}$, we have that

• •

either $\rho_{1}$ and $\rho_{2}$ are non-productive

• •

or $\rho_{1}$ and $\rho_{2}$ are productive and weakly-balanced, and there exists $x\in B^{+}$ primitive such that $\rho_{1}$ and $\rho_{2}$ are $x\mathrm{-commuting}$,

• •

or $\rho_{1}$ and $\rho_{2}$ are productive, strongly-balanced and non-commuting, and there exists $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that $\rho_{1}$ and $\rho_{2}$ are $(f,w)\mathrm{-aligned}$.

In order to prove \Crefr:ctp-concrete, we first need some preliminary combinatorial results.

Lemma B.8**.**

\cref@constructprefix

page\cref@result Let $c_{1},c_{2},d_{1},d_{2}\in\mathcal{C}(B)$. If for all $i\in\mathbb{N}$, $\textsl{dist}_{f}(d_{1}^{i}c_{1}[\varepsilon],d_{2}^{i}c_{2}[\varepsilon])\leqslant L$, then there exist $e_{1},e_{2}\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}$, $e_{1}d_{1}^{i}c_{1}[\varepsilon]=e_{2}d_{2}^{i}c_{2}[\varepsilon]$.

Proof B.9**.**

Suppose that for all $i\in\mathbb{N}$, $\textsl{dist}_{f}(d_{1}^{i}c_{1}[\varepsilon],d_{2}^{i}c_{2}[\varepsilon])\leqslant L$. Then for all $i\in\mathbb{N}$, there exist $f_{1},f_{2}\in\mathcal{C}(B)$, such that $f_{1}^{-1}d_{1}^{i}c_{1}[\varepsilon]=f_{2}^{-1}d_{2}^{i}c_{2}[\varepsilon]$ and $|f_{1}|+|f_{2}|\leqslant L$. Let $C_{L}=\{(f_{1},f_{2})\mid|f_{1}|+|f_{2}|\leqslant L\}$. $C_{L}$ is finite. Thus there exist some $(f_{1},f_{2})\in C_{L}$ such that there is an infinite number of $i\in\mathbb{N}$ such that $f_{1}^{-1}d_{1}^{i}c_{1}[\varepsilon]=f_{2}^{-1}d_{2}^{i}c_{2}[\varepsilon]$. Furthermore, there exists $i_{0}$ such that for all $i\geqslant i_{0}$, there exists $e_{1},e_{2}\in\mathcal{C}(B)$ and $e_{1}d_{1}^{i-i_{0}}c_{1}[\varepsilon]=e_{2}d_{2}^{i-i_{0}}c_{2}[\varepsilon]$. Finally, by \Crefr:saarela, we obtain that for all $i\in\mathbb{N}$, $e_{1}d_{1}^{i}c_{1}[\varepsilon]=e_{2}d_{2}^{i}c_{2}[\varepsilon]$.

Lemma B.10**.**

\cref@constructprefix

page\cref@result Let $u_{1},w_{1},u_{2},w_{2}\in B^{*}$ and $v_{1},v_{2}\in B^{+}$ such that $|v_{1}|=|v_{2}|$ and $u_{1}v_{1}^{i}w_{1}=u_{2}v_{2}^{i}w_{2}$ for all $i\in\mathbb{N}$. Then there exists $x\in B^{+}$ and $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i\geqslant 1$, there exist $k\in\mathbb{N}$ such that $v_{1}^{i}=f_{1}[x^{k}]$ and $v_{2}^{i}=f_{2}[x^{k}]$.

Proof B.11**.**

There exists $i_{0}$ sufficiently large so that $v_{1}^{i_{0}}$ and $v_{2}^{i_{0}}$ overlap with a common factor of length greater than $|v_{1}|+|v_{2}|-\gcd(|v_{1}|,|v_{2}|)$. Thus, by \Crefr:fw, $\rho(v_{1})\sim\rho(v_{2})$.

Let $t,t^{\prime}\in B^{*}$ and $\alpha,\beta\geqslant 1$ such that $v_{1}=(tt^{\prime})^{\alpha}$ and $v_{2}=(t^{\prime}t)^{\beta}$. We choose $x=\rho(v_{1})=tt^{\prime}$, $f_{1}=(tt^{\prime},\varepsilon)$ and $f_{2}=(t^{\prime},t)$. Then for all $i\geqslant 1$, let $k=\alpha i-1\geqslant 0$, and we have $v_{1}^{i}=(tt^{\prime})^{\alpha i}=tt^{\prime}(tt^{\prime})^{\alpha i-1}=f_{1}[x^{k}]$ and $v_{2}^{i}=(t^{\prime}t)^{\alpha i}=t^{\prime}x^{\alpha i-1}t=f_{2}[x^{k}]$.

Lemma B.12**.**

\cref@constructprefix

page\cref@result Let $u_{1},w_{1},u_{2},w_{2},y_{2}\in B^{*}$ and $v_{1},v_{2},x_{2}\in B^{+}$ such that $|v_{1}|=|v_{2}|+|x_{2}|$ and $u_{1}v_{1}^{i}w_{1}=u_{2}v_{2}^{i}w_{2}x_{2}^{i}y_{2}$ for all $i\in\mathbb{N}$. Then there exists $x\in B^{+}$ and $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i\geqslant 1$, there exist $k\in\mathbb{N}$ such that $v_{1}^{i}=f_{1}[x^{k}]$ and $v_{2}^{i}w_{2}x_{2}^{i}=f_{2}[x^{k}]$.

Proof B.13**.**

There exists $m_{0}$ sufficiently large so that $v_{1}^{m_{0}}$ overlap with both $v_{2}^{m_{0}}$ and $x_{2}^{m_{0}}$ with common factors of length greater than $|v_{1}|+|v_{2}|-\gcd(|v_{1}|,|v_{2}|)$ and $|v_{1}|+|x_{2}|-\gcd(|v_{1}|,|x_{2}|)$. Thus, by \Crefr:fw, $\rho(v_{1})\sim\rho(v_{2})$ and $\rho(v_{1})\sim\rho(x_{2})$. As $|v_{1}|=|v_{2}|+|x_{2}|$, we have that $|v_{2}w_{2}x_{2}|\geqslant|v_{1}|$. Yet $v_{2}w_{2}x_{2}$ is a factor of $v_{1}^{*}$, then, by \Crefr:prootf-of-factor, $\bar{\rho}(v_{2}w_{2}x_{2})\sim\rho(v_{1})$.

Let $t_{1},t_{1}^{\prime},t_{2},t_{2}^{\prime}\in B^{*}$ and $\alpha,\beta,\gamma\geqslant 1$ such that $t_{1}t_{1}^{\prime}=t_{2}t_{2}^{\prime}$, and $v_{1}=(t_{1}t_{1}^{\prime})^{\alpha}$, $v_{2}=(t_{1}^{\prime}t_{1})^{\beta}$ and $x_{2}=(t_{2}^{\prime}t_{2})^{\gamma}$. Note that $\alpha=\beta+\gamma$. Also we have $v_{2}w_{2}x_{2}=t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta}t_{2}$, for some $\theta\geqslant 0$. We choose $x=\rho(v_{1})=t_{1}t_{1}^{\prime}$, $f_{1}=((t_{1}t_{1}^{\prime})^{\alpha},\varepsilon)$ and $f_{2}=(t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta},t_{2})$. Then for all $i\geqslant 1$, let $k=\alpha(i-1)\geqslant 0$, and we have $v_{1}^{i}=(t_{1}t_{1}^{\prime})^{\alpha i}=(t_{1}t_{1}^{\prime})^{\alpha}x^{\alpha(i-1)}=f_{1}[x^{k}]$ and $v_{2}^{i}w_{2}x_{2}^{i}=v_{2}^{i-1}v_{2}w_{2}x_{2}x_{2}^{i-1}=(t_{1}^{\prime}t_{1})^{\beta(i-1)}t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta}t_{2}(t_{2}^{\prime}t_{2})^{\gamma(i-1)}=t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta}x^{\alpha(i-1)}t_{2}=f_{2}[x^{k}]$.

Lemma B.14**.**

\cref@constructprefix

page\cref@result Let $u_{1},w_{1},y_{1},u_{2},w_{2},y_{2}\in B^{*}$ and $v_{1},x_{1},v_{2},x_{2}\in B^{+}$ such that $|v_{1}|+|x_{1}|=|v_{2}|+|x_{2}|$ and $u_{1}v_{1}^{i}w_{1}x_{1}^{i}y_{1}=u_{2}v_{2}^{i}w_{2}x_{2}^{i}y_{2}$ for all $i\in\mathbb{N}$.

• •

If $|v_{1}|\neq|v_{2}|$ and $|x_{1}|\neq|x_{2}|$ then there exists $x\in B^{+}$ and $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i\geqslant 1$, there exist $k\in\mathbb{N}$ such that $v_{1}^{i}w_{1}x_{1}^{i}=f_{1}[x^{k}]$ and $v_{2}^{i}w_{2}x_{2}^{i}=f_{2}[x^{k}]$.

• •

If $|v_{1}|=|v_{2}|$ and $|x_{1}|=|x_{2}|$ then there exist $w\in B^{*}$ and $f,g_{1},g_{2}\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}$, $v_{1}^{i}w_{1}x_{1}^{i}=g_{1}f^{i}[w]$ and $v_{2}^{i}w_{2}x_{2}^{i}=g_{2}f^{i}[w]$.

Proof B.15**.**

If $|v_{1}|\neq|v_{2}|$ and $|x_{1}|\neq|x_{2}|$, suppose $|v_{1}|>|v_{2}|$. There exists $i_{0}$ sufficiently large so that $v_{1}^{i_{0}}$ overlap with both $x_{2}^{i_{0}}$ with a common factor of length greater than $|v_{1}|+|x_{2}|-\gcd(|v_{1}|,|x_{2}|)$. Thus, by \Crefr:fw, $\rho(v_{1})\sim\rho(x_{2})$. Using the same argument, we have that $\rho(v_{1})\sim\rho(v_{2})$ and $\rho(x_{1})\sim\rho(x_{2})$. As $|v_{1}|+|x_{1}|=|v_{2}|+|x_{2}|$, we have that $|v_{2}|+|x_{2}|\geqslant|v_{1}|$ and thus $|v_{2}w_{2}x_{2}|\geqslant|v_{1}|$. Yet $v_{2}w_{2}x_{2}$ is a factor of $v_{1}^{*}$, then, by \Crefr:prootf-of-factor, $\bar{\rho}(v_{2}w_{2}x_{2})\sim\rho(v_{1})$. Symmetrically, $\bar{\rho}(v_{1}w_{1}x_{1})\sim\rho(x_{2})$.

Let $t_{1},t_{1}^{\prime},t_{2},t_{2}^{\prime},t_{3},t_{3}^{\prime}\in B^{*}$ and $\alpha,\beta,\gamma,\delta\geqslant 1$ such that $t_{1}t_{1}^{\prime}=t_{2}t_{2}^{\prime}=t_{3}t_{3}^{\prime}$, and $v_{1}=(t_{1}t_{1}^{\prime})^{\alpha}$, $x_{1}=(t_{2}^{\prime}t_{2})^{\beta}$, $v_{2}=(t_{1}^{\prime}t_{1})^{\gamma}$ and $x_{2}=(t_{3}^{\prime}t_{3})^{\delta}$. Note that $\alpha+\beta=\gamma+\delta$. Also we have $v_{1}w_{1}x_{1}=(t_{1}t_{1}^{\prime})^{\theta_{1}}t_{2}$ and $v_{2}w_{2}x_{2}=t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta_{2}}t_{3}$, for some $\theta_{1},\theta_{2}\geqslant 0$. We choose $x=\rho(v_{1})=t_{1}t_{1}^{\prime}$ $f_{1}=((t_{1}t_{1}^{\prime})^{\theta_{1}},t_{2})$ and $f_{2}=(t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta_{2}},t_{3})$. Then for all $i\geqslant 1$, let $k=(\alpha+\beta)(i-1)\geqslant 0$, and we have $v_{1}^{i}w_{1}x_{1}^{i}=v_{1}^{i-1}v_{1}w_{1}x_{1}x_{1}^{i-1}=(t_{1}t_{1}^{\prime})^{\theta_{1}}(t_{1}t_{1}^{\prime})^{(\alpha+\beta)(i-1)}t_{2}=f_{1}[x^{k}]$ and $v_{2}^{i}w_{2}x_{2}^{i}=v_{2}^{i-1}v_{2}w_{2}x_{2}x_{2}^{i-1}=t_{1}^{\prime}(t_{1}t_{1}^{\prime})^{\theta_{2}}(t_{1}t_{1}^{\prime})^{(\gamma+\delta)(i-1)}t_{3}=f_{2}[x^{k}]$.

If $|v_{1}|<|v_{2}|$, we obtain the same result.

If $|v_{1}|=|v_{2}|$ and $|x_{1}|=|x_{2}|$, we only have that $\rho(v_{1})\sim\rho(v_{2})$ and $\rho(x_{1})\sim\rho(x_{2})$. If $|u_{1}|<|u_{2}|$, let $v$ such that we have $u_{2}=u_{1}v$ and $v_{1}v=vv_{2}$; if $|u_{1}|=|u_{2}|$, let $v=\varepsilon$ and we have $u_{1}=u_{2}$ and $v_{1}=v_{2}$; if $|u_{1}|>|u_{2}|$, let $v$ such that we have $u_{1}=u_{2}v$ and $v_{2}v=vv_{1}$. Similarly, if $|y_{1}|<|y_{2}|$, let $x$ such that we have $y_{2}=xy_{1}$ and $x_{2}x=xx_{1}$; if $|y_{1}|=|y_{2}|$, let $x=\varepsilon$ and we have $y_{1}=y_{2}$ and $x_{1}=x_{2}$; if $|y_{1}|>|y_{2}|$, let $x$ such that we have $y_{1}=xy_{2}$ and $x_{1}x=xx_{2}$.

Finally, from $v$ and $y$, we obtain that

• •

$v_{1}v=vv_{2}$, $w_{1}=vw_{2}x$, $x_{2}x=xx_{1}$, or

• •

$v_{1}v=vv_{2}$, $w_{1}x=vw_{2}$, $x_{1}x=xx_{2}$, or

• •

$v_{2}v=vv_{1}$, $vw_{1}=w_{2}x$, $x_{2}x=xx_{1}$, or

• •

$v_{2}v=vv_{1}$, $vw_{1}x=w_{2}$, $x_{1}x=xx_{2}$.

We handle the first case. The others are similar. We choose $f=(v_{2},x_{2})$, $w=w_{2}$, and $g_{1}=(v,x)$ and $g_{2}=(\varepsilon,\varepsilon)$. Then for all $i\in\mathbb{N}$, $v_{1}^{i}w_{1}x_{1}^{i}=v_{1}^{i}w_{1}x_{1}^{i}=v_{1}^{i}vw_{2}xx_{1}^{i}=vv_{2}^{i}w_{2}x_{2}^{i}x=v(f^{i}[w])x=g_{1}f^{i}[w]$ and $v_{2}^{i}w_{2}x_{2}^{i}=g_{2}f^{i}[w]$.

Lemma B.16**.**

\cref@constructprefix

page\cref@result Let $f\in\mathcal{C}(B)$, $w\in B^{*}$, and $x\in B^{+}$ a primitive word. Let $\rho_{1}$ and $\rho_{2}$ be two synchronised, productive, strongly-balanced and $(f,w)\mathrm{-aligned}$ lassos. If $\rho_{1}$ is $x\mathrm{-commuting}$, then $\rho_{2}$ is $x\mathrm{-commuting}$.

Proof B.17**.**

Let \rho_{1}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \rho_{2}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2}.

We have that $\|e_{1}\|=\|e_{2}\|$, and $\rho_{1}$ and $\rho_{2}$ are $(f,w)\mathrm{-aligned}$. By \Crefd:aligned-lasso, there exist some contexts $g_{1},g_{2}\in\mathcal{C}(B)$ such that for all $i\geqslant 0$, $e_{1}^{i}d_{1}c_{1}[\varepsilon]=g_{1}f^{i}w$ and $e_{2}^{i}d_{2}c_{2}[\varepsilon]=g_{2}f^{i}w$. If $\rho_{1}$ is $x\mathrm{-commuting}$ then, by \Crefd:commuting-lasso, there exist $h\in\mathcal{C}(B)$ such that for all $i\geqslant 0$, there exists $j\geqslant 0$ such that $e_{1}^{i}d_{1}c_{1}[\varepsilon]=h[x^{j}]$. Hence, there exists $k\geqslant 0$ and $h^{\prime}\in\mathcal{C}(B)$ such that $f^{i}w=h^{\prime}[x^{k}]$ and then $g_{2}f^{i}[w]=g_{2}h^{\prime}[x^{k}]$. Therefore, by \Crefd:commuting-lasso, $\rho_{2}$ is $x\mathrm{-commuting}$.

We can now prove \Crefr:ctp-concrete.

Proof B.18** (Proof of \Crefr:ctp-concrete).**

Let \rho_{1}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \rho_{2}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2}. By \Crefr:dist-to-equation, there exist $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}$ $f_{1}e_{1}^{i}d_{1}c_{1}[\varepsilon]=f_{2}e_{2}^{i}d_{2}c_{2}[\varepsilon]$. Then we have that $|e_{1}|=|e_{2}|$. We observe 10 cases.

If $|e_{1}|=0$ or $|e_{2}|=0$ then $|e_{1}|=|e_{2}|=0$ and $\rho_{1},\rho_{2}$ are not productive.

If $e_{1},e_{2}\in B^{+}\times\{\varepsilon\}$, then by \Crefr:comb-1-1, there exists $x\in B^{+}$ such that both $\rho_{1}$ and $\rho_{2}$ are productive, weakly-balanced and $x\mathrm{-commuting}$. The same holds for the other three cases where exactly two of the four components of $e_{1}$ and $e_{2}$ are empty.

If $e_{1}\in B^{+}\times\{\varepsilon\}$ and $e_{2}\in B^{+}\times B^{+}$, then by \Crefr:comb-1-2, there exists $x\in B^{+}$ such that both $\rho_{1}$ and $\rho_{2}$ are productive, weakly-balanced and $x\mathrm{-commuting}$. The same holds for the other three cases where exactly one of the four components of $e_{1}$ and $e_{2}$ is empty.

If $e_{1},e_{2}\in B^{+}\times B^{+}$, then by \Crefr:comb-2-2, there are two cases. Firstly, if $\|e_{1}\|\neq\|e_{2}\|$ then there exists $x\in B^{+}$ such that both $\rho_{1}$ and $\rho_{2}$ are productive, weakly-balanced and $x\mathrm{-commuting}$. Secondly, if $\|e_{1}\|=\|e_{2}\|$ then there exist $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that both $\rho_{1}$ and $\rho_{2}$ are productive, strongly-balanced, and $(f,w)\mathrm{-aligned}$. However, by \Crefr:aligned-can-be-com, if it still happens that either one of $\rho_{1}$ and $\rho_{2}$ is $x\mathrm{-commuting}$, then both $\rho_{1}$ and $\rho_{2}$ are $x\mathrm{-commuting}$. If not, then they both are non-commuting.

Lemma B.19**.**

\cref@constructprefix

page\cref@result Let $\rho$ be a productive lasso, and $x,x^{\prime}\in B^{+}$ be two primitive words.

• •

If $x\sim x^{\prime}$ and $\rho$ is $x\mathrm{-commuting}$ then $\rho$ is $x^{\prime}\mathrm{-commuting}$

• •

If $\rho$ is $x\mathrm{-commuting}$ and $x^{\prime}\mathrm{-commuting}$ then $x\sim x^{\prime}$

Proof B.20**.**

Let \rho:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q and $x,x^{\prime}\in B^{+}$ be two primitive words.

Firstly, suppose that $x\sim x^{\prime}$ and $\rho$ is $x\mathrm{-commuting}$. By definition, there exists $f\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}$ there exists $k\in\mathbb{N}$ such that $e^{i}dc[\varepsilon]=f[x^{k}]$. As $x\sim x^{\prime}$, there exist $t,t^{\prime}\in B^{*}$ such that $x=tt^{\prime}$ and $x^{\prime}=t^{\prime}t$. Let $g=(t,t^{\prime})$. We obtain that for all $i\in\mathbb{N}$ there exists $k\in\mathbb{N}$ such that $e^{i}dc[\varepsilon]=f[x^{k}]=fg[x^{\prime k-1}]$.

Secondly, suppose that $\rho$ is both $x\mathrm{-commuting}$ and $x^{\prime}\mathrm{-commuting}$. By definition, there exists $f,f^{\prime}\in\mathcal{C}(B)$ such that for all $i\in\mathbb{N}$ there exists $k,k^{\prime}\in\mathbb{N}$ such that $e^{i}dc[\varepsilon]=f[x^{k}]=f^{\prime}[x^{\prime k^{\prime}}]$. As $x$ and $x^{\prime}$ are primitive, we have that $k=k^{\prime}$ and thus $|x|=|x^{\prime}|$. By \Crefr:comb-1-1, we obtain that $x\sim x^{\prime}$.

Lemma B.21**.**

\cref@constructprefix

page\cref@result Let $x\in B^{+}$ a primitive word. Let $\rho_{1}$ and $\rho_{2}$ be two synchronised productive lassos. If $\rho_{1}$ is $x\mathrm{-commuting}$, then $\rho_{2}$ is $x\mathrm{-commuting}$.

Proof B.22**.**

Let \rho_{1}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \rho_{2}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2}. By \Crefr:ctp-concrete, we observe two cases. First, consider that $\|e_{1}\|=\|e_{2}\|$ and that there exists $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that $\rho_{1}$ and $\rho_{2}$ are $(f,w)\mathrm{-aligned}$. By \Crefr:aligned-can-be-com, if $\rho_{1}$ is $x\mathrm{-commuting}$, then $\rho_{2}$ is $x\mathrm{-commuting}$. Otherwise, we only have that $|e_{1}|=|e_{2}|$, and there exists $x^{\prime}\in B^{+}$ primitive such that $\rho_{1}$ and $\rho_{2}$ are $x^{\prime}\mathrm{-commuting}$. If $\rho_{1}$ is $x\mathrm{-commuting}$ then, by \Crefr:conjugated-com, we have that $x^{\prime}\sim x$ and, by \Crefr:conjugated-com again, that $\rho_{2}$ is $x\mathrm{-commuting}$.

Lemma B.23**.**

\cref@constructprefix

page\cref@result Let $\rho_{1},\dots,\rho_{k}$ be $k$ synchronised productive lassos that are pairwise aligned, strongly balanced and not commuting. Then there exist $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that they are all $(f,w)\mathrm{-aligned}$.

Proof B.24**.**

Let \rho_{i}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{i}}}p_{i}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{i}}}q_{i}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{i}}}q_{i} for $i\in\{1,\dots,k\}$. As $\rho_{1},\dots,\rho_{k}$ are pairwise aligned, there exist $f_{2},\dots,f_{k}\in\mathcal{C}(B)$ and $w_{2},\dots,w_{k}\in B^{*}$ such that for all $i\in\{2,\dots,k\}$, $\rho_{1}$ and $\rho_{i}$ are $(f_{i},w_{i})\mathrm{-aligned}$. Then for all $i\in\{2,\dots,k\}$, there exist $f_{i},g_{i},h_{i}\in\mathcal{C}(B)$ and $w_{i}\in B^{*}$ such that for all $j\in\mathbb{N}$, $e_{1}^{j}d_{1}c_{1}[\varepsilon]=g_{i}f_{i}^{j}[w_{i}]$ and $e_{i}^{j}d_{i}c_{i}[\varepsilon]=h_{i}f_{i}^{j}[w_{i}]$.

Let $\ell,r\in\{2,\dots,k\}$ such that |\overleftarrow{\makebox{g_{\ell}\vphantom{x}}}|=max\{|\overleftarrow{\makebox{g_{i}\vphantom{x}}}|\mid i\in\{2,\dots,k\}\} and |\overrightarrow{\makebox{g_{r}\vphantom{x}}}|=max\{|\overrightarrow{\makebox{g_{i}\vphantom{x}}}|\mid i\in\{2,\dots,k\}\}. Let g=(\overleftarrow{\makebox{g_{\ell}\vphantom{x}}},\overrightarrow{\makebox{g_{r}\vphantom{x}}}), f=(\overleftarrow{\makebox{f_{\ell}\vphantom{x}}},\overrightarrow{\makebox{f_{r}\vphantom{x}}}), and $w=g^{-1}d_{1}c_{1}[\varepsilon]$. By definition, for all $i\in\{2,\dots,k\}$, |\overleftarrow{\makebox{g_{i}\vphantom{x}}}|\leqslant|\overleftarrow{\makebox{g_{\ell}\vphantom{x}}}| and |\overleftarrow{\makebox{g_{i}\vphantom{x}}}|\leqslant|\overleftarrow{\makebox{g_{r}\vphantom{x}}}|. Thus, for all $i\in\{2,\dots,k\}$, $g_{i}^{-1}g\in\mathcal{C}(B)$. We have that $|g|>|d_{1}c_{1}|$, otherwise it would contradict that the lassos are all non-commuting. Thus, $w\in B^{*}$.

By \Crefr:aligned-means-conjugated, we have that \rho(\overleftarrow{\makebox{e\vphantom{x}}})\sim\rho(\overleftarrow{\makebox{f_{\ell}\vphantom{x}}}) and \rho(\overrightarrow{\makebox{e\vphantom{x}}})\sim\rho(\overrightarrow{\makebox{f_{r}\vphantom{x}}}). Therefore, we can show that for all $j\in\mathbb{N}$, $e_{1}^{j}d_{1}c_{1}[\varepsilon]=gf^{j}[w]$. Then, for all $i\in\{2,\dots,k\}$ and $j\in\mathbb{N}$, $e_{i}^{j}d_{i}c_{i}[\varepsilon]=h_{i}f_{i}^{j}[w_{i}]=h_{i}g_{i}^{-1}e_{1}^{j}d_{1}c_{1}[\varepsilon]=h_{i}g_{i}^{-1}gf^{j}[w]$.

Proof B.25** (Proof of \Crefr:all-commuting-or-aligned).**

The length of the contexts labelling the loops must be equal, as the outputs must grow at the same pace when the loops are pumped. By \Crefr:com-means-com, if one of the lassos is $x\mathrm{-commuting}$ then they are all $x\mathrm{-commuting}$. Otherwise, none of them are commuting. Then, by \Crefr:ctp-concrete, they are also all pairwise aligned and strongly balanced. Therefore by \Crefr:all-correctly-aligned, there exists $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that they are all $(f,w)\mathrm{-aligned}$.

B.3 Lassos Consecutive to a Commuting Lasso

We can now state the following Lemma.

Lemma B.26**.**

\cref@constructprefix

page\cref@result Let $x\in B^{+}$ a primitive word and let $\Delta=\mathsf{split}_{c}(x,H_{1},H_{2})$ for some $H_{1}H_{2}$ an $x\mathrm{-commuting}$ lasso in $\mathcal{T}^{k}$. For any two synchronised lassos $\rho_{1}$ and $\rho_{2}$ in $\mathcal{T}_{\Delta}$, we have that

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either $\rho_{1}$ and $\rho_{2}$ are non-productive,

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or $\rho_{1}$ and $\rho_{2}$ are productive, weakly-balanced, and $\mathrm{strongly-}x\mathrm{-commuting}$,

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or $\rho_{1}$ and $\rho_{2}$ are productive, strongly-balanced, non-commuting, and there exists $g,f\in\mathcal{C}(B)$ such that $\rho_{1}$ and $\rho_{2}$ are $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$.

In order to prove \Crefr:strongly-commuting-or-aligned, we first need some additional combinatorial results.

Lemma B.27**.**

\cref@constructprefix

page\cref@result Let $c_{1},c_{2},d_{1},d_{2}\in\mathcal{C}(B)$ and $x\in B^{+}$ a primitive word. If for all $i,j\in\mathbb{N}$, $\textsl{dist}_{f}(d_{1}^{j}c_{1}[x^{i}],d_{2}^{j}c_{2}[x^{i}])\leqslant L$, then there exist $e_{1},e_{2}\in\mathcal{C}(B)$ such that for all $i,j\in\mathbb{N}$, $e_{1}d_{1}^{j}c_{1}[x^{i}]=e_{2}d_{2}^{j}c_{2}[x^{i}]$.

Proof B.28**.**

Suppose that for all $i,j\in\mathbb{N}$, $\textsl{dist}_{f}(d_{1}^{j}c_{1}[x^{i}],d_{2}^{j}c_{2}[x^{i}])\leqslant L$. Then for all $i,j\in\mathbb{N}$, there exist $f_{1},f_{2}\in\mathcal{C}(B)$, such that $f_{1}^{-1}d_{1}^{j}c_{1}[x^{i}]=f_{2}^{-1}d_{2}^{j}c_{2}[x^{i}]$ and $|f_{1}|+|f_{2}|\leqslant L$. Let $C_{L}=\{(f_{1},f_{2})\mid|f_{1}|+|f_{2}|\leqslant L\}$. $C_{L}$ is finite. Thus there exist some $(f_{1},f_{2})\in C_{L}$ such that there is an infinite number of $i,j\in\mathbb{N}$ such that $f_{1}^{-1}d_{1}^{j}c_{1}[x^{i}]=f_{2}^{-1}d_{2}^{j}c_{2}[x^{i}]$. Furthermore, there exists $i_{0},j_{0}$ such that for all $i\geqslant i_{0}$, $j\geqslant j_{0}$, there exists $e_{1},e_{2}\in\mathcal{C}(B)$ and $e_{1}d_{1}^{j-j_{0}}c_{1}[x^{i-i_{0}}]=e_{2}d_{2}^{j-j_{0}}c_{2}[x^{i-i_{0}}]$. Finally, by applying \Crefr:saarela two times, we obtain that for all $i,j\in\mathbb{N}$, $e_{1}d_{1}^{j}c_{1}[x^{i}]=e_{2}d_{2}^{j}c_{2}[x^{i}]$.

Lemma B.29**.**

\cref@constructprefix

page\cref@result Let $s_{1},u_{1},w_{1},u_{2},w_{2},y_{2}\in B^{*}$ and $t_{1},x_{2},v\in B^{+}$ such that $v$ is primitive, $|t_{1}|=|x_{2}|$ and for all $i,j\in\mathbb{N}$, $s_{1}t_{1}^{j}u_{1}v^{i}w_{1}=u_{2}v^{i}w_{2}x_{2}^{j}y_{2}$. Then there exist $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i,j\geqslant 1$ there exists $k\in\mathbb{N}$ such that $t_{1}^{j}u_{1}v^{i}=f_{1}[v^{k}]$ and $v^{i}w_{2}x_{2}^{j}=f_{2}[v^{k}]$.

Proof B.30**.**

We can find sufficiently large $i_{0}$ and $j_{0}$ such that $t_{1}^{j_{0}}$ and $x_{2}^{j_{0}}$ both overlap with $v^{i_{0}}$ with a common factor of length greater than $|t_{1}|+|v|-\gcd(|t_{1}|,|v|)=|x_{2}|+|v|-\gcd(|x_{2}|,|v|)$. Thus by \Crefr:fw, $\rho(t_{1})\sim\rho(x_{2})\sim\rho(v)$. As $|t_{1}|=|x_{2}|$, $|t_{1}u_{1}v|\geqslant|x_{2}|$. Yet $t_{1}u_{1}v$ is a factor of $x_{2}^{*}$ and, by \Crefr:prootf-of-factor, $\bar{\rho}(t_{1}u_{1}v)\sim\rho(x_{2})$. Similarly, $\bar{\rho}(vw_{2}x_{2})\sim\rho(t_{1})$. Thus $\rho(t_{1})\sim\rho(x_{2})\sim\bar{\rho}(t_{1}u_{1}v)\sim\bar{\rho}(vw_{2}x_{2})\sim\rho(v)$.

Let $z_{1},z_{1}^{\prime},z_{2},z_{2}^{\prime}\in B^{*}$ and $\alpha\geqslant 1$ such that $z_{1}z_{1}^{\prime}=z_{2}z_{2}^{\prime}$, and $v=z_{1}z_{1}^{\prime}$, $t_{1}=(z_{1}^{\prime}z_{1})^{\alpha}$ and $x_{2}=(z_{2}^{\prime}z_{2})^{\alpha}$. Also we have $t_{1}u_{1}v=z_{1}^{\prime}(z_{1}z_{1}^{\prime})^{\theta_{1}}$, $vw_{2}x_{2}=(z_{1}z_{1}^{\prime})^{\theta_{2}}z_{2}$, for some $\theta_{1},\theta_{2}\geqslant 0$.

We choose $f_{1}=(z_{1}^{\prime},(z_{1}z_{1}^{\prime})^{\theta_{1}})$, and $f_{2}=((z_{1}z_{1}^{\prime})^{\theta_{2}},z_{2})$. Then, for all $i,j\geqslant 1$, we have $t_{1}^{j}u_{1}v^{i}=z_{1}^{\prime}(z_{1}z_{1}^{\prime})^{\alpha(j-1)+\theta_{1}+(i-1)}$ and $v^{i}w_{2}x_{2}^{j}=(z_{1}z_{1}^{\prime})^{(i-1)+\theta_{2}+\alpha(j-1)}z_{2}$. Let $k=(i-1)+\alpha(j-1)\geqslant 0$. And we obtain $t_{1}^{j}u_{1}v^{i}=f_{1}[v^{k}]$ and $v^{i}w_{2}x_{2}^{j}=f_{2}[v^{k}]$.

Lemma B.31**.**

\cref@constructprefix

page\cref@result Let $s_{1},u_{1},w_{1},s_{2},u_{2},w_{2}\in B^{*}$ and $t_{1},t_{2},v\in B^{+}$ such that $|t_{1}|=|t_{2}|$ and for all $i,j\in\mathbb{N}$, $s_{1}t_{1}^{j}u_{1}v^{i}w_{1}=s_{2}t_{2}^{j}u_{2}v^{i}w_{2}$. Then there exist some contexts $f,g,h_{1},h_{2}\in\mathcal{C}(B)$ such that for all $i,j\in\mathbb{N}$, $t_{1}^{j}u_{1}v^{i}=h_{1}g^{j}f[v^{i}]$ and $t_{2}^{j}u_{2}v^{i}=h_{2}g^{j}f[v^{i}]$.

Proof B.32**.**

There exists $j_{0}$ sufficiently large such that $t_{1}^{j_{0}}$ overlap with $t_{2}^{j_{0}}$ with a common factor of length greater than $|t_{1}|+|t_{2}|-\gcd(|t_{1}|,|t_{2}|)=|t_{1}|=|t_{2}|$. Thus by \Crefr:fw, $\rho(t_{1})\sim\rho(t_{2})$. If $|s_{1}|\leqslant|s_{2}|$, let $t$ such that $s_{1}t=s_{2}$ and $t_{1}t=tt_{2}$. If $|s_{1}|\geqslant|s_{2}|$, let $t$ such that $s_{1}=s_{2}t$ and $tt_{1}=t_{2}t$. If $|w_{1}|\leqslant|w_{2}|$, let $i_{0}$ such that $v^{i_{0}}w_{1}=w_{2}$ and $v^{i_{0}}\preceq_{s}u_{1}$. If $|w_{1}|\geqslant|w_{2}|$, let $i_{0}$ such that $w_{1}=v^{i_{0}}w_{2}$ and $v^{i_{0}}\preceq_{s}u_{2}$. We obtain that

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$s_{1}t=s_{2}$, $t_{1}t=tt_{2}$, $u_{1}=tu_{2}v^{i_{0}}$, $v^{i_{0}}w_{1}=w_{2}$, or

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$s_{1}=s_{2}t$, $tt_{1}=t_{2}t$, $tu_{1}=u_{2}v^{i_{0}}$, $v^{i_{0}}w_{1}=w_{2}$, or

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$s_{1}t=s_{2}$, $t_{1}t=tt_{2}$, $u_{1}v^{i_{0}}=tu_{2}$, $w_{1}=v^{i_{0}}w_{2}$, or

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$s_{1}=s_{2}t$, $tt_{1}=t_{2}t$, $tu_{1}v^{i_{0}}=u_{2}$, $w_{1}=v^{i_{0}}w_{2}$.

We handle the first case. The others are similar. We choose $h_{1}=(t,v^{i_{0}})$, $h_{2}=(\varepsilon,\varepsilon)$, $g=(t_{2},\varepsilon)$, and $f=(u_{2},\varepsilon)$. Then for all $i\in\mathbb{N}$, $t_{1}^{j}u_{1}v^{i}=t_{1}^{j}tu_{2}v^{i_{0}}v^{i}=tt_{2}^{j}u_{2}v^{i}v^{i_{0}}=h_{1}g^{j}f[v^{i}]$ and $t_{2}^{j}u_{2}v^{i}=h_{2}g^{j}f[v^{i}]$.

Lemma B.33**.**

\cref@constructprefix

page\cref@result Let $s_{1},u_{1},w_{1},s_{2},u_{2},w_{2},y_{2}\in B^{*}$ and $t_{1},t_{2},x_{2},v\in B^{+}$ such that $|t_{1}|=|t_{2}|+|x_{2}|$ and for all $i,j\in\mathbb{N}$, $s_{1}t_{1}^{j}u_{1}v^{i}w_{1}=s_{2}t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}y_{2}$. Then there exist $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i,j\geqslant 1$ there exists $k\in\mathbb{N}$ such that $t_{1}^{j}u_{1}v^{i}=f_{1}[v^{k}]$ and $t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}=f_{2}[v^{k}]$.

Proof B.34**.**

We can find sufficiently large $i_{0}$ and $j_{0}$ such that $t_{1}^{j_{0}}$ overlap with $t_{2}^{j_{0}}$, $v^{i_{0}}$ and $x_{2}^{j_{0}}$ with common factors of respective length greater than $|t_{1}|+|t_{2}|-\gcd(|t_{1}|,|t_{2}|)$, $|t_{1}|+|v|-\gcd(|t_{1}|,|v|)$, and $|t_{1}|+|x_{2}|-\gcd(|t_{1}|,|x_{2}|)$. Thus by \Crefr:fw, $\rho(t_{1})\sim\rho(t_{2})\sim\rho(x_{2})\sim\rho(v)$. As $|t_{1}|=|x_{2}|$, $|t_{1}u_{1}v|\geqslant|x_{2}|$. Similarly to \Crefr:comb-com-2-2-rev, we can show that $\bar{\rho}(t_{1}u_{1}v)\sim\bar{\rho}(t_{2}u_{2}v)\sim\bar{\rho}(vw_{2}x_{2})\sim\rho(v)$, and reconstruct the words to obtain the result.

Lemma B.35**.**

\cref@constructprefix

page\cref@result Let $s_{1},u_{1},w_{1},y_{1},s_{2},u_{2},w_{2},y_{2}\in B^{*}$ and $t_{1},x_{1},t_{2},x_{2},v\in B^{+}$ such that $|t_{1}|+|t_{2}|=|x_{1}|+|x_{2}|$, $|t_{1}|\neq|x_{1}|$ and $|t_{2}|\neq|x_{2}|$, and for all $i,j\in\mathbb{N}$, $s_{1}t_{1}^{j}u_{1}v^{i}w_{1}x_{1}^{j}y_{1}=s_{2}t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}y_{2}$. Then there exist $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i,j\geqslant 1$ there exists $k\in\mathbb{N}$ such that $t_{1}^{j}u_{1}v^{i}w_{1}x_{1}^{j}=f_{1}[v^{k}]$ and $t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}=f_{2}[v^{k}]$.

Proof B.36**.**

Without loss of generality, consider that $|t_{1}|>|t_{2}|$. We can find sufficiently large $i_{0}$ and $j_{0}$ such that $t_{1}^{j_{0}}$ overlap with $t_{2}^{j_{0}}$, $v^{i_{0}}$ and $x_{2}^{j_{0}}$ with common factors of respective length greater than $|t_{1}|+|t_{2}|-\gcd(|t_{1}|,|t_{2}|)$, $|t_{1}|+|v|-\gcd(|t_{1}|,|v|)$, and $|t_{1}|+|x_{2}|-\gcd(|t_{1}|,|x_{2}|)$. Thus by \Crefr:fw, $\rho(t_{1})\sim\rho(t_{2})\sim\rho(x_{2})\sim\rho(v)$. Similarly, we can show that $\rho(x_{1})\sim\rho(x_{2})$. Finally, similarly to \Crefr:comb-com-2-2-rev, we can show that $\bar{\rho}(t_{1}u_{1}v)\sim\bar{\rho}(t_{2}u_{2}v)\sim\bar{\rho}(vw_{1}x_{1})\sim\bar{\rho}(vw_{2}x_{2})\sim\rho(v)$, and reconstruct the words to obtain the result.

Lemma B.37**.**

\cref@constructprefix

page\cref@result Let $s_{1},u_{1},w_{1},y_{1},s_{2},u_{2},w_{2},y_{2}\in B^{*}$ and $t_{1},x_{1},t_{2},x_{2},v\in B^{+}$ such that $|t_{1}|=|x_{1}|$, $|t_{2}|=|x_{2}|$, and for all $i,j\in\mathbb{N}$, $s_{1}t_{1}^{j}u_{1}v^{i}w_{1}x_{1}^{j}y_{1}=s_{2}t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}y_{2}$. Then there exist some contexts $f,g,h_{1},h_{2}\in\mathcal{C}(B)$ such that for all $i,j\in\mathbb{N}$, $t_{1}^{j}u_{1}v^{i}w_{1}x_{1}^{j}=h_{1}g^{j}f[v^{i}]$ and $t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}=h_{2}g^{j}f[v^{i}]$.

Proof B.38**.**

There exists $j_{0}$ sufficiently large such that $t_{1}^{j_{0}}$ overlap with $t_{2}^{j_{0}}$ with a common factor of length greater than $|t_{1}|+|t_{2}|-\gcd(|t_{1}|,|t_{2}|)=|t_{1}|=|t_{2}|$. Thus by \Crefr:fw, $\rho(t_{1})\sim\rho(t_{2})$. The same applies to $x_{1}^{j_{0}}$ and $x_{2}^{j_{0}}$, and $\rho(x_{1})\sim\rho(x_{2})$. By a reasoning similar to \Crefr:comb-com-2-2, based on the length of $s_{1}$, $s_{2}$, $y_{1}$ and $y_{2}$, we can show that there exists $t,x\in B^{*}$ such that we obtain

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$s_{1}t=s_{2}$, $t_{1}t=tt_{2}$, $u_{1}v^{i}w_{1}=tu_{2}v^{i}w_{2}x$, $xx_{1}=x_{2}x$, $xy_{1}=y_{2}$, or

• •

$s_{1}t=s_{2}$, $t_{1}t=tt_{2}$, $u_{1}v^{i}w_{1}x=tu_{2}v^{i}w_{2}$, $x_{1}x=xx_{2}$, $y_{1}=xy_{2}$, or

• •

$s_{1}=s_{2}t$, $tt_{1}=t_{2}t$, $tu_{1}v^{i}w_{1}=u_{2}v^{i}w_{2}x$, $xx_{1}=x_{2}x$, $xy_{1}=y_{2}$, or

• •

$s_{1}=s_{2}t$, $tt_{1}=t_{2}t$, $tu_{1}v^{i}w_{1}x=u_{2}v^{i}w_{2}$, $x_{1}x=xx_{2}$, $y_{1}=xy_{2}$.

Again similarly to \Crefr:comb-com-2-2, we can reconstruct the words $t_{1}^{j}u_{1}v^{i}w_{1}x_{1}^{j}$ and $t_{2}^{j}u_{2}v^{i}w_{2}x_{2}^{j}$ to obtain the result.

Proof B.39** (Proof of \Crefr:strongly-commuting-or-aligned).**

Let $x\in B^{+}$ a primitive word, $H_{1}H_{2}$ be an $x\mathrm{-commuting}$ in $\mathcal{T}^{k}$, and let $\Delta=\mathsf{split}_{c}(x,H_{1},H_{2})$. Let \rho_{1}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{1} and \rho_{2}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{2} be two lassos in $\mathcal{T}_{\Delta}$.

As $H_{1}H_{2}$ is $x\mathrm{-commuting}$, we can find two synchronised $x\mathrm{-commuting}$ lassos in $\mathcal{T}$ around $p_{1}$ and $p_{2}$. Let \rho_{1}^{\prime}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{0,1}}}i_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1,1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2,1}}}p_{1} and \rho_{2}^{\prime}:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{0,2}}}i_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1,2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2,2}}}p_{2} be those two lassos. Furthermore, by definition of $\mathsf{split}_{c}$, we have that $\Delta(p_{1})=c_{1}$ and $\Delta(p_{2})=c_{2}$, and for all $i\in\mathbb{N}$, there exists $k\in\mathbb{N}$ such that $c_{2,1}^{i}c_{1,1}c_{0,1}[\varepsilon]=c_{1}[x^{k}]$ and $c_{2,2}^{i}c_{1,2}c_{0,2}[\varepsilon]=c_{2}[x^{k}]$. By \Crefd:ctp, for all $i,j\in\mathbb{N}$, $\textsl{dist}_{f}(e_{1}^{j}d_{1}c_{1}[x^{i}],e_{2}^{j}d_{2}c_{2}[x^{i}])\leqslant L$.

By \Crefr:com-dist-to-equation, there exist $f_{1},f_{2}\in\mathcal{C}(B)$ such that for all $i,j\in\mathbb{N}$, $f_{1}e_{1}^{j}d_{1}c_{1}[x^{i}]=f_{2}e_{2}^{j}d_{2}c_{2}[x^{i}]$. Then we have that $|e_{1}|=|e_{2}|$. We observe 10 cases.

If $|e_{1}|=0$ or $|e_{2}|=0$ then $|e_{1}|=|e_{2}|=0$ and $\rho_{1},\rho_{2}$ are non-productive.

If $e_{1}\in\{\varepsilon\}\times B^{+}$ and $e_{2}\in B^{+}\times\{\varepsilon\}$, then by \Crefr:comb-com-2-2-rev, both $\rho_{1}$ and $\rho_{2}$ are productive and $\mathrm{strongly-}x\mathrm{-commuting}$. The same holds for the symmetrical case where $e_{1}\in B^{+}\times\{\varepsilon\}$ and $e_{2}\{\varepsilon\}\times B^{+}$.

If $e_{1},e_{2}\in B^{+}\times\{\varepsilon\}$, then by \Crefr:comb-com-2-2, there exist $f,g\in\mathcal{C}(B)$ such that both $\rho_{1}$ and $\rho_{2}$ are productive and $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$. However, if it still happens that either one of $\rho_{1}$ and $\rho_{2}$ is $\mathrm{strongly-}x\mathrm{-commuting}$ (implying that \rho(\overleftarrow{\makebox{e_{1}\vphantom{x}}})\sim\rho(\overrightarrow{\makebox{e_{1}\vphantom{x}}})\sim x or \rho(\overleftarrow{\makebox{e_{2}\vphantom{x}}})\sim\rho(\overrightarrow{\makebox{e_{2}\vphantom{x}}})\sim x), then both $\rho_{1}$ and $\rho_{2}$ are $\mathrm{strongly-}x\mathrm{-commuting}$. If not, then they both are non-commuting. The same holds for the symmetrical case where $e_{1},e_{2}\in\{\varepsilon\}\times B^{+}$.

If $e_{1}\in B^{+}\times\{\varepsilon\}$ and $e_{2}\in B^{+}\times B^{+}$, then by \Crefr:comb-com-2-3, both $\rho_{1}$ and $\rho_{2}$ are productive and $\mathrm{strongly-}x\mathrm{-commuting}$. The same holds for the other three cases where exactly one of the four components of $e_{1}$ and $e_{2}$ is empty.

If $e_{1},e_{2}\in B^{+}\times B^{+}$ and $\|e_{1}\|\neq\|e_{2}\|$, then by \Crefr:comb-com-3-3-weak, both $\rho_{1}$ and $\rho_{2}$ are productive and $\mathrm{strongly-}x\mathrm{-commuting}$.

If $e_{1},e_{2}\in B^{+}\times B^{+}$ and $\|e_{1}\|=\|e_{2}\|$, then by \Crefr:comb-com-3-3-strong, there exist $f,g\in\mathcal{C}(B)$ such that both $\rho_{1}$ and $\rho_{2}$ are productive and $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$. However, if it still happens that either one of $\rho_{1}$ and $\rho_{2}$ is $\mathrm{strongly-}x\mathrm{-commuting}$ (implying that \rho(\overleftarrow{\makebox{e_{1}\vphantom{x}}})\sim\rho(\overrightarrow{\makebox{e_{1}\vphantom{x}}})\sim x or \rho(\overleftarrow{\makebox{e_{2}\vphantom{x}}})\sim\rho(\overrightarrow{\makebox{e_{2}\vphantom{x}}})\sim x), then both $\rho_{1}$ and $\rho_{2}$ are $\mathrm{strongly-}x\mathrm{-commuting}$. If not, then they both are non-commuting.

Proof B.40** (Proof of \Crefr:all-strongly-commuting-or-aligned).**

By \Crefr:strongly-commuting-or-aligned, if not productive, two lassos following a commuting lasso are either strongly commuting or strongly aligned. This result can be lifted to $k$ runs in a similar way to \Crefr:all-commuting-or-aligned.

B.4 Lassos Consecutive to a Non-Commuting Lasso

We finally study the properties of lassos that are consecutive to a non-commuting lasso. The following Lemma shows that only a non-commuting lasso can follow a non-commuting lasso.

Lemma B.41**.**

\cref@constructprefix

page\cref@result Let $f\in\mathcal{C}(B)$, $w\in B^{*}$, such that there exists no $x\in B^{+}$ and $g\in\mathcal{C}(B)$ such that, for all $i\in\mathbb{N}$, there exists $k\in\mathbb{N}$ such that $f^{i}[w]=g[x^{k}]$. Let $c,d\in\mathcal{C}(B)$, and $i\geqslant 1$. Then there exists no $x\in B^{+}$ and $g\in\mathcal{C}(B)$ such that, for all $j\in\mathbb{N}$, there exists $k\in\mathbb{N}$ such that $d^{j}cf^{i}[w]=g[x^{k}]$.

Proof B.42**.**

This can easily be proved by contradiction.

The lassos after a non-commuting lasso are then always fully-aligned. Given a string-to-context transducer that satisfies the contextual twinning property, we can thus view its restriction after an aligned lasso as the pair of two classical finite state transducers that both satisfy the classical twinning property.

Lemma B.43**.**

\cref@constructprefix

page\cref@result Let $f\in\mathcal{C}(B)$ and $w\in B^{*}$ and let $\Delta=\mathsf{split}_{nc}(f,w,H_{1},H_{2})$, for some $H_{1}H_{2}$ a productive, non-commuting and $(f,w)-aligned$ lasso in $\mathcal{T}^{\leqslant|Q|}$. Then \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} and \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} both satisfy the TP.

Proof B.44**.**

We show the result for \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}}. The proof for \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} is symmetrical.

Let \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}y}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}z}}q and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x^{\prime}}}p^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}y^{\prime}}}q^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}z^{\prime}}}q^{\prime} two lassos in \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}}. By the definition of \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}}, there exist $(p,c),(p^{\prime},c^{\prime})\in\Delta$ such that \overrightarrow{\makebox{c\vphantom{x}}}=x and \overrightarrow{\makebox{c^{\prime}\vphantom{x}}}=x^{\prime}, and p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q and p^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d^{\prime}}}q^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e^{\prime}}}q^{\prime} in $\mathcal{T}_{\Delta}$ such that \overrightarrow{\makebox{d\vphantom{x}}}=y, \overrightarrow{\makebox{d^{\prime}\vphantom{x}}}=y^{\prime}, \overrightarrow{\makebox{e\vphantom{x}}}=z, \overrightarrow{\makebox{e^{\prime}\vphantom{x}}}=z^{\prime}.

From $\Delta=\mathsf{split}_{nc}(f,w,H_{1},H_{2})$ and $H_{1}H_{2}$ being non-commuting, we know that there exist $i\geqslant 1$ and two lassos \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}o\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{3}}}p and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}^{\prime}}}o^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}^{\prime}}}p^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{3}^{\prime}}}p^{\prime} in $\mathcal{T}$, such that $c_{3}c_{2}c_{1}[\varepsilon]=cf^{i}[w]$ and $c_{3}^{\prime}c_{2}^{\prime}c_{1}^{\prime}[\varepsilon]=c^{\prime}f^{i}[w]$. Thus we can build two lassos in $\mathcal{T}$: \rho_{1}:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}o\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}dc_{3}c_{2}}}q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}q and \rho_{2}:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}^{\prime}}}o^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d^{\prime}c_{3}^{\prime}c_{2}^{\prime}}}q^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e^{\prime}}}q^{\prime}.

*By \Crefd:ctp, for all $j\in\mathbb{N}$, $\textsl{dist}_{f}(e^{j}dcf^{i}[w],e^{\prime j}d^{\prime}c^{\prime}f^{i}[w])\leqslant L$. As $H_{1}H_{2}$ is non-commuting, by \Crefr:nc-nc, $\rho_{1}$ and $\rho_{2}$ must also be non-commuting, and thus strongly-balanced. As $H_{1}H_{2}$ and $\rho_{1}$ and $\rho_{2}$ are strongly-balanced and non-commuting the $f^{i}[w]$ part can only overlap with itself in the words $e^{j}dcf^{i}[w]$ and $e^{\prime j}d^{\prime}c^{\prime}f^{i}[w]$. Therefore, we can derive that for all $j\in\mathbb{N}$, $\textsl{dist}_{p}(xyz^{j},x^{\prime}y^{\prime}z^{\prime j})\leqslant L$. *

Lemma B.45**.**

\cref@constructprefix

page\cref@result Let $x\in B^{+}$ a primitive word and let $\Delta_{0}=\mathsf{split}_{c}(x,H_{1},H_{2})$, for some $H_{1}H_{2}$ an productive and $x\mathrm{-commuting}$ lasso in $\mathcal{T}^{\leqslant|Q|}$. Let $g,f\in\mathcal{C}(B)$ and let $\Delta=\mathsf{extract}_{nc}(g,f,x,\Delta_{0},H_{3},H_{4})$, for some $H_{3}H_{4}$ a productive, non-commuting and $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$ lasso in $\mathcal{T}_{\Delta_{0}}^{\leqslant|Q|}$. Then \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} and \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} both satisfy the TP.

Proof B.46**.**

We show the result for \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}}. The proof for \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} is symmetrical.

Let \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}v_{0}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}v_{1}}}q_{3}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}v_{2}}}q_{3} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}v_{0}^{\prime}}}q_{2}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}v_{1}^{\prime}}}q_{3}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}v_{2}^{\prime}}}q_{3}^{\prime} two lassos in \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}}. By the definition of \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}}, there exist $(q_{2},e_{0}),(q_{2}^{\prime},e_{0}^{\prime})\in\Delta$ such that \overrightarrow{\makebox{e_{0}\vphantom{x}}}=v_{0} and \overrightarrow{\makebox{e_{0}^{\prime}\vphantom{x}}}=v_{0}^{\prime}, and q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}}q_{3}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{3} and q_{2}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}^{\prime}}}q_{3}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}^{\prime}}}q_{3}^{\prime} in $\mathcal{T}_{\Delta}$ such that \overrightarrow{\makebox{e_{1}\vphantom{x}}}=v_{1}, \overrightarrow{\makebox{e_{1}^{\prime}\vphantom{x}}}=v_{1}^{\prime}, \overrightarrow{\makebox{e_{2}\vphantom{x}}}=v_{2}, \overrightarrow{\makebox{e_{2}^{\prime}\vphantom{x}}}=v_{2}^{\prime}.

From $\Delta=\mathsf{extract}_{nc}(g,f,x,\Delta_{0},H_{3},H_{4})$ and $H_{3}H_{4}$ being $\mathrm{strongly-}(g,f)\mathrm{-aligned}$, we know that there exist $i\geqslant 1$ and two lassos \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{0}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{0}^{\prime}}}q_{1}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}^{\prime}}}q_{2}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}^{\prime}}}q_{2}^{\prime} in $\mathcal{T}_{\Delta_{0}}$, such that $d_{2}d_{1}d_{0}=e_{0}g^{i}f$ and $d_{2}^{\prime}d_{1}^{\prime}d_{0}^{\prime}=e_{0}^{\prime}g^{i}f$. Thus we can build two lassos in $\mathcal{T}_{\Delta_{0}}$: \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{0}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{1}t_{2}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}d_{2}d_{1}}}q_{3}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{3} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{0}^{\prime}}}q_{1}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}t_{1}t_{2}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}^{\prime}d_{2}^{\prime}d_{1}^{\prime}}}q_{3}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}^{\prime}}}q_{3}^{\prime}.

From $\Delta_{0}=\mathsf{split}_{c}(x,H_{1},H_{2})$ and $H_{1}H_{2}$ being $x\mathrm{-commuting}$, we know that there exist $k\geqslant 1$, $c\in\mathcal{C}(B)$ and two lassos \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{0}}}q_{0}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}q_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}q_{1} and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{0}^{\prime}}}q_{0}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}^{\prime}}}q_{1}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}^{\prime}}}q_{1}^{\prime} in $\mathcal{T}$, such that $c_{2}c_{1}c_{0}[\varepsilon]=d_{0}c[x^{k}]$ and $c_{2}^{\prime}c_{1}^{\prime}c_{0}^{\prime}[\varepsilon]=d_{0}^{\prime}c[w^{k}]$. Thus we can build two lassos in $\mathcal{T}$: \rho_{1}:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{0}}}q_{0}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s_{1}s_{2}t_{1}t_{2}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}d_{2}d_{1}c_{2}c_{1}}}q_{3}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}q_{3} and \rho_{2}:\ \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{0}^{\prime}}}q_{0}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}s_{1}s_{2}t_{1}t_{2}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}^{\prime}d_{2}^{\prime}d_{1}^{\prime}c_{2}^{\prime}c_{1}^{\prime}}}q_{3}^{\prime}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}^{\prime}}}q_{3}^{\prime}.

By \Crefd:ctp, for all $j\in\mathbb{N}$, $\textsl{dist}_{f}(e_{2}^{j}e_{1}e_{0}g^{i}fc[x^{k}],e_{2}^{\prime j}e_{1}^{\prime}e_{0}^{\prime}g^{i}fc[x^{k}])\leqslant L$. As $H_{1}H_{2}$ is non-commuting, by \Crefr:nc-nc, $\rho_{1}$ and $\rho_{2}$ must also be non-commuting, and thus strongly-balanced. As $H_{1}H_{2}$ and $\rho_{1}$ and $\rho_{2}$ are strongly-balanced and non-commuting the $g^{i}fc[x^{k}]$ part can only overlap with itself in the words $e_{2}^{j}e_{1}e_{0}g^{i}fc[x^{k}]$ and $e_{2}^{\prime j}e_{1}^{\prime}e_{0}^{\prime}g^{i}fc[x^{k}]$. Therefore, we can derive that for all $j\in\mathbb{N}$, $\textsl{dist}_{p}(v_{0}v_{1}v_{2}^{j},v_{0}^{\prime}v_{1}^{\prime}v_{2}^{\prime j})\leqslant L$.

Proof B.47** (Proof of \Crefr:all-fully-aligned).**

By \Crefr:split-nc-tp,r:extract-nc-tp, \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} and \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} both satisfy the TP.

Appendix C Proofs of \Crefsec:construction: Determinisation

C.1 Observation on the $\mathsf{choose}$ operator

We have considered a functional transducer $\mathcal{T}$, and have assumed that $\mathcal{T}$ is trim. As a consequence, if we consider two runs \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q and \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q, then there exists a run q\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e}}f\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}g}} where $f$ is final. By functionality, we have $ged_{1}c_{1}[\varepsilon]=ged_{2}c_{2}[\varepsilon]$, hence $d_{1}c_{1}[\varepsilon]=d_{2}c_{2}[\varepsilon]$. This implies that even if the $\mathsf{choose}$ operator may select different contexts corresponding to different runs leading to the same state, when they are applied to $\varepsilon$, they yield the same word.

Similarly, if we consider two runs p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q and p\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q, and a word $w$ such that \xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}i\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}v}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}q with $w=dc[\varepsilon]$, then we have $d_{1}[w]=d_{2}[w]$. As a consequence, the choice realised by $\mathsf{choose}$ has no impact as soon as one compares the contexts applied to a possible output word produced before.

Last, using a similar reasoning, we can prove that when considering two runs p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q and p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q such that $p_{1}$ and $p_{2}$ appear in some $\Delta:Q\hookrightarrow\mathcal{C}(B)$, obtained after a non-commuting lasso, then we have $d_{1}\Delta(p_{1})=d_{2}\Delta(p_{2})$. Hence, the choice realised by $\mathsf{choose}$ has actually no impact.

In the sequel, we will often write equalities involving partial functions $\Delta:Q\hookrightarrow\mathcal{C}(B)$. When these equalities are in one of the three above situations, we will thus omit the operator $\mathsf{choose}$, for simplicity of the writing.

C.2 Additional Notations

For a run H in $\mathcal{T}^{k}$, we denote by $\textsf{word}(H)$ the word read by $H$.

Given a state $\overline{p}=(x,\Delta,H)\in\overline{Q}$, we let $x_{\overline{p}}$ be equal to $x$.

Given a state $\overline{p}=(x,\Delta,H)\in\overline{Q}$ and some run $H^{\prime}$ in $\mathcal{T}^{k}$ such that the start state of $H^{\prime}$ is the end state of $H$, we let $\overline{p}\bullet H^{\prime}=(x,\Delta,HH^{\prime})$.

When necessary, we extend the run notation to include the transducer name. For instance, \overline{p}\xrightarrow[\mathcal{D}]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q} depicts a transition in $\mathcal{D}$ from state $\overline{p}$ to state $\overline{q}$ reading the word $u$ and outputting the context $c$.

C.3 Correctness

We start with the following easy observation:

Lemma C.1**.**

\cref@constructprefix

page\cref@result Let $\Delta:Q\hookrightarrow\mathcal{C}(B)$ and $H$ a history in $\mathcal{T}_{\Delta}^{\leqslant|Q|}$. If $H=H_{1}H_{2}$, with $H_{1}H_{2}$ a non-productive lasso. then $\Delta\bullet H_{1}=(\Delta\bullet H_{1})\bullet H_{2}$.

Proof C.2** (Proof of \Crefr:corr-extend).**

We proceed to a case analysis.

$x_{\overline{p}}=\varepsilon$*  Let $\overline{p}=(\varepsilon,t_{\textsl{init}},H_{1})$. Either $H_{2}$ is non-productive, $\overline{q}=\overline{p}$, $c=c_{\varepsilon}$ and, by \Crefr:act-non-prod, $\Delta_{\overline{p}}\bullet H_{2}[\varepsilon]=t_{\textsl{init}}\bullet H_{1}H_{2}[\varepsilon]=t_{\textsl{init}}\bullet H_{1}[\varepsilon]=\Delta_{\overline{p}}[\varepsilon]=\Delta_{\overline{q}}c[\varepsilon]$.*

Otherwise, $H_{2}$ is productive and we pass through the else if block at Line 4.

If $H_{1}H_{2}$ is $x\mathrm{-commuting}$ for some $x\in B^{+}$, then $\Delta=\mathsf{split}_{c}(x,H_{1},H_{2})$, $k=\mathsf{pow}_{c}(x,H_{1},H_{2})$, $\overline{q}=(x,\Delta,id_{\Delta})$ and $c=(\varepsilon,x^{k})$. Thus, $\Delta_{\overline{p}}\bullet H_{2}[\varepsilon]=t_{\textsl{init}}\bullet H_{1}H_{2}[\varepsilon]=\Delta[x^{k}]=\Delta_{\overline{q}}c[\varepsilon]$.

If $H_{1}H_{2}$ is $(f,w)\mathrm{-aligned}$ for some $f\in\mathcal{C}(B)$ and $w\in B^{*}$, then $\Delta=\mathsf{split}_{nc}(f,w,H_{1},H_{2})$, $\overline{q}=(\bot,\Delta,id_{\Delta})$ and $c=f\cdot(\varepsilon,w)$. Thus, $\Delta_{\overline{p}}\bullet H_{2}[\varepsilon]=t_{\textsl{init}}\bullet H_{1}H_{2}[\varepsilon]=\Delta f[w]=\Delta_{\overline{q}}c[\varepsilon]$.

$x_{\overline{p}}\in B^{+}$*  Let $\overline{p}=(x,\Delta_{0},H_{1})$ and $j\in\mathbb{N}$. Either $H_{2}$ is non-productive, $\overline{q}=\overline{p}$, $c=c_{\varepsilon}$ and, by \Crefr:act-non-prod, $\Delta_{\overline{p}}\bullet H_{2}[x^{j}]=\Delta_{0}\bullet H_{1}H_{2}[x^{j}]=\Delta_{0}\bullet H_{1}[x^{j}]=\Delta_{\overline{p}}[x^{j}]=\Delta_{\overline{q}}c[x^{j}]$.*

Otherwise, $H_{2}$ is productive and we pass through the else if block at Line 12.

If $H_{1}H_{2}$ is $\mathrm{strongly-}x\mathrm{-commuting}$, then $k=|\textsf{out}(H_{2})|/|x|$, $\overline{q}=\overline{p}$ and $c=(\varepsilon,x^{k})$. Thus, $\Delta_{\overline{p}}\bullet H_{2}[x^{j}]=\Delta_{0}\bullet H_{1}H_{2}[x^{j}]=\Delta_{0}\bullet H_{1}[x^{j+k}]=\Delta_{\overline{q}}c[x^{j}]$.

If $H_{1}H_{2}$ is $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$ for some $g,f\in\mathcal{C}(B)$, then $\Delta=\linebreak\mathsf{extract}_{nc}(g,f,x,\Delta_{0},H_{1},H_{2})$, $\overline{q}=(\bot,\Delta,id_{\Delta})$ and $c=gf$. Thus, $\Delta_{\overline{p}}\bullet H_{2}[x^{j}]=\Delta_{0}\bullet H_{1}H_{2}[x^{j}]=\Delta gf[x^{j}]=\Delta_{\overline{q}}c[x^{j}]$.

Proof C.3** (Proof of \Crefr:corr-simplify).**

The result follows from \Crefr:corr-simplify-nc,r:corr-simplify-s&c.

Lemma C.4** (Correctness of $\textsc{simplify}($) for non-commuting states).**

\cref@constructprefix

page\cref@result Let $\overline{p}\in\overline{Q}_{\infty}$ such that $x_{\overline{p}}=\bot$ and $(\overline{q},c)=\textsc{simplify}(\overline{p})$. Then $\overline{q}\in\overline{Q}$ and $\Delta_{\overline{p}}=\Delta_{\overline{q}}c$.

Proof C.5**.**

Let $\overline{p}=(\bot,\Delta,H)$. The fact that $\overline{q}\in\overline{Q}$ is trivial. As $x_{\overline{p}}=\bot$, we only pass through the if block at Line 23. Let $\Delta^{\prime}=\Delta\bullet H$ and $c=\textsf{lcc}(\Delta^{\prime})$. and $\overline{q}=(\bot,\Delta^{\prime}\cdot c^{-1},id_{\Delta^{\prime}})$. Thus $\Delta_{\overline{q}}c=(\Delta\bullet H)\cdot c^{-1}c=\Delta_{\overline{p}}$.

Lemma C.6** (Correctness of $\textsc{simplify}($) for startup and commuting states).**

\cref@constructprefix

page\cref@result Let $\overline{p}\in\overline{Q}_{\infty}$ such that $x_{\overline{p}}\neq\bot$ and $(\overline{q},c)=\textsc{simplify}(\overline{p})$. Then $\overline{q}\in\overline{Q}$ and we have:

• •

If $x_{\overline{p}}=\varepsilon$ then $\Delta_{\overline{p}}[\varepsilon]=\Delta_{\overline{q}}c[\varepsilon]$.

• •

If $x_{\overline{p}}\in B^{+}$ then for all $k\in\mathbb{N}$, $\Delta_{\overline{p}}[x_{\overline{p}}^{k}]=\Delta_{\overline{q}}c[x_{\overline{p}}^{k}]$.

Proof C.7**.**

First observe that the fact that $\overline{q}\in\overline{Q}$ can be proven using a simple induction.

We now consider the second property, and proceed by strong induction on $|H_{\overline{p}}|$. If $H_{\overline{p}}=id$ then, as $x_{\overline{p}}\neq\bot$, we only pass through the else statement at Line 31. Then the result is trivially obtained.

Otherwise $|H_{\overline{p}}|>0$. If $H_{\overline{p}}$ doesn’t contain a loop then, again, we only pass through the else statement at Line 31, and the result is trivially obtained.

If $H_{\overline{p}}$ contains a loop, we pass through the else if block at Line 26. Let $\overline{p}=(x,\Delta,H)$ and $H=H_{1}H_{2}H_{3}$ where $H_{1}H_{2}$ is the first lasso in $H_{\overline{p}}$. Let $\overline{q}=(x,\Delta,H_{1})$, $(\overline{r},c)=\textsc{extend\_with\_loop}(\overline{q},H_{2})$, and $(\overline{s},d)=\textsc{simplify}(\overline{r}\bullet H_{3})$. We observe two cases.

$x_{\overline{p}}=\varepsilon$*  By \Crefr:corr-extend, we have that $t_{\textsl{init}}\bullet H_{1}\bullet H_{2}[\varepsilon]=\Delta_{\overline{q}}\bullet H_{2}[\varepsilon]=\Delta_{\overline{r}}c[\varepsilon]$.*

If $x_{\overline{r}}=\bot$ then, by \Crefr:corr-simplify-nc, $\Delta_{\overline{r}}\bullet H_{3}=\Delta_{\overline{s}}d$. We obtain that

[TABLE]

If $x_{\overline{r}}\neq\bot$, as $|H_{\overline{r}\bullet H_{3}}|\leqslant|H_{1}|+|H_{3}|<|H_{\overline{p}}|$, then, by the induction hypothesis, \Crefr:corr-simplify-s&c holds for $\textsc{simplify}(\overline{r}\bullet H_{3})$.

If $x_{\overline{r}}=\varepsilon$, we have that $c=c_{\varepsilon}$ and $\Delta_{\overline{r}}\bullet H_{3}[\varepsilon]=\Delta_{\overline{s}}d[\varepsilon]$. We obtain that

[TABLE]

If $x_{\overline{r}}\in B^{+}$, we have that $c=(\varepsilon,x^{\ell})$ for some $\ell\in\mathbb{N}$ and for all $k\in\mathbb{N}$, $\Delta_{\overline{r}}\bullet H_{3}[x^{k}]=\Delta_{\overline{s}}d[x^{k}]$. We obtain that

[TABLE]

$x_{\overline{p}}\in B^{+}$*  By \Crefr:corr-extend, we have that for all $k\in\mathbb{N}$, $t_{\textsl{init}}\bullet H_{1}\bullet H_{2}[x^{k}]=\Delta_{\overline{q}}\bullet H_{2}[x^{k}]=\Delta_{\overline{r}}c[x^{k}]$. Let $j\in\mathbb{N}$.*

If $x_{\overline{r}}=\bot$, then, by \Crefr:corr-simplify-nc, $\Delta_{\overline{r}}\bullet H_{3}=\Delta_{\overline{s}}d$. We obtain that

[TABLE]

If $x_{\overline{r}}\neq\bot$, as $|H_{\overline{r}\bullet H_{3}}|=|H_{1}|+|H_{3}|<|H_{\overline{p}}|$, then, by the induction hypothesis, \Crefr:corr-simplify-s&c holds for $\textsc{simplify}(\overline{r}\bullet H_{3})$.

Also, as $x_{\overline{q}}=x_{\overline{p}}\in B^{+}$, by construction, it can only happen that $x_{\overline{r}}\in B^{+}$. Thus we have that $c=(\varepsilon,x^{\ell})$ for some $\ell\in\mathbb{N}$ and for all $k\in\mathbb{N}$, $\Delta_{\overline{r}}\bullet H_{3}[x^{k}]=\Delta_{\overline{s}}d[x^{k}]$. We obtain that

[TABLE]

Lemma C.8** (Correctness of transitions).**

\cref@constructprefix

page\cref@result For all $\overline{q}\in\overline{Q}$ such that \overline{i}\xrightarrow[\mathcal{D}]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q}, $\Delta_{\overline{q}}c[\varepsilon]=t_{\textsl{init}}\bullet u[\varepsilon]$.

Proof C.9**.**

We proceed by induction on $|u|$. If $u=\varepsilon$, the result is obtained trivially. If $u=u^{\prime}a$ with $a\in A$, let $\overline{p}=(x,\Delta,H)\in\overline{Q}$ such that \overline{i}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u^{\prime}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{p}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}a}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}\overline{q}, and $(\overline{q},d)=\textsc{simplify}((x,\Delta,H\bullet a))$. By the hypothesis of induction, we obtain that $t_{\textsl{init}}\bullet u^{\prime}[\varepsilon]=\Delta_{\overline{p}}c[\varepsilon]$. By extending with $a$, we get that $t_{\textsl{init}}\bullet u[\varepsilon]=t_{\textsl{init}}\bullet u^{\prime}\bullet a[\varepsilon]=\Delta_{\overline{p}}c\bullet a[\varepsilon]=(\Delta_{\overline{p}}\bullet a)c[\varepsilon]$. We observe three cases. If $\overline{p}\in\overline{Q}_{\textsf{start}}$ then $c=c_{\varepsilon}$ and, by \Crefr:corr-simplify-s&c, $(\Delta_{\overline{p}}\bullet a)c[\varepsilon]=\Delta_{\overline{p}}\bullet a[\varepsilon]=\Delta_{\overline{q}}d[\varepsilon]=\Delta_{\overline{q}}dc[\varepsilon]$. If $\overline{p}\in\overline{Q}_{\textsf{com}}$ then $c[\varepsilon]=x^{k}$, for some $k\in\mathbb{N}$, and, by \Crefr:corr-simplify-s&c, $(\Delta_{\overline{p}}\bullet a)c[\varepsilon]=\Delta_{\overline{p}}\bullet a[x^{k}]=\Delta_{\overline{q}}d[x^{k}]=\Delta_{\overline{q}}dc[\varepsilon]$. Finally, if $\overline{p}\in\overline{Q}_{\neg\textsf{com}}$ then, by \Crefr:corr-simplify-nc, $(\Delta_{\overline{p}}\bullet a)c[\varepsilon]=\Delta_{\overline{q}}dc[\varepsilon]$.

C.4 Boundedness

Lemma C.10**.**

\cref@constructprefix

page\cref@result Let $\overline{p}=(x,\Delta,H_{1})$ and $\overline{q}=(x^{\prime},\Delta^{\prime},H^{\prime})$, $c\in\mathcal{C}(B)$ and $H_{2}\in\mathcal{R}(\mathcal{T}_{\Delta\bullet H_{1}}^{\leqslant|Q|})$, such that $(\overline{q},c)=\textsc{extend\_with\_loop}(\overline{p},H_{2})$. Then $|word(H^{\prime})|\leqslant|word(H_{1})|$.

Proof C.11**.**

We obtain the result by a trivial case analysis of extend_with_loop.

Lemma C.12**.**

Let $\overline{p}=(x,\Delta,H)$ and $\overline{s}=(x^{\prime},\Delta^{\prime},H^{\prime})$, and $c\in\mathcal{C}(B)$ such that $(\overline{s},c)=\textsc{simplify}(\overline{p})$. Then $|word(H^{\prime})|<|Q|^{|Q|}$.

Proof C.13**.**

We proceed by strong induction on the length of $word(H)$.

If $H=id$ then we can only pass through the if block at Line 23 or the else if block at Line 31. In both cases, the result is obtained trivially.

Otherwise, let $n=|word(H)|$, and we observe three cases. Firstly, if $x=\bot$, then we pass through the else if block at Line 31, and the result is obtained trivially. Secondly, if there is a loop in $H$, then we pass through the else if block at Line 26. Let $H=H_{1}H_{2}H_{3}$ where $H_{1}H_{2}$ is the first lasso in $H_{\overline{p}}$. Let $\overline{q}=(x,\Delta,H_{1})$, $(\overline{r},c)=\textsc{extend\_with\_loop}(\overline{q},H_{2})$, and $(\overline{s},d)=\textsc{simplify}(\overline{r}\bullet H_{3})$. By \Crefr:ewl-shorter-history, we have that $|word(H_{\overline{r}})|\leqslant|word(H_{1})|$. Thus $|word(H_{\overline{r}\bullet H_{3}})|=|word(H_{\overline{r}}H_{3})|\leqslant|word(H_{1}H_{3})|<|word(H)|$, and, by the hypothesis of induction applied on $\overline{r}\bullet H_{3}$, we obtain the result. Thirdly, if there is no loop in $H$, then we pass through the else if block at Line 31 and we have $|word(H^{\prime})|=|word(H)|<|Q|^{|Q|}$. Indeed, suppose we had that $|word(H)|\geqslant|Q|^{|Q|}$, then $H$ must contain a loop, which is a contradiction.

Lemma C.14**.**

\cref@constructprefix

page\cref@result Let $\overline{p}=(x,\Delta,H_{1})$ and $\overline{q}=(x^{\prime},\Delta^{\prime},H^{\prime})$, $c\in\mathcal{C}(B)$ and $H_{2}\in\mathcal{R}(\mathcal{T}_{\Delta\bullet H_{1}}^{\leqslant|Q|})$, such that $(\overline{q},c)=\textsc{extend\_with\_loop}(\overline{p},H_{2})$. We assume that $|word(H_{1}H_{2})|\leqslant|Q|^{|Q|}$, $|x|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ and for all $(q,d)\in\Delta$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$. Then we distinguish two cases:

• •

if $x^{\prime}\neq\bot$, then $|x^{\prime}|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ and for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$,

• •

if $x^{\prime}=\bot$, then for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant 2M_{\mathcal{T}}|Q|^{|Q|}$.

Proof C.15**.**

We proceed by case analysis of extend_with_loop. If $H_{2}$ is non-productive then we pass through the if block at Line 2, the returned state is $\overline{p}$ and the result is obtained trivially. Otherwise, $H_{2}$ is productive.

If $x_{\overline{p}}=\varepsilon$ then we pass through the else if block at Line 4. Either $H_{1}H_{2}$ is $x\mathrm{-commuting}$, for some $x\in B^{+}$, and we let $\Delta^{\prime}=\mathsf{split}_{c}(x,H_{1},H_{2})$, or $H_{1}H_{2}$ is $(f,w)\mathrm{-aligned}$, for some $f\in\mathcal{C}(B)$ and $w\in B^{*}$, and we let $\Delta^{\prime}=\mathsf{split}_{nc}(f,w,H_{1},H_{2})$. In both cases, by definition of $\Delta^{\prime}$, we have that, for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant M_{\mathcal{T}}+|\textsf{out}(H_{1})|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$, because $|\textsf{word}(H_{2})|\geqslant 1$ and $|word(H_{1}H_{2})|\leqslant|Q|^{|Q|}$.

If $x_{\overline{p}}\in B^{+}$ then we pass through the else if block at Line 4. If $H_{1}H_{2}$ is $\mathrm{strongly-}x\mathrm{-commuting}$, then the returned state is $\overline{p}$ and the result is obtained trivially. If $H_{1}H_{2}$ is $\mathrm{strongly-}(g,f,x)\mathrm{-aligned}$, for some $f,g\in\mathcal{C}(B)$, then we let $\Delta^{\prime}=\mathsf{extract}_{nc}(g,f,x,\Delta,H_{1},H_{2})$. By definition of $\Delta^{\prime}$, we have that, for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant|\Delta(q)|+|\textsf{out}(H_{1})|\leqslant 2M_{\mathcal{T}}|Q|^{|Q|}$, because $|\Delta(q)|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ and $|word(H_{1})|\leqslant|Q|^{|Q|}$.

Lemma C.16**.**

\cref@constructprefix

page\cref@result Let $\overline{p}=(x,\Delta,H)$ and $\overline{s}=(x^{\prime},\Delta^{\prime},H^{\prime})$, and $c\in\mathcal{C}(B)$ such that $(\overline{s},c)=\textsc{simplify}(\overline{p})$ and $x,x^{\prime}\in B^{*}$. If $|word(H)|\leqslant|Q|^{|Q|}$, $|x|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ and for all $(q,d)\in\Delta$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ then $|x^{\prime}|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ and for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$.

Proof C.17**.**

We proceed by strong induction on the length of $word(H)$. Let $n=|word(H)|$, we observe two cases. First, if there is a loop in $H$, then we pass through the else if block at Line 26. Let $H=H_{1}H_{2}H_{3}$ where $H_{1}H_{2}$ is the first lasso in $H_{\overline{p}}$. Let $\overline{q}=(x,\Delta,H_{1})$, $(\overline{r},c)=\textsc{extend\_with\_loop}(\overline{q},H_{2})$, $\overline{r}=(x^{\prime\prime},\Delta^{\prime\prime},H^{\prime\prime})$, and $(\overline{s},d)=\textsc{simplify}(\overline{r}\bullet H_{3})$. We have that $|word(H_{1}H_{2})|\leqslant|word(H)|\leqslant|Q|^{|Q|}$. Then, by \Crefr:ewl-small-x-and-delta, we have that $|x^{\prime\prime}|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$ and for all $(q,d)\in\Delta^{\prime\prime}$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$. Again, by \Crefr:ewl-shorter-history, we have that $|word(H_{\overline{r}})|\leqslant|word(H_{1})|$. Thus $|word(H_{\overline{r}\bullet H_{3}})|=|word(H_{\overline{r}}H_{3})|\leqslant|word(H_{1}H_{3})|<|word(H)|\leqslant|Q|^{|Q|}$, and, by the hypothesis of induction applied on $\overline{r}\bullet H_{3}$, we obtain the result. Second, if there is no loop in $H$, then we pass through the else if block at Line 31 and the result is obtained trivially.

Lemma C.18** (Boundedness of $\mathcal{D}$).**

\cref@constructprefix

page\cref@result For all $\overline{q}=(x^{\prime},\Delta^{\prime},H^{\prime})\in\overline{Q}$ such that \overline{i}\xrightarrow[D]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q}, the following assertions are satisfied:

• •

$|x^{\prime}|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$,

• •

$|word(H^{\prime})|<|Q|^{|Q|}$,

• •

if $x^{\prime}\neq\bot$, then for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$,

• •

if $x^{\prime}=\bot$, then for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant 4M_{\mathcal{T}}|Q|^{|Q|+2}$.

Proof C.19**.**

We distinguish two cases, whether $x^{\prime}=\bot$ or not.

We start with the case $x^{\prime}\neq\bot$ and proceed by induction on $|u|$. If $u=\varepsilon$, the result is obtained trivially. If $u=u^{\prime}a$ with $a\in A$, let $\overline{p}=(x,\Delta,H)\in\overline{Q}$ such that \overline{i}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u^{\prime}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{p}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}a}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d}}\overline{q}, and $(\overline{q},d)=\textsc{simplify}((x,\Delta,H\bullet a))$. By the hypothesis of induction, we have that $|x|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$, for all $(q,d)\in\Delta$, $|d|\leqslant M_{\mathcal{T}}|Q|^{|Q|}$, and $|word(H)|<|Q|^{|Q|}$. By extending with $a$, we have that $|word(H\bullet a)|\leqslant|Q|^{|Q|}$. Then by \Crefr:simplify-small-x-and-delta, we obtain the result.

We now consider that $x^{\prime}=\bot$. The execution \overline{i}\xrightarrow[D]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q} can be decomposed as \overline{i}\xrightarrow[D]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}\overline{p}\xrightarrow[D]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}a}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}\overline{p}^{\prime}\xrightarrow[D]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{3}}}\overline{q}, with $x_{\overline{p}}\neq\bot$ and $x_{\overline{p}^{\prime}}=\bot$. The transition from $\overline{p}$ to $\overline{p}^{\prime}$ involves the removal of a loop, by extend_with_loop, which is non-commuting. As a consequence of the 2-loop property, we have that some intermediate state $\overline{p}^{\prime\prime}=(\bot,\Delta,id_{\Delta})$ is computed, and that \overleftarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} and \overrightarrow{\makebox{\mathcal{T}{\Delta}\vphantom{x}}} both satisfy the classical twinning property. In addition, thanks to the first case of this proof, and to \Crefr:ewl-small-x-and-delta, we also have that for all $(q,d)\in\Delta^{\prime}$, $|d|\leqslant 2M_{\mathcal{T}}|Q|^{|Q|}$. The behaviour of our procedure starting from this intermediate state $\overline{p}^{\prime\prime}$ is exactly the one of the determinisation procedure of Choffrut (performed on the two sides of the context). See for details the Line 24 of \Crefsimplify. Thanks to results of [10, 7], we know that delays stored in the determinisation construction of Choffrut have size at most $2n^{2}M$, where $M$ is the size of the largest output of the transducer, and $n$ is the number of states. As a consequence, we obtain that for all $(q,d)\in\Delta^{\prime}$, we have :

[TABLE]

C.5 Proof of \Creft:determinisation

Proof C.20** (Proof of \Creft:determinisation).**

The fact that $\mathcal{D}$ is deterministic is direct by an observation of its definition.

To prove the equivalence between $\mathcal{D}$ and $\mathcal{T}$, we consider a word $u\in A^{*}$, and the run \overline{i}\xrightarrow[\mathcal{D}]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c}}\overline{q} of $\mathcal{D}$ on $u$. By \Crefr:corr-trans, we have $\Delta_{\overline{q}}c[\varepsilon]=t_{\textsl{init}}\bullet u[\varepsilon]$. This entails:

[TABLE]

The definition of $\overline{t_{\textsl{final}}}$ then directly implies $[\![T]\!]=[\![\mathcal{D}]\!]$.

Last, the boundedness of $\mathcal{D}$ is a consequence of \Crefr:boundedness.

Appendix D Proofs of \Crefsec:decision: Decision

Lemma D.1**.**

\cref@constructprefix

page\cref@result If a string-to-context transducer $\mathcal{T}$ satisfies the small-2-loop property then the simplify procedure is well-defined for arbitrarily-long histories in $\mathcal{T}^{2}$.

Proof D.2**.**

From an arbitrary history in $\mathcal{T}^{2}$, it is always possible to find lassos for which both the initial part and the loop part have lengths less than $|Q|^{2}$.

We can then prove that if a string-to-context transducer satisfies the small-2-loop property then the function it realizes satisfies the contextual Lipschitz property.

Lemma D.3**.**

\cref@constructprefix

page\cref@result If a string-to-context transducer $\mathcal{T}$ satisfies the small-2-loop property then for all runs \rho_{1}:\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}i_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}q_{1} and \rho_{2}:\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}i_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}q_{2}, with $i_{1},i_{2}$ initial states, we have $\textsl{dist}_{f}(d_{1}c_{1}[\varepsilon],d_{2}c_{2}[\varepsilon])\leqslant 10M_{\mathcal{T}}|Q|^{|Q|+2}$.

Proof D.4**.**

Let $H$ be the history containing only the two runs $\rho_{1}$ and $\rho_{2}$, and let $(\overline{q},c)=\textsc{simplify}(\overline{i}\bullet H)$. By \Crefr:corr-simplify-s&c and \Crefr:small-2-loop-can-simplify, we have that $t_{\textsl{init}}\bullet H[\varepsilon]=\Delta_{\overline{i}\bullet H}[\varepsilon]=\Delta_{\overline{q}}c[\varepsilon]$. Let $e_{1},e_{2}\in\mathcal{C}(B)$ are such that $\Delta_{\overline{q}}=\{(q_{1},e_{1}),(q_{2},e_{2})\}$. We thus have $d_{1}c_{1}[\varepsilon]=e_{1}c[\varepsilon]$ and $d_{2}c_{2}[\varepsilon]=e_{2}c[\varepsilon]$. Therefore $\textsl{dist}_{f}(d_{1}c_{1}[\varepsilon],d_{2}c_{2}[\varepsilon])\leqslant|e_{1}|+|e_{2}|$. Let $\overline{q}=(x_{\overline{q}},\Delta,H_{\overline{q}})$. By definition, we know that $\Delta_{\overline{q}}=\Delta\bullet H_{\overline{q}}$ and that, by \Crefr:boundedness, for all $(q,d)\in\Delta$, $|d|\leqslant 4M_{\mathcal{T}}|Q|^{|Q|+2}$, and $|word(H_{\overline{q}})|<|Q|^{|Q|}$. Thus $|e_{1}|\leqslant 5M_{\mathcal{T}}|Q|^{|Q|+2}$ and $|e_{2}|\leqslant 5M_{\mathcal{T}}|Q|^{|Q|+2}$ and we obtain the result.

Proof D.5** (Proof of \Crefr:small-2-loop-implies-lip).**

Let $\mathcal{T}$ be a string-to-context transducer. Assume that $\mathcal{T}$ satisfies the small-2-loop property and let $u_{1},u_{2}\in\textsf{dom}([\![\mathcal{T}]\!])$. We want to prove that there exists $K\in\mathbb{N}$ such that $\textsl{dist}_{f}([\![\mathcal{T}]\!](u_{1}),[\![\mathcal{T}]\!](u_{2}))\leqslant K\textsl{dist}_{p}(u_{1},u_{2})$. If $u_{1}=u_{2}$ then $\textsl{dist}_{f}([\![\mathcal{T}]\!](u_{1}),[\![\mathcal{T}]\!](u_{2}))=0$, and the result is trivially obtained, whatever the value of $K$ is.

In the following, we assume that $u_{1}\neq u_{2}$ and thus $\textsl{dist}_{p}(u_{1},u_{2})\geqslant 1$. Let \rho_{1}:\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}i_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}}}f_{1}\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}} and \rho_{2}:\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}i_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}}}f_{2}\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}} be the corresponding runs in $\mathcal{T}$. Let $u=\textsf{lcp}(u_{1},u_{2})$ and $u_{1}^{\prime},u_{2}^{\prime}$ such that $u_{1}=uu_{1}^{\prime}$ and $u_{2}=uu_{2}^{\prime}$. Let $p_{1},p_{2}\in Q$ the states that $\rho_{1}$ and $\rho_{2}$ reach after having read $u$. That is \rho_{1}:\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{1}}}i_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}^{\prime}}}p_{1}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}^{\prime}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{1}^{\prime\prime}}}f_{1}\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{1}}} and \rho_{2}:\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}c_{2}}}i_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}^{\prime}}}p_{2}\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}^{\prime}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}d_{2}^{\prime\prime}}}f_{2}\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}e_{2}}}.

By \Crefr:bounded-fork, $\textsl{dist}_{f}(d_{1}^{\prime}c_{1}[\varepsilon],d_{2}^{\prime}c_{2}[\varepsilon])\leqslant 10M_{\mathcal{T}}|Q|^{|Q|+2}$. Therefore,

[TABLE]

Proof D.6** (Proof of \Crefr:decision-det).**

By \Creft:main and \Crefr:small-2-loop-implies-lip, $\mathcal{T}$ admits an equivalent sequential S2C transducer iff $\mathcal{T}$ satisfies the small-2-loop property (see also the figure below). Because of this equivalence, we give a procedure to decide whether $\mathcal{T}$ satisfies the small-2-loop property.

The procedure first non-deterministically guesses a counter-example to the small-2-loop property and then verifies that it is indeed a counter-example. By definition of the small-2-loop property, the counter-example can have one of the following four shapes:

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page\cref@result* a run H:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{0},d_{0})}}(i_{1},i_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{1},d_{1})}}(p_{1},p_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(p_{1},p_{2})}}\overline{p} in $\mathcal{T}^{2}$, with $|u_{1}|<|Q|^{2}$ and $|u_{2}|<|Q|^{2}$, that is a productive lasso neither commuting nor aligned.* 2. 2.

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page\cref@result* a run H:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{0},d_{0})}}(i_{1},i_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{1},d_{1})}}(p_{1},p_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{2},d_{2})}}(p_{1},p_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{3}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{3},d_{3})}}(q_{1},q_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{4}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{4},d_{4})}}(q_{1},q_{2}) in $\mathcal{T}^{2}$, with $|u_{i}|<|Q|^{2}$, for all $i\in\{1,\dots,4\}$, such that the first lasso is a productive $x\mathrm{-commuting}$ lasso, for some $x\in B^{+}$, and the second lasso is a productive lasso neither $\mathrm{strongly-}x\mathrm{-commuting}$ nor strongly aligned.* 3. 3.

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page\cref@result* a run H:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{0},d_{0})}}(i_{1},i_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{1},d_{1})}}(p_{1},p_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{2},d_{2})}}(p_{1},p_{2}) in $\mathcal{T}^{2}$, with $|u_{1}|<|Q|^{2}$ and $|u_{2}|<|Q|^{2}$, that is a productive aligned lasso but, for $\Delta$ appropriately obtained with $\mathsf{split}_{nc}$, \overleftarrow{\makebox{\mathcal{T}_{\Delta}\vphantom{x}}} and/or \overrightarrow{\makebox{\mathcal{T}_{\Delta}\vphantom{x}}} do not satisfy the twinning property.* 4. 4.

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page\cref@result* a run H:\;\xrightarrow[]{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{0},d_{0})}}(i_{1},i_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{1}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{1},d_{1})}}(p_{1},p_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{2}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{2},d_{2})}}(p_{1},p_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{3}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{3},d_{3})}}(q_{1},q_{2})\xrightarrow[]{{\color[rgb]{1,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,1}\pgfsys@color@cmyk@stroke{0}{1}{0}{0}\pgfsys@color@cmyk@fill{0}{1}{0}{0}u_{4}}|{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(c_{4},d_{4})}}(q_{1},q_{2}) in $\mathcal{T}^{2}$, with $|u_{i}|<|Q|^{2}$, for all $i\in\{1,\dots,4\}$, such that the first lasso is a productive aligned lasso, the second lasso is productive and strongly aligned but, for $\Delta$ appropriately obtained with $\mathsf{extract}_{nc}$, \overleftarrow{\makebox{\mathcal{T}_{\Delta}\vphantom{x}}} and/or \overrightarrow{\makebox{\mathcal{T}_{\Delta}\vphantom{x}}} do not satisfy the twinning property.*

*Verifying that a lasso in $\mathcal{T}^{2}$ is not commuting (resp. not aligned) boils down to checking whether there exists no $x\in B^{+}$ such that the lasso is $x\mathrm{-commuting}$ (resp. no $f\in\mathcal{C}(B)$ and $w\in B^{*}$ such that the lasso is $(f,w)\mathrm{-aligned}$). In both cases, the search space for the words $x,w$ and context $f$ can be narrowed down to factors of the output contexts of the given lasso. Thus the verification for shape 1 can be done in polynomial time. Similarly, the verification for shapes 2, 3 and 3 can be done in polynomial time. Furthermore, all three shapes are of polynomial size, by definition of the small-2-loop property, yielding the result. *

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