Rejecting inadmissible rules in reduced normal forms in S4

TL;DR

This paper explores methods to identify and reject inadmissible rules in the modal logic S4 by analyzing reduced normal form rules and their associated rejection substitutions, extending previous approaches to rule sets.

Contribution

It introduces a generalized method for rejecting inadmissible rules in S4, linking reduced normal form techniques with rejection substitutions and extending to rule sets.

Findings

Established relations between reduced normal form rules and rejection substitutions.

Generalized the method to handle sets of rules in S4.

Provided a framework for rejecting inadmissible rules in modal logic S4.

Abstract

Several methods for checking admissibility of rules in the modal logic $S4$ are presented in [1], [15]. These methods determine admissibility of rules in $S4$, but they don't determine or give substitutions rejecting inadmissible rules. In this paper, we investigate some relations between one of the above methods, based on the reduced normal form rules, and sets of substitutions which reject them. We also generalize the method in [1], [15] for one rule to admissibility of a set of rules.

Figures

Equations

$$r_i=\{\begin{array}{ll}\frac{W_i}{J_i}& \text{if }\alpha (i-1)=0\\ \frac{W_i^{\prime }}{W_i}& \text{if }\alpha (i-1)=1\end{array}$$

$$r_i=\{\begin{array}{ll}\frac{W_i}{J_i^{\prime }}& \text{if }\alpha (i-1)=0\\ \frac{J_i^{\prime }}{j_i}& \text{if }\alpha (i-1)=1\end{array}$$

$$r_i=\{\begin{array}{ll}\frac{W_i}{J_i\cup J}& \text{if }\alpha (i-1)=0\\ \frac{J_i\cup J}{J_i}& \text{if }\alpha (i-1)=1\end{array}$$

Full text

Keywords: sequent calculus, admissible rule, modal logic, S4, substitution, complexity.

2010 Mathematics Subject Classification: 03B47, 03D15.

1. Introduction

111This research was in part supported by a grant from IPM (NO)

Logical admissible rules were studied by Lorenzen [10], Harrop [7], Mints [11].The question wether algorithms exist for recognising admissibility of rules, posed by Friedman [10], is affirmatively solved by Rybakov [13], [14] for the modal logic $S4$, for a broad range of propositional modal logics, for example $K4$ and $GL$ in [15], and by Roziere [12] for IPC using methods of proof theory.

Algorithms deciding admissibility for some transitive modal logics and IPC, based on projective formulae and unification, are described in Ghilardi [3], [4],[5],[6]. They combine resolution and tableau approaches for finding projective approximations of a formula and rely on the existence of an algorithm for theorem proving. Lemhoff and Metcalfe [8] introduced a Gentzen-style system for analytic proof systems that derive admissible rules of non-classical logics.

A practically feasible realisation for S4 built on the algorithm for IPC in [5] is described in [16]. These algorithms were specifically designed for finding general solutions for matching and unification problems. In contrast, the original algorithm of [13] can be used to find only some solution of such problems in S4. In [1] are presented some methods, specially a tableau method, for checking rule admissibility in $S4$. For more references for inadmissible rules in $S4$ see [2], [9], [13], [15].

In this paper, in section $2$, we present deduction systems for $S4$ and some results in this system useful for producing substitutions rejecting some inadmissible rules. In section $3$, using Kripke models based on $S4$-formulas in the normal form as their nodes, we provide necessary and sufficient conditions to the determine validity or admissibility of one or several rules (with the same substitution rejecting them) in the normal forms. A way to build the sets used in this conditions and the relations between them and validity or admissibility of rules is presented. In section $4$, the relations between sets of substitutions rejecting sets of rules, and based on them an algorithm to decompose them to their components, are presented. We conclude this section by applying the algorithm on an example. The different ways to decompose the sets make different branches of trees for which examples are presented and show complexity of the problem of producing substitutions rejecting a set of rules.

2. Deduction systems for $S4$

To study substitutions rejecting admissibility of rules in $S4$, specially those reject admissibility of the rules $\cfrac{\diamond p}{p}$ and $\cfrac{\diamond p,p\leftrightarrow\Box p}{p}$, we introduce deduction systems for $S4$ and some useful results in $S4$. A Hilbert system for $S4$ is obtained by adding to the language, axiom schemas and rules for classical logic, the modal $\Box$ and the schemas:

$K:\Box(A\rightarrow B)\rightarrow(\Box A\rightarrow\Box B)$

$T:\Box A\rightarrow A$

$4:\Box A\rightarrow\Box\Box A$

and the necessitation rule, If $\vdash A$ then $\vdash\Box A$.

$\diamond A$ is defined as $\neg\Box\neg A$.

The Gentzen sequent system $G1s$ for $S4$ is [17]:

$\hskip 31.2982ptAx:\hskip 28.45274ptA\vdash A\hskip 162.18062pt\perp L:\hskip 34.14322pt\perp\vdash$

$\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \wedge L:\hskip 28.45274pt\frac{A,B,\Gamma\vdash\Delta}{A\wedge B,\Gamma\vdash\Delta}\hskip 139.41832pt\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \wedge R:\hskip 28.45274pt\frac{\Gamma\vdash A,\Delta\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \Gamma\vdash B,\Delta}{\Gamma\vdash A\wedge B,\Delta}$

$\hskip 31.2982pt\vee L:\hskip 28.45274pt\frac{A,\Gamma\vdash\Delta\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ B,\Gamma\vdash\Delta}{A\vee B,\Gamma\vdash\Delta}$                                    $\vee R:\hskip 28.45274pt\frac{\Gamma\vdash A,B,\Delta}{\Gamma\vdash A\vee B,\Delta}$

$\hskip 25.6073pt\rightarrow L:\hskip 28.45274pt\frac{\Gamma\vdash A,\Delta\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ B,\Gamma\vdash\Delta}{A\rightarrow B,\Gamma\vdash\Delta}\hskip 99.58464pt\rightarrow R:\hskip 25.6073pt\frac{A,\Gamma\vdash B,\Delta}{\Gamma\vdash A\rightarrow B,\Delta}$

$\hskip 34.14322pt\neg L:\hskip 28.45274pt\frac{\Gamma\vdash A,\Delta}{\neg A,\Gamma\vdash\Delta}\hskip 170.71652pt\neg R:\hskip 25.6073pt\frac{A,\Gamma\vdash\Delta}{\Gamma\vdash\neg A,\Delta}$

$\hskip 31.2982ptLW:\hskip 28.45274pt\frac{\Gamma\vdash\Delta}{A,\Gamma\vdash\Delta}$                                                    $RW:\hskip 25.6073pt\frac{\Gamma\vdash\Delta}{\Gamma\vdash\Delta,A}$

$Cut:\hskip 28.45274pt\frac{\Gamma\vdash A,\Delta\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ A,\Gamma\vdash\Delta}{\Gamma\vdash\Delta}$

$\hskip 42.67912pt\Box L:\hskip 28.45274pt\frac{A,\leavevmode\nobreak\ ,\leavevmode\nobreak\ \Box A\leavevmode\nobreak\ \Gamma\vdash\Delta}{\Box A,\leavevmode\nobreak\ \Gamma\vdash\Delta}\hskip 128.0374pt\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \Box R:\hskip 28.45274pt\frac{\Box\Gamma\vdash A,\leavevmode\nobreak\ \diamond\Delta}{\Box\Gamma,\leavevmode\nobreak\ \Gamma^{{}^{\prime}}\vdash\Box A,\leavevmode\nobreak\ \diamond\Delta,\leavevmode\nobreak\ \Delta^{{}^{\prime}}}$

$\hskip 48.36958pt\diamond L:\hskip 28.45274pt\frac{A,\leavevmode\nobreak\ \Box\Gamma\vdash\diamond\Delta}{\diamond A,\leavevmode\nobreak\ \Box\Gamma,\leavevmode\nobreak\ \Gamma^{{}^{\prime}}\vdash\diamond\Delta,\leavevmode\nobreak\ \Delta^{{}^{\prime}}}\hskip 125.19194pt\diamond R:\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \frac{\Gamma\vdash A,\leavevmode\nobreak\ \diamond A,\leavevmode\nobreak\ \Delta}{\Gamma\vdash\diamond A,\Delta}$

In this paper, to produce substitutions rejecting admissibility of rules, we need formula $A$ with $\nvdash A$ and $\vdash\diamond A$, and/or $\vdash A\leftrightarrow\Box A$; See Table $1$ and Table $2$.

Lemma 2.1**.**

The following hold in $S4$:

$(1)\vdash\Box(A\rightarrow B)\rightarrow(\diamond A\rightarrow\diamond B)$**

$(2)$* If $\vdash A\rightarrow B$ then $\vdash\Box A\rightarrow\Box B$*

$(3)$* If $\vdash A\rightarrow B$ then $\vdash\diamond A\rightarrow\diamond B$*

$(4)\vdash\Box\Box A\leftrightarrow\Box A$**

$(5)\vdash\diamond\diamond A\leftrightarrow\diamond A$**

$(6)\vdash\Box\diamond\Box\diamond A\leftrightarrow\Box\diamond A$**

$(7)\vdash\diamond\Box\diamond\Box A\leftrightarrow\diamond\Box A$.

$(8)\vdash\Box A\rightarrow\circ A$* where $\circ$ is a sequence of the modals $\Box$ and $\diamond$.*

Lemma 2.2**.**

The following are provable in $S4$:

$(1)\vdash(\Box A\rightarrow B)\rightarrow\diamond(A\rightarrow B)$**

$(2)\vdash\diamond(A\rightarrow\diamond B)\rightarrow(\Box A\rightarrow\diamond B)$**

$(3)\vdash(\diamond A\rightarrow\Box B)\leftrightarrow\Box(\diamond A\rightarrow\Box B)$**

$(4)\vdash(\diamond A\rightarrow\Box B)\rightarrow\Box(A\rightarrow\Box B)$**

$(5)\vdash(A\leftrightarrow\diamond A)\wedge(B\leftrightarrow\Box B)\rightarrow((A\rightarrow B)\leftrightarrow\Box(A\rightarrow B))$**

$(6)\vdash(A\leftrightarrow\Box A)\wedge(B\leftrightarrow\Box B)\rightarrow((A\vee B)\leftrightarrow\Box(A\vee B))$**

$(7)\vdash(\Box A\vee\Box B)\leftrightarrow\Box(\Box A\vee\Box B)$**

$(8)\vdash(\Box A\vee\Box B)\rightarrow\Box(A\vee\Box B)$**

Proof.

$(1)$ We prove $\vdash\Box\neg(A\rightarrow B)\rightarrow\neg(\Box A\rightarrow B)$

$1.\Box\neg(A\rightarrow B)\hskip 128.0374pt$    assumption

$2.\neg(A\rightarrow B)\rightarrow A$                                     tautology

$3.\Box(\neg(A\rightarrow B))\rightarrow\Box A$                              by $2,2.1(2)$

$4.\Box A\hskip 190.63338pt1,3,$ MP

$5.\neg(A\rightarrow B)\rightarrow\neg B$                                    tautology

$6.\Box\neg(A\rightarrow B)\rightarrow\Box\neg B$                              $5,2.1(2)$

$7.\Box\neg B$                                                       $1,6$, MP

$8.\Box\neg B\rightarrow\neg B$                                            by axiom $T$

$9.\neg B$                                                          by $9,10$, MP

$10.\Box A\rightarrow(\neg B\rightarrow\neg(\Box A\rightarrow B))$                  tautology

$11.\neg(\Box A\rightarrow B)$                                          $4,9,10$, MP

$(2)$ We prove $\vdash\neg(\Box A\rightarrow\diamond B)\rightarrow\Box\neg(A\rightarrow\diamond B)$

$1.\neg(\Box A\rightarrow\diamond B)$                                    assumption

$2.\neg(\square A\rightarrow\diamond B)\rightarrow\square A$                         tautology

$3.\square A$                                                    by $1,2,MP$

$4.\neg(\square A\rightarrow\diamond B)\rightarrow\neg\diamond B$                      tautology

$5.\neg\diamond B$                                                by $1,4,MP$

$6.\Box\neg B$                                                 by $5$

$7.\Box\Box\neg B$                                               by $6$ and by axiom $4$

$8.\Box\neg\diamond B$                                                  by $7$

$9.A\rightarrow(\neg\diamond B\rightarrow\neg(A\rightarrow\diamond B))$                  tautology

$10.\Box A\rightarrow\Box(\neg\diamond B\rightarrow\neg(A\rightarrow\diamond B))$           by $9,2.1(2)$

$11.\Box(\neg\diamond B\rightarrow\neg(A\rightarrow\diamond B))$                      by $10,3,MP$

$12.\Box\neg\diamond B\rightarrow\square\neg(A\rightarrow\diamond B)$                      by $11$ and axiom $K$

$13.\Box\neg(A\rightarrow\diamond B)$                                        by $8,12,MP$

$(3)$ We get $\vdash\neg\Box(\diamond A\rightarrow\Box B)\rightarrow\neg(\diamond A\rightarrow\Box B)$

$1.\neg\Box(\diamond A\rightarrow\Box B)$                                        assumption

$2.\diamond\neg(\diamond A\rightarrow\Box B)$                                       by $1$

$3.\neg(\diamond A\rightarrow\Box B)\rightarrow\diamond A$                                 tautology

$4.\diamond\neg(\diamond A\rightarrow\Box B)\rightarrow\diamond\diamond A$                          by $3,2.1(3)$

$5.\diamond\diamond A$                                                        by $2$ and $4$ and MP

$6.\diamond\diamond A\rightarrow\diamond A$                                              by axiom $4$

$7.\diamond A$                                                          by $5$ and $6$ and MP

$8.\neg(\diamond A\rightarrow\Box B)\rightarrow\neg\Box B$                              tautology

$9.\diamond\neg(\diamond A\rightarrow\Box B)\rightarrow\diamond\neg\Box B$                         by $8,2.1(3)$

$10.\diamond\neg\Box B$                                                   by $2$ and $9$ and MP

$11.\neg\Box\Box B$                                                    by $10$

$12.\neg\Box\Box B\rightarrow\neg\Box B$                                      by axiom $4$

$13.\neg\Box B$                                                       by $11$ and $12$ and MP

$14.\diamond A\rightarrow(\neg\Box B\rightarrow\neg(\diamond A\rightarrow\Box B))$              tautology

$15.\neg(\diamond A\rightarrow\Box B)$                                           by $7,13,14$

$(4)$

$1.(A\rightarrow\diamond A)\rightarrow((\diamond A\rightarrow\Box B)\rightarrow(A\rightarrow\Box B))$                   tautology

$2.\Box(A\rightarrow\diamond A)\rightarrow\Box((\diamond A\rightarrow\Box B)\rightarrow(A\rightarrow\Box B))$              by $1,2.1(2)$

$3.A\rightarrow\diamond A$                                                                        by axiom $4$

$4.\Box(A\rightarrow\diamond A)$                                                                  by $3$ and necessitation rule

$5.\Box((\diamond A\rightarrow\Box B)\rightarrow(A\rightarrow\Box B))$                                       by $2$ and $4$ and MP

$6.\Box(\diamond A\rightarrow\Box B)\rightarrow\Box(A\rightarrow\Box B)$                                       by $5$ and axiom $K$

$7.(\diamond A\rightarrow\Box B)\leftrightarrow\Box(\diamond A\rightarrow\Box B)$                                        by part $(3)$

$8.(\diamond A\rightarrow\Box B)\rightarrow\Box(A\rightarrow\Box B)$                                          by $6$ and $7$ ∎

For formula $A$ in $S4$, let the property $(*)$ be $\nvdash A$ and $\vdash\diamond A$ and the property $(**)$ be the property $(*)$ in addition $\vdash A\leftrightarrow\Box A$.

Let $\sigma(p)=A$. If $A$ has the property $(*)$ then $\sigma$ rejects the rule $r=\cfrac{\diamond p}{p}$ and if $A$ has the property $(**)$ then $\sigma$ rejects the rule $r=\cfrac{\diamond p,p\leftrightarrow\Box p}{p}$.

Lemma 2.3**.**

Let $A$ and $B$ formulas in $S4$ with the property $(**)$ then $A\vee B$ has the property $(**)$.

Proof.

Let $M_{1}\nvDash A,M_{2}\nvDash B$ and $\vdash A\vee B$. Let $M=(M_{1}+M_{2})^{\prime}$ obtained by adding a new root [math] below $M_{1}$ and $M_{2}$. Then $M,0\vDash A\vee B$. Let $M,0\vDash A$, since $\vdash A\leftrightarrow\Box A$ then $M,0\vDash\Box A$ and then $M_{1}\vDash A$, contradiction: Then $\nvdash A\vee B$. By $\vdash\diamond A\vee\diamond B\leftrightarrow\diamond(A\vee B)$ we get $\vdash\diamond(A\vee B)$, and $\vdash A\vee B\leftrightarrow\Box(A\vee B)$ by Lemma $2.2(6)$. ∎

We can get formulas $A$ with $\vdash\diamond A$ as follows. By Lemma $2.1(8)$, Lemma $2.2(1)$ and $2.2(4)$, and replacing $p$ with $\Box p$ or $\diamond p$ and simplifying modals by Lemma $2.1$, we get formulas $A$ with $\vdash\diamond A$ as in Lemma $2.4$. By some replacements we get $\vdash A$ and we remove them from the list. Note that if $\vdash\diamond A$ then $\vdash\diamond(A\vee B)$ and $\vdash\diamond(B\rightarrow A)$ and if $B$ is obtained from $A$ by replacement then $\vdash\diamond B$.

Lemma 2.4**.**

$\vdash\diamond A$* for the following formulas $A$ in $S4$*

$(1)(p\rightarrow\Box p)$*              $(7)\diamond p\rightarrow\diamond\Box\diamond p$               $(13)\Box(\diamond p\rightarrow\Box\diamond p)$

$(2)(p\rightarrow\Box\diamond p)$*           $(8)\diamond\Box p\rightarrow\Box\diamond\Box p$            $(14)\Box(\diamond p\rightarrow\diamond\Box\diamond p)$

$(3)(p\rightarrow\diamond\Box p)$*            $(9)\diamond\Box\diamond p\rightarrow\Box\diamond p$           $(15)\Box(\diamond\Box p\rightarrow\Box\diamond\Box p)$

$(4)(p\rightarrow\Box\diamond\Box p)$*        $(10)\Box(p\rightarrow\Box\diamond p)$             $(16)\Box(\diamond\Box\diamond p\rightarrow\Box\diamond p)$

$(5)(p\rightarrow\diamond\Box\diamond p)$*         $(11)\Box(p\rightarrow\diamond\Box\diamond p)$            $(17)\Box\diamond(p\rightarrow\Box\diamond\Box p)$

$(6)\diamond p\rightarrow\Box\diamond p$*          $(12)\Box(\Box p\rightarrow\Box\diamond\Box p)$          $(18)\Box\diamond(\diamond p\rightarrow\Box\diamond p)$

Proof.

$(1)-(5)$ by Lemma $2.1(8)$ and $2.2(1)$, $(6)$ and $(7)$ from $(1)-(5)$, $(8)$ from $(6)$, $(9)$ from $(8)$ by replacement, $(10)-(12)$ from $(6)-(8)$ by Lemma $2.2(1)$, $(13)-(14)$ from $(10)-(12),\leavevmode\nobreak\ (15)$ from$(13),\leavevmode\nobreak\ (16)$ form $(15)$ and $(18)$ from $(17)$ by replacement, and $(17)$ from $(12)$ by Lemma $2.2(1)$. ∎

Remark 2.1**.**

By Lemma $2.2$, except for $(1)-(5)$ and $(7)$ in Lemma $2.4$, we get $\vdash A\leftrightarrow\Box A$. In the lemma note that $(12)$ and then $(17)$ and $(18)$ are theorems in $S4$. The other parts of the lemma are not theorems. Consider $S4$-models $M_{1}$ and $M_{2}$. $(1)-(7),(10),(11),(13),(14)$ are not valid in $M_{1}$; $(8),(9),(15),(16)$ are not valid in $M_{2}$.

Example 2.1**.**

The proofs of $\vdash\diamond A$ in Lemma $2.4$ in the sequent calculus are interesting. For example we prove some cases.

$\cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ p\vdash p}{p\vdash p,\Box p}RW}{\vdash p,p\rightarrow\Box p}\rightarrow R}{\vdash p,\diamond(p\rightarrow\Box p)}\diamond R,RW}{p\vdash\Box p,\diamond(p\rightarrow\Box p)}\Box R}{\vdash p\rightarrow\Box p,\diamond(p\rightarrow\Box p)}\rightarrow R}{\vdash\diamond(p\rightarrow\Box p)}\diamond R$**

$\cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ p\vdash p}{p\vdash\diamond p}\diamond R,RW}{p\vdash\Box\diamond p,\diamond p}LW}{\vdash\diamond p,p\rightarrow\Box\diamond p}\rightarrow R}{\vdash\diamond p,\Box(p\rightarrow\Box\diamond p),\diamond\Box(p\rightarrow\Box\diamond p)}\Box R,RW}{\vdash\diamond p,\diamond\Box(p\rightarrow\Box\diamond p)}\diamond R}{p\vdash\Box\diamond p,\diamond\Box(p\rightarrow\Box\diamond p)}\Box R}{\vdash p\rightarrow\Box\diamond p,\diamond\Box(p\rightarrow\Box\diamond p)}\rightarrow R}{\vdash\Box(p\rightarrow\Box\diamond p),\diamond\Box(p\rightarrow\Box\diamond p)}\Box R}{\vdash\diamond\Box(p\rightarrow\Box\diamond p)}\diamond R$**

$\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ p\vdash p\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ }{p\vdash p,\Box\diamond\Box p}RW}{\vdash p,p\rightarrow\Box\diamond\Box p}\rightarrow R}{\vdash p,\diamond(p\rightarrow\Box\diamond\Box p)}\diamond R,RW}{\vdash\Box p,\diamond(p\rightarrow\Box\diamond\Box p)}\Box R}{\vdash\diamond\Box p,\diamond(p\rightarrow\Box\diamond\Box p)}\diamond R}{\vdash\diamond\Box p,\Box\diamond(p\rightarrow\Box\diamond\Box p)}\Box R}{\vdash\diamond\Box p,\diamond\Box\diamond(p\rightarrow\Box\diamond\Box p)}\diamond R,RW}{p\vdash\Box\diamond\Box p,\diamond\Box\diamond(p\rightarrow\Box\diamond\Box p)}\Box R,LW}{\vdash(p\rightarrow\Box\diamond\Box p),\diamond\Box\diamond(p\rightarrow\Box\diamond\Box p)}\rightarrow R}{\vdash\diamond(p\rightarrow\Box\diamond\Box p),\diamond\Box\diamond(p\rightarrow\Box\diamond\Box p)}\diamond R,RW}{\vdash\Box\diamond(p\rightarrow\Box\diamond\Box p),\diamond\Box\diamond(p\rightarrow\Box\diamond\Box p)}\Box R}{\vdash\diamond\Box\diamond(p\rightarrow\Box\diamond\Box p)}\diamond R$**

3. Kripke models and reduce normal forms

A Kripke frame is a pair $(W,R)$ where $W$ is a non-empty set of words and $R$ is a relation on $W$, i.e. $R\subseteq W\times W$. A Kripke model is a triple $(W,R,V)$, where $(W,R)$ is a frame and $V$ is a valuation that assigns sets of worlds to propositional variables, i.e. $V:P\rightarrow P(W)$, where $P$ is the set of propositional variables. $V(p)$ is interpreted as the set of worlds where $p$ is true. An $S4$-model is a Kripke model $(W,R,V)$ in which $R$ is reflexive and transitive.

$M,w\models\psi$ is defined as usual for an $S4$-formula $\psi$, an $S4$-model $M$ and a $w$ in $M$. $\psi$ is valid in a model $M$, denoted $M\models\psi$, if $M,w\models\psi$ for all worlds $w$ in $M$.

A rule of inference is as the following: $r=\frac{\alpha_{1},...,\alpha_{n}}{\beta}$ that $\alpha_{1},...,\alpha_{n}$ and $\beta$ are $S4$-formulae. $r$ is valid in an $S4$-model $M$, if from $M\models\alpha_{1},...,M\models\alpha_{n}$ follows $M\models\beta$. The rule $r$ is valid if it is valid in each $S4$-model $M$.

For a rule of inference $r=\frac{\alpha_{1},...,\alpha_{n}}{\beta}$, a logic $\lambda$ and substitution $\sigma$ we use $\lambda\vdash_{\sigma}r$ if from $\lambda\vdash\sigma(\alpha_{1}),...,\lambda\vdash\sigma(\alpha_{n})$ it follows that $\lambda\vdash\sigma(\beta)$. If $\lambda\nvdash_{\sigma}r$ we say $\sigma$ rejects $r$ in $\lambda$. $r$ is admissible in $\lambda$ if $\lambda\vdash_{\sigma}r$ for every substitution $\sigma$.

Definition 3.1**.**

A Kripke model $K_{n}$ is called $n$-characterizing for a modal logic $\lambda$ (any normal modal logic, not necessarily an extension of the system $K4$) if the domain of the valuation $V$ from $K_{n}$ is the set $P$ which consists of $n$ different propositional variables, and if the following holds. For any formula $\alpha$ which is build up of variables from $P$

$\alpha\in\lambda\Leftrightarrow K_{n}\models\alpha$**

We use $Ch_{\lambda}(n)$ for n-characterizing Kripke models $K_{n}$.

Lemma 3.1**.**

*Let $\mathcal{M}=(M,V)$ be a $\lambda$-model.

$(1)$* If $\sigma$ is a substitution and $S(x_{i})=V(\sigma(x_{i}))$, then $S$ is a definable valuation for which $S(\alpha)=V(\sigma(\alpha))$ for each $\lambda$ formula $\alpha$, that is $(M,S),w\vDash\alpha$ iff $(M,V),w\vDash\sigma(\alpha)$ for each $w\in M$.

$(2)$* If $S$ is a definable valuation in $M$, that is for each variable $x_{i}$ there is a formula $\phi_{i}$ such that $(M,S),w\vDash x_{i}$ iff $(M,V),w\vDash\phi_{i}$, and if $\sigma(x_{i})=\phi_{i}$ is a substitution then $S(\alpha)=V(\sigma(\alpha))$ for each $\lambda$-formula $\alpha$, that is $(M,S),w\vDash\alpha$ iff $(M,V),w\vDash\sigma(\alpha)$ for each $w\in M$.

$(3)$* If $\sigma$ is a substitution, $S$ is a definable valuation for which $S(\alpha)=V(\sigma(\alpha))$ for each $\lambda$ formula $\alpha$, $r:=\frac{\alpha_{1},...,\alpha_{m}}{\beta}$ is a rule and $\sigma(r):=\frac{\sigma(\alpha_{1}),...,\sigma(\alpha_{m})}{\sigma(\beta)}$ then $r$ is valid in $(M,S)$ iff $\sigma(r)$ is valid in $(M,V)$.*

Proof.

$(1)$ and $(2)$ by easy induction on $\alpha$ and $(3)$ by $(1)$ and $(2)$.

∎

Theorem 3.1**.**

*Let $(K_{n},V_{n})$, $n\in N$, be a sequence of $n$-characterizing models for a modal logic $\lambda$. Inference rules $r_{1}:=\frac{\alpha_{11},...,\alpha_{1m_{1}}}{\beta_{1}},...,r_{k}:=\frac{\alpha_{k1},...,\alpha_{km_{k}}}{\beta_{k}}$ are inadmissible in $\lambda$ with the same substitution $\sigma$ iff $r_{1},...,r_{k}$ are invalid in $(K_{n},S)$ for some $n\in N$ and some definable valuation $S$ of variables from $r_{1},...,r_{k}$ in $K_{n}$ (that is, If $S(\alpha_{ij})=K_{n}$ and $S(\beta_{i})\neq K_{n}$ for $i=1,...,k$ and $j=1,...,m_{i}$).

Proof.

$(\Leftarrow)$ If $r_{1},...,r_{k}$ are not admissible in $\lambda$ with the same substitution $\sigma$ then for $i=1,...,k$ and $j=1,...,m_{i}$, we get $\displaystyle\vdash_{\lambda}\sigma(\alpha_{ij})$ and $\displaystyle\nvdash_{\lambda}\sigma(\beta_{i})$ then $(K_{n},V_{n})\displaystyle\vDash_{\lambda}\sigma(\alpha_{ij})$ and $(K_{n},V_{n})\displaystyle\nvDash_{\lambda}\sigma(\beta_{i})$ for $n$ the number of variables in $\sigma(r_{1}),...,\sigma(r_{k})$, thus $\sigma(r_{1}),...,\sigma(r_{k})$ is invalid in $(K_{n},V_{n})$, so $r_{1},...,r_{k}$ are invalid in $(K_{n},S)$ with the definable valuation $S(x_{i})=V_{n}(\sigma(x_{i}))$ for each variable $x_{i}$ free in $r$ by Lemma 3.1.

$(\Rightarrow)$ If $r_{1},...,r_{k}$ is invalid in $(K_{n},S)$ with the definable valuation $S$ for each variable $x_{i}$ free in $r_{1},...,r_{k}$ then by Lemma $3.1$, there is a substitution $\sigma$ for each free variable $x_{i}$ in $r_{1},...,r_{k}$ such that $S(x_{i})=V_{n}(\sigma(x_{i}))$ and $\sigma(r_{1}),...,\sigma(r_{k})$ are invalid in $(K_{n},V_{n})$, that is, for $i=1,...,k$ and $j=1,...,m_{i}$ we get $(K_{n},V_{n})\displaystyle\vDash_{\lambda}\sigma(\alpha_{ij})$ and $(K_{n},V_{n})\displaystyle\nvDash_{\lambda}\sigma(\beta_{i})$, then $\displaystyle\vdash_{\lambda}\sigma(\alpha_{ij})$ and $\displaystyle\nvdash_{\lambda}\sigma(\beta_{i})$, therefore $r_{1},...,r_{k}$ are not admissible in $\lambda$.

∎

Definition 3.2**.**

A rule $r$ is said to be in reduced normal form if it has the form

$(rnf)\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ r=\frac{\displaystyle\bigvee_{j\in I}\phi_{j}}{\displaystyle\bigvee_{j\in J}\phi_{j}}$*      or      $r=\frac{\displaystyle\bigvee_{j\in I}\phi_{j}}{\displaystyle p_{i}}$*

and each disjunct $\phi_{j}$ has the form

$\displaystyle\bigwedge_{0\leq i\leq n}p_{i}^{t(i,j,0)}\wedge\displaystyle\bigwedge_{0\leq i\leq n}\diamond p_{i}^{t(i,j,1)}$,

where ($i$) all $\phi_{j}$ are different $(ii)\leavevmode\nobreak\ p_{0},...,p_{n}$ denote propositional variables, $(iii)$ $t$ is a Boolean function $t:\{0,...,n\}\times\{1,...,s\}\times\{0,1\}\rightarrow\{0,1\}$ and ($iv$) $\alpha^{0}=\neg\alpha$ and $\alpha^{1}=\alpha$ for any formula $\alpha$.

Remark 3.1**.**

For the rule $r=\frac{\bigvee_{i\in I}\phi_{i}}{\bigvee_{j\in J}\phi_{j}}$ in reduced normal form we can suppose $J\subseteq I$, because let $r^{\prime}=\frac{\bigvee_{i\in I}\phi_{i}}{\bigvee_{j\in J-\{k\}}\phi_{j}},k\in I-J,$ then $r$ and $r^{\prime}$ are rejected by the same substitutions. Let a substitution $\sigma$ rejects $r^{\prime}$, i.e., $\vdash\sigma(\bigvee_{i\in I}\phi_{i})$ and $\nvdash\sigma(\bigvee_{j\in J-\{k\}}\phi_{j})$. For each $i\in I,\leavevmode\nobreak\ \leavevmode\nobreak\ \phi_{i}\rightarrow\neg\phi_{k}$ and then $\bigvee_{i\in I}\phi_{i}\rightarrow\neg\phi_{k}$ are tautologies, so $\vdash\neg\sigma(\phi_{k})$ and therefore $\nvdash\sigma(\bigvee_{j\in J}\phi_{j})$. Then $\sigma$ rejects $r$. The other side is obvious. For simplicity, we use $r=\frac{I}{J}$

Using the renaming technique any modal rule can be transformed into an equivalent rule in reduced normal form [1], [15].

Theorem 3.2**.**

*Any rule $r=\frac{\alpha}{\beta}$ can be transformed in exponential time to an equivalent rule in reduced normal form.

Proof.

see [1], [15].

∎

Let $\Theta_{n}=\{\phi_{1},...,\phi_{s}\}$ be the set of all disjuncts in $n$ variables $p_{1},...,p_{n}$. For every $\phi_{j}\in\Theta_{n},let$

$\theta(\phi_{j})=\{p_{i}\leavevmode\nobreak\ |\leavevmode\nobreak\ t(i,j,0)=1\}\hskip 56.9055pt\theta_{\diamond}(\phi_{j})=\{p_{i}\leavevmode\nobreak\ |\leavevmode\nobreak\ t(i,j,1)=1\}$.

For every subset of disjuncts $W\subseteq\Theta_{n}$, let $M(\Theta_{n},W)$ denote the Kripke model in which

$W^{M}=W$, $R^{M}=\{(\phi_{1},\phi_{2})\leavevmode\nobreak\ |\leavevmode\nobreak\ \theta_{\diamond}(\phi_{2})\subseteq\theta_{\diamond}(\phi_{1})\}$, $(p_{i},\phi_{j})\in\nu^{m}\Leftrightarrow p_{i}\in\theta(\phi_{j})$.

Example 3.1**.**

In Figure $2$ and Figure $3$ we show $M(\Theta_{2},\Theta_{2})$ and $M(\Theta_{3},\Theta_{3})$, in which $i$ is used for $\phi_{i}$ for which $\Theta(\phi_{i})$ and $\Theta_{\diamond}(\phi_{i})$ are determined as they are showed in the figures. For example in Figure $2$:

$\phi_{1}:=p_{1}\wedge p_{2}\wedge\diamond p_{1}\wedge\diamond p_{2}$*, $\phi_{6}:=\neg p_{1}\wedge p_{2}\wedge\diamond p_{1}\wedge\neg\diamond p_{2}$, $\phi_{16}:=\neg p_{1}\wedge\neg p_{2}\wedge\neg\diamond p_{1}\wedge\neg\diamond p_{2}$.

If $\Theta_{\diamond}(\phi_{i})\subseteq\Theta(\phi_{i})$ we show the node $i$ by $\bullet$ and otherwise we show it by $\circ$ for which $\vdash\neg\phi_{i}$ and then we can remove it from disjunctions in the rules. The right figures only show the nodes $\bullet$.

*The first column in Table $1$ and Table $2$ show all the sets $W$, as in Theorem $3.5$, for which if $W\subseteq I$, the rule $r=\cfrac{I}{p_{1}}$ is inadmissible. The second column shows the simplified form of $\displaystyle\bigvee_{i\in W}\phi_{i}$, and the third shows the conditions on formulas $A,B$ and $C$ for them the substitution $\sigma$ with $\sigma(p_{1})=A,\sigma(p_{2})=B$ and $\sigma(p_{3})=C$ rejects the rule $r$. We suppose in the tables $\vdash\diamond A$ except if $\neg\diamond p_{1}$ occurs in the simplified form. In the last three cases in Table $2$, the substitution $\sigma$ has not been found yet.

Theorem 3.3**.**

Let ${\mathcal{N}}=(N,R)$ and ${\mathcal{N}}\models\displaystyle\bigvee_{i\in I}\phi_{i}$ and $W=\{\phi_{i}\in\Theta_{n}\mid\exists x\in N\leavevmode\nobreak\ \leavevmode\nobreak\ {\mathcal{N}},x\models\phi_{i}\}$. Then

$(1)$* If ${\mathcal{N}},x\models\phi_{i}$ then ${\mathcal{N}},x\models\phi$ iff ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\models\phi$ for each formula $\phi$.

$(2)$* $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I,{\mathcal{M}}(\Theta_{n},W),\phi_{i}\models\phi_{i}\}$.

$(3)$* ${\mathcal{M}}(\Theta_{n},W)\models\displaystyle\bigvee_{i\in I^{\prime}}\phi_{i}$ iff $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I^{\prime}\}$.

$(4)$* ${\mathcal{N}}\vDash\displaystyle\bigvee_{i\in I^{\prime}}\phi_{i}$ iff $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I^{\prime}\}$.

$(5)$* If ${\mathcal{N}}\nvDash p_{i}$ iff ${\mathcal{M}}(\Theta_{n},W)\nvDash p_{i}$ for $i=1,...,n$.

$(6)$* If for each subset $D$ of $N$ there exists $a\in N$ such that*

$\theta_{\diamond}(a)=\theta(a)\cup\displaystyle\bigcup_{d\in D}\theta_{\diamond}(d)$**

then for each subset $D$ of $W$ there exists $\phi_{j}\in W$ such that

$\theta_{\diamond}(\phi_{j})=\theta(\phi_{j})\cup\displaystyle\bigcup_{\phi\in D}\theta_{\diamond}(\phi)$**

*where $\Theta(a)=\{p_{i}\mid N,a\vDash p_{i}\}$ and $\Theta_{\diamond}(a)=\{p_{i}\mid N,a\vDash\diamond p_{i}\}$.

Proof.

First note that if ${\mathcal{N}},x\models\phi_{i}$ and ${\mathcal{N}},x\models\phi_{j}$ then $\phi_{i}=\phi_{j}$. Define $f:N\displaystyle\longrightarrow_{onto}W$ with $f(x)=\phi_{i}$ iff ${\mathcal{N}},x\models\phi_{i}$. $f$ is a homomorphism, because for $x,y\in N$ if $xRy$, $f(x)=\phi_{i},f(y)=\phi_{j}$ and $p_{k}\in\Theta_{\diamond}(f(y))=\Theta_{\diamond}(\phi_{j})$, since ${\mathcal{N}},y\models\phi_{j}$ then ${\mathcal{N}},y\models\diamond p_{k}$, thus ${\mathcal{N}},x\models\diamond p_{k}$ and since ${\mathcal{N}},x\models\phi_{i}$, so $p_{k}\in\Theta_{\diamond}(f(x))=\Theta_{\diamond}(\phi_{i})$. Therefore $\Theta_{\diamond}(f(y))\subseteq\Theta_{\diamond}(f(x))$, then $f(x)R^{\mathcal{M}}f(y)$.

$(1)$ Since $f$ is a surjective homomorphism then ${\mathcal{N}},x\models\phi$ iff ${\mathcal{M}}(\Theta_{n},W),f(x)\models\phi$ iff ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\models\phi$, if ${\mathcal{N}},x\models\phi_{i}$.

$(2)$ By $(1)$ for $\phi=\phi_{i}\in W$ let ${\mathcal{N}},x\models\phi_{i}$ then ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\models\phi_{i}$. Again $W\subseteq\{\phi_{i}\mid i\in I\}$; Because, if $j\notin I$ and $\phi_{j}\in W$, let ${\mathcal{N}},x\vDash\phi_{j}$, since $\phi_{j}\rightarrow\neg\phi_{i}$ for each $i\in I$ and thus $\phi_{j}\rightarrow\neg\displaystyle\bigvee_{i\in I}\phi_{i}$ are tautologies then ${\mathcal{N}},x\vDash\neg\displaystyle\bigvee_{i\in I}\phi_{i}$, contradicting ${\mathcal{N}}\vDash\displaystyle\bigvee_{i\in I}\phi_{i}$.

$(3)(\Rightarrow)$ Let $\phi_{j}\in W$and $j\in I^{\prime}$. Since ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\models\phi_{j}$ and $\models\phi_{j}\rightarrow\neg\phi_{i}$ for $i\neq j$ and then $\models\phi_{j}\rightarrow\neg\displaystyle\bigvee_{i\in I^{\prime}}\phi_{i}$ so ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\models\neg\displaystyle\bigvee_{i\in I^{\prime}}\phi_{i}$.

$(\Leftarrow)$ By $(2)$.

$(4)$ By $(1)$ and $(3)$.

$(5)$ Let ${\mathcal{N}},w\nvDash p_{i}$ and $f(w)=\phi_{j}$ then ${\mathcal{M}}(\Theta_{n},W),f(w)\nvDash p_{i}$ therefore ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\nvDash p_{i}$.

$(6)$ Let $D\subseteq W$. Since $f$ is surjective, for each $\phi_{i}\in D$ pick and fix a representative $d_{i}$ such that $f(d_{i})=\phi_{i}$ and take $D^{\prime}=\{d_{i}\in N\mid\phi_{i}\in D\}$. Let $a\in N,f(a)=\phi_{j}$ and $\theta_{\diamond}(a)=\theta(a)\cup\displaystyle\bigcup_{d\in D^{\prime}}\theta_{\diamond}(d)=\theta(a)\cup\displaystyle\bigcup_{\phi_{i}\in D}\theta_{\diamond}(d_{i})$, then by $\Theta(a)=\Theta(f(a))=\Theta(\phi_{j})$ and $\Theta_{\diamond}(a)=\Theta_{\diamond}(f(a))=\Theta_{\diamond}(\phi_{j})$ and $\Theta_{\diamond}(d_{i})=\Theta_{\diamond}(f(d_{i}))=\theta_{\diamond}(\phi_{i}))$, we get $\Theta_{\diamond}(\phi_{j})=\theta(\phi_{j})\cup\displaystyle\bigcup_{\phi_{i}\in D}\theta_{\diamond}(\phi_{i}))$.

∎

Theorem 3.4**.**

A rule $r=\frac{\displaystyle\bigvee_{i\in I}\phi_{i}\vee\displaystyle\bigvee_{j\in J}\phi_{j}}{\displaystyle\bigvee_{j\in J}\phi_{j}}$ is invalid for $S4$ models iff there is a set $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I\cup J\}$ such that

$(1)$* ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\vDash\phi_{j}$ for all $\phi_{j}\in W$.

$(2)$* ${\mathcal{M}}(\Theta_{n},W)\vDash\displaystyle\bigvee_{i\in I}\phi_{i}\vee\displaystyle\bigvee_{j\in J}\phi_{j}$.

$(3)$* There is $i\in I$ such that $\phi_{i}\in W$ and ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\nvDash\displaystyle\bigvee_{j\in J}\phi_{j}$.

$(4)$* $r$ is invalid in ${\mathcal{M}}(\Theta_{n},W)$.

Proof.

Let $r$ is invalid in ${\mathcal{N}}(N,R)$ then ${\mathcal{N}}\vDash\displaystyle\bigvee_{i\in I}\phi_{i}\vee\displaystyle\bigvee_{j\in J}\phi_{j}$ and ${\mathcal{N}}\nvDash\displaystyle\bigvee_{j\in J}\phi_{j}$. Let $W=\{\phi_{i}\in\Theta_{n}\mid\exists w\in N\leavevmode\nobreak\ \leavevmode\nobreak\ {\mathcal{N}},w\models\phi_{i}\}$. By Theorem 3.3, we get $(1)$ and $(2)$.

$(3)$ Since ${\mathcal{N}}\nvDash\displaystyle\bigvee_{j\in J}\phi_{j}$, then $W\nsubseteq\{\phi_{i}\in\Theta_{n}\mid i\in J\}$ and since $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I\cup J\}$ then there is an $i\in I-J$ such that $\phi_{i}\in W$. By $\vDash\phi_{i}\rightarrow\neg\displaystyle\bigvee_{j\in J}\phi_{j}$ and ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\vDash\phi_{i}$ we get ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\vDash\neg\displaystyle\bigvee_{j\in J}\phi_{j}$, then ${\mathcal{M}}(\Theta_{n},W),\phi_{i}\nvDash\displaystyle\bigvee_{j\in J}\phi_{j}$.

$(4)$ By $(2)$ and $(3)$.

∎

Theorem 3.5**.**

*A rule $r=\cfrac{\displaystyle\bigvee_{i\in I}\phi_{i}}{p_{1}}$ is inadmissible for $S4$ iff there is a set $W\subseteq\{\phi_{i}\mid i\in I\}$ such that

$(1)$ There is $\phi_{j}\in W$ such that ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\nvDash p_{1}$.

$(2)$* ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\vDash\phi_{j}$ for all $\phi_{j}\in W$.

$(3)$* For each subset $D$ of $W$ there exists $\phi_{j}\in W$ such that

$\theta_{\diamond}(\phi_{j})=\theta(\phi_{j})\cup\displaystyle\bigcup_{\phi\in{\mathcal{D}}}\theta_{\diamond}(\phi)$.

*Note that in (3),${\mathcal{D}}$ can be empty.

Proof.

$(\Rightarrow)$ Let $r$ be inadmissible for $S4$, then by Theorem 3.1 there is an $n$-characterizing $S4$ model $Ch_{S4}(n)$ in which $r$ is invalid and by its way of construction the above condition $(3)$ holds for it. Let $W=\{\phi_{i}\in\Theta_{n}|\exists w\in Ch_{S4}(n)\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ Ch_{S4}(n),w\models\phi_{i}\}$. Then the conditions $(1)-(3)$ hold for ${\mathcal{M}}(\Theta_{n},W)$ by Theorem 3.3.

$(\Leftarrow)$ By $(1)$ and $(2)$, the rule $r$ is invalid in ${\mathcal{M}}(\Theta_{n},W)$, and by the proof of Theorem 3.4.10 [1] there is an extension of ${\mathcal{M}}(\Theta_{n},W)$ to an $n$-characterizing $S4$ model $Ch_{S4}(n)$ with effectively contractible definable valuation $S$ of the variables $p_{1},...,p_{n}$ which coincides with the original valuation of ${\mathcal{M}}(\Theta_{n},W)$ and $Ch_{S4}(n)\models\displaystyle\bigvee_{i\in I}\phi_{i}$. Since ${\mathcal{M}}(\Theta_{n},W)\nvDash p_{1}$ and $Ch_{S4}(n)$ is an extension of ${\mathcal{M}}(\Theta_{n},W)$ then $Ch_{S4}(n)\nvDash p_{1}$. Then $Ch_{S4}(n)\nvDash r$, therefore $r$ is inadmissible by Theorem 3.1 ∎

Theorem 3.6**.**

*A rule $r=\frac{\displaystyle\bigvee_{i\in I}\phi_{i}\vee\displaystyle\bigvee_{j\in J}\phi_{j}}{\displaystyle\bigvee_{j\in J}\phi_{j}}$ is inadmissible for $S4$ iff there is a set $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I\cup J\}$ such that

$(1)$ $\phi_{i}\in W$ for some $i\in I$.

$(2)$* ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\vDash\phi_{j}$ for all $\phi_{j}\in W$.

$(3)$* For each subset $\mathcal{(}D)$ of ${\mathcal{M}}$ there exists $\phi_{j}\in W$ such that

$\theta_{\diamond}(\phi_{j})=\theta(\phi_{j})\cup\displaystyle\bigcup_{\phi\in{\mathcal{D}}}\theta_{\diamond}(\phi)$.

Proof.

$(\Rightarrow)$ Similar to the previous theorem.

$(\Leftarrow)$ Similar to the previous theorem except $Ch_{S4}(n)\models\displaystyle\bigvee_{i\in I}\phi_{i}\vee\displaystyle\bigvee_{j\in J}\phi_{j}$ and since ${\mathcal{M}}(\Theta_{n},W)\nvDash\displaystyle\bigvee_{j\in J}\phi_{j}$ and $Ch_{S4}(n)$ is an extension of ${\mathcal{M}}(\Theta_{n},W)$ then $Ch_{S4}(n)\nvDash\displaystyle\bigvee_{j\in J}\phi_{j}$.

∎

Theorem 3.7**.**

*Rules $r_{1}=\frac{\displaystyle\bigvee_{i\in I_{1}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{1}}\phi_{j}}{\displaystyle\bigvee_{j\in J_{1}}\phi_{j}},...,r_{m}=\frac{\displaystyle\bigvee_{i\in I_{m}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}{\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}$ are invalid in the same $S4$ model iff there is a set $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I_{k}\cup J_{k}\}$ for $k=1,...,m$ such that

$(1)$* $\phi_{i}\in W$ for some $i\in I_{k}$ for $k=1,...,m$.

$(2)$* ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\models\phi_{j}$ for all $\phi_{j}\in W$.

then $r_{1},...,r_{m}$ are invalid ${\mathcal{M}}(\Theta_{n},W)$.*

Theorem 3.8**.**

*Rules $r_{1}=\frac{\displaystyle\bigvee_{i\in I_{1}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{1}}\phi_{j}}{\displaystyle\bigvee_{j\in J_{1}}\phi_{j}},...,r_{m}=\frac{\displaystyle\bigvee_{i\in I_{m}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}{\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}$ are inadmissible for $S4$ with the same substitution $\sigma$ iff there is a set $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I_{k}\cup J_{k}\}$ for $k=1,...,m$ such that

$(1)$* $\phi_{i}\in W$ for some $i\in I_{k}$ for $k=1,...,m$.

$(2)$* ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\models\phi_{j}$ for all $\phi_{j}\in W$.

$(3)$* For each subset $\mathcal{(}D)$ of ${\mathcal{M}}$ there exists $\phi_{j}\in W$ such that

$\theta_{\diamond}(\phi_{j})=\theta(\phi_{j})\cup\displaystyle\bigcup_{\phi\in{\mathcal{D}}}\theta_{\diamond}(\phi)$.

Proof.

$(\Rightarrow)$ Let $r_{1},...,r_{m}$ are inadmissible with the same $\sigma$. $r_{1},...,r_{m}$ are invalid in an $n$-characterizing $S4$ model $Ch_{S4}(n)$ by Theorem $3.1$. Let $W=\{\phi_{i}\in\Theta_{n}\mid\exists x\in Ch_{S4}(n)\leavevmode\nobreak\ \leavevmode\nobreak\ Ch_{S4}(n),x\vDash\phi_{i}\}$, then $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I_{k}\cup J_{k}\}$ for $k=1,...,m$. Similar to the previous theorems, the conditions $(1)-(3)$ hold for ${\mathcal{M}}(\Theta_{n},W)$.

$(\Leftarrow)$ By $(1)$ and $(2)$, the rule $r^{\prime}_{k}=\frac{\displaystyle\bigvee_{i\in W}\phi_{i}}{\displaystyle\bigvee_{j\in J_{k}}\phi_{j}}$ is invalid in ${\mathcal{M}}(\Theta_{n},W)$, and by the proof of Theorem 3.4.10 [1] there is an extension of ${\mathcal{M}}(\Theta_{n},W)$ to an $n$-characterizing $S4$ model $Ch_{S4}(n)$ with effectively contractible definable valuation $S$ of the variables $p_{1},...,p_{n}$ which coincides with the original valuation of ${\mathcal{M}}(\Theta_{n},W)$ and $Ch_{S4}(n)\models\displaystyle\bigvee_{i\in W}\phi_{i}$ then $Ch_{S4}(n)\models\displaystyle\bigvee_{i\in I_{k}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{k}}\phi_{j}$. Since ${\mathcal{M}}(\Theta_{n},W)\nvDash\displaystyle\bigvee_{j\in J_{k}}\phi_{j}$ and $Ch_{S4}(n)$ is an extension of ${\mathcal{M}}(\Theta_{n},W)$ then $Ch_{S4}(n)\nvDash\displaystyle\bigvee_{j\in J_{k}}\phi_{j}$. Then $Ch_{S4}(n)\nvDash r_{1},...,Ch_{S4}(n)\nvDash r_{m}$, therefore $r_{1},...,r_{m}$ are inadmissible by the same $\sigma$ by Theorem $3.1$. ∎

Theorem 3.9**.**

*Rules $r_{1}=\frac{\displaystyle\bigvee_{i\in I_{1}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{1}}\phi_{j}}{\displaystyle\bigvee_{j\in J_{1}}\phi_{j}},...,r_{m}=\frac{\displaystyle\bigvee_{i\in I_{m}}\phi_{i}\vee\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}{\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}$ are inadmissible for $S4$ with the same substitution $\sigma$ iff there is a set $W\subseteq\{\phi_{i}\in\Theta_{n}\mid i\in I_{k}\cup J_{k}\}$ for $k=1,...,m$ such that $r^{\prime}_{1}=\frac{\displaystyle\bigvee_{\phi_{i}\in W}\phi_{i}}{\displaystyle\bigvee_{j\in J_{1}}\phi_{j}},...,r^{\prime}_{m}=\frac{\displaystyle\bigvee_{\phi_{i}\in W}\phi_{i}}{\displaystyle\bigvee_{j\in J_{m}}\phi_{j}}$ are inadmissible for $S4$ with the same substitution $\sigma$.

Proof.

$(\Leftarrow)$ is trivial.

$(\Rightarrow)$ Let $\phi_{j}\notin\{\phi_{i}\in\Theta_{n}\mid i\in I_{l}\cup J_{l}\}$ then for $i\in I_{l}\cup J_{l}$ we get $\vdash\phi_{i}\rightarrow\neg\phi_{j}$ then $\vdash\displaystyle\bigvee_{i\in I_{l}\cup J_{l}}\sigma(\phi_{i})\rightarrow\neg\sigma(\phi_{j})$ and since $\vdash\displaystyle\bigvee_{i\in I_{l}\cup J_{l}}\sigma(\phi_{i})$ so $\vdash\neg\sigma(\phi_{j})$. For $k=1,...,m$, since $\vdash\displaystyle\bigvee_{i\in I_{k}\cup J_{k}}\sigma(\phi_{i})$ then $\vdash\displaystyle\bigvee_{i\in I_{k}\cup J_{k}-\{j\}}\sigma(\phi_{i})$ therefore the rule $\frac{\displaystyle\bigvee_{i\in I_{k}\cup J_{k}-\{j\}}\sigma(\phi_{i})}{\displaystyle\bigvee_{j\in J_{k}}\phi_{j}}$ is inadmissible for $S4$ with the substitution $\sigma$. ∎

In the previous theorems, in the sets $W$ there must be appropriate formula $\phi_{i}$ to support conditions such as ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\vDash\phi_{j}$ for all $\phi_{j}\in W$. In the end of this section we study the construction and properties of these sets.

Definition 3.3**.**

$Supp_{1}\subseteq{\mathcal{P}}(\Theta_{n})$* is the smallest set where*

$(1)$* If $C_{1},...,C_{m}$ are clusters of ${\mathcal{M}}_{n}$, that is, $\forall x,y\in C_{i}\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \Theta_{\diamond}(x)=\Theta_{\diamond}(y)=\Theta_{\diamond}(C_{i})$ for each $i=1,...,m$, and $X_{i}\subseteq C_{i}$ where $\displaystyle\bigcup_{x\in X_{i}}\Theta(x)=\Theta_{\diamond}(C_{i})$ for each $i=1,...,m$, then $\displaystyle\bigcup_{i=1}^{m}X_{i}\in Supp_{1}$.

$(2)$* If $Y_{i}\subseteq W\in Supp_{1}$ and $X_{i}\subseteq C_{i}$ where $C_{i}$ is a cluster, $\forall y\in Y_{i}\leavevmode\nobreak\ \leavevmode\nobreak\ \Theta_{\diamond}(y)\subseteq\Theta_{\diamond}(C_{i})$ and $\Theta_{\diamond}(C_{i})=\displaystyle\bigcup_{y\in Y_{i}}\Theta(y)\cup\displaystyle\bigcup_{x\in X_{i}}\Theta(x)$ for $i=1,...,n$, then $W\cup\displaystyle\bigcup_{i=1}^{n}X_{i}\in Supp_{1}$.

$Supp_{2}\subseteq Supp_{1}$* is defined as $W\in Supp_{2}$ if in addition*

$(3)$* For each $\mathcal{D}\subseteq W$ there is $z\in W$ such that $\Theta_{\diamond}(z)=\Theta(z)\cup\displaystyle\bigcup_{x\in\mathcal{D}}\Theta_{\diamond}(x)$.*

Example 3.2**.**

Let $W_{1}=\{55\}$, $W_{2}=\{46\}$, $W_{3}=\{55,46\}$, $W_{4}=\{55\}\cup\{13,14\}$

$W_{5}=\{30,31\}\cup\{21,23\}\cup\{13,14\}$, $W_{6}=\{21,55\}$, $W_{7}=\{21,30,55\}$,

$W_{8}=\{8,21,30,55\}$*.

These sets are in $Supp_{1}$ and $W_{1},W_{2},W_{6},W_{8}\in Supp_{2}$ but $W_{3},W_{4},W_{5},W_{7}\notin Supp_{2}$.

Remark 3.2**.**

*If $n\leq 3$, the condition $(3)$ can be replaced with

$(4)$ $\exists x\in W$ where $\Theta(x)=\Theta_{\diamond}(x)$.

$(5)$ If $x,y\in W$ then $zRx$ and $zRy$ for some $z\in W$ with $\Theta_{\diamond}(z)\subseteq\Theta(z)\cup\Theta_{\diamond}(x)\cup\Theta_{\diamond}(y)$ .

But for $n\geq 4$, let $W=\{x_{1},x_{2},x_{3},y_{1},y_{2},y_{3},z\}$ with $\Theta(x_{i})=\Theta_{\diamond}(x_{i})=\{p_{i}\},\leavevmode\nobreak\ i=1,2,3,\leavevmode\nobreak\ \Theta_{\diamond}(y_{1})=\{p_{1},p_{2},p_{4}\},\leavevmode\nobreak\ \Theta_{\diamond}(y_{2})=\{p_{1},p_{3},p_{4}\},\leavevmode\nobreak\ \Theta_{\diamond}(y_{3})=\{p_{2},p_{3},p_{4}\},\leavevmode\nobreak\ \Theta_{\diamond}(z)=\{p_{1},p_{2},p_{3},p_{4}\}$ and $\Theta(y_{1})=\Theta(y_{2})=\Theta(y_{3})=\{p_{4}\},\Theta(z)=\emptyset.$ Then $W\in Supp_{1}$ and the conditions $(4),(5)$ are true for $W$ but $(3)$ is false for $\mathcal{D}=\{x_{1},x_{2},x_{3}\}$.

Theorem 3.10**.**

$(1)$* If $x\in W\in Supp_{1}$ and $W^{\prime}=\{w\in W\mid xRw\}$ then $W^{\prime}\in Supp_{1}$.

$(2)$* If $x\in W\in Supp_{1}$ and $W^{\prime}=\{w\in W\mid xRw\}$ and $\Theta_{\diamond}(x)=\Theta(x)$ then $W^{\prime}\in Supp_{2}$.

$(3)$* If $x\in W\in Supp_{1}$ and $\Theta_{\diamond}(x)=\Theta_{\diamond}(y)$, then $W\cup\{y\}\in Supp_{1}$.

$(4)$* If $W_{1},W_{2}\in Supp_{1}$ then $W_{1}\cup W_{2}\in Supp_{1}$.

$(5)$* $W\in Supp_{1}$ iff ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\vDash\phi_{j}$ for all $\phi_{j}\in W$.

$(6)$* $W\in Supp_{2}$ iff*

$(i)$* ${\mathcal{M}}(\Theta_{n},W),\phi_{j}\vDash\phi_{j}$ for all $\phi_{j}\in W$ and*

$(ii)$* For each subset $\mathcal{(}D)$ of $W$ there exists $\phi_{j}\in W$ such that*

$\theta_{\diamond}(\phi_{j})=\theta(\phi_{j})\cup\displaystyle\bigcup_{\phi\in{\mathcal{D}}}\theta_{\diamond}(\phi)$.

$(7)$* $r=\frac{I,J}{J}$ is invalid iff there is $W\in Supp_{1}$ such that $W\cap I\neq\emptyset$ and $W\subseteq I\cup J$.

$(8)$* $r=\frac{I,J}{J}$ is inadmissible iff there is $W\in Supp_{2}$ such that $W\cap I\neq\emptyset$ and $W\subseteq I\cup J$.

$(9)$* If $r=\frac{I,J}{J}$ is valid then $r=\frac{I,J}{J}$ is admissible.

$(10)$* If $r=\frac{I,J}{J}$ is invalid then $r^{\prime}=\frac{x,I,J}{x,J}$ is inadmissible for some $x$ with $\theta(x)=\theta_{\diamond}(x)$.

$(11)$* If $r_{1}=\frac{I_{1},J_{1}}{J_{1}},...,r_{k}=\frac{I_{k},J_{k}}{J_{k}}$ are invalid in the same $S4$-model iff there is $W\in Supp_{1}$ such that $W\cap I_{i}\neq\emptyset$ and $W\subseteq I_{i}\cup J_{i}$ for $i=1,...,k$.

$(12)$* If $r_{1}=\frac{I_{1},J_{1}}{J_{1}},...,r_{k}=\frac{I_{k},J_{k}}{J_{k}}$ are inadmissible with the same $\sigma$ iff there is $W\in Supp_{2}$ such that $W\cap I_{i}\neq\emptyset$ and $W\subseteq I_{i}\cup J_{i}$ for $i=1,...,k$.

$(13)$* If $r_{1}=\frac{I_{1},J_{1}}{J_{1}},...,r_{k}=\frac{I_{k},J_{k}}{J_{k}}$ are invalid in the same $S4$-model then $r^{\prime}_{1}=\frac{x,I_{1},J_{1}}{x,J_{1}},...,r^{\prime}=\frac{x,I_{1},J_{1}}{x,J_{1}}$ are inadmissible with the same $\sigma$ for some $x$ with $\theta(x)=\theta_{\diamond}(x)$.

Proof.

$(1)$ $(i)$ Let $C_{1},...,C_{m}$ be clusters of ${\mathcal{M}}_{n}$ and $X_{i}\subseteq C_{i}$ where $\displaystyle\bigcup_{x\in X_{i}}\Theta(x)=\Theta_{\diamond}(C_{i})$ for each $i=1,...,m$, and $x\in W=\displaystyle\bigcup_{i=1}^{m}X_{i}$. Let $x\in X_{i}\subseteq C_{i}$ Then $W^{\prime}=\displaystyle\bigcup_{\forall y\in X_{j}xRy}X_{j}\in Supp_{1}$.

$(ii)$ Let $Y_{i}\subseteq W_{1}\in Supp_{1}$ and $X_{i}\subseteq C_{i}$ where $C_{i}$ is a cluster, $\forall y\in Y_{i}\leavevmode\nobreak\ \leavevmode\nobreak\ \Theta_{\diamond}(y)\subseteq\Theta_{\diamond}(C_{i})$ and $\Theta_{\diamond}(C_{i})=\displaystyle\bigcup_{y\in Y_{i}}\Theta(y)\cup\displaystyle\bigcup_{x\in X_{i}}\Theta(x)$, for $i=1,...,n$, and $x\in W=W_{1}\cup\displaystyle\bigcup_{i=1}^{n}X_{i}$. Then $W^{\prime}=W^{\prime}_{1}\cup\displaystyle\bigcup_{\forall y\in X_{j}xRy}X_{j}$ and by induction hypothesis $W^{\prime}_{1}\in Supp_{1}$, and if $x$ is in some $C_{i}$, then $\forall y\in Y_{i}\leavevmode\nobreak\ \leavevmode\nobreak\ xRy$ and $Y_{i}\subseteq W^{\prime}_{1}$, so $W^{\prime}\in Supp_{1}$ by Definition $3.3(2)$.

$(2)$ By $(1),W^{\prime}\in Supp_{1}$ and if $\mathcal{D}\subseteq W^{\prime}$ and $y\in\mathcal{D}$ then $\Theta_{\diamond}(y)\subseteq\Theta_{\diamond}(x)$ then $\displaystyle\bigcup_{y\in\mathcal{D}}\Theta_{\diamond}(y)\subseteq\Theta_{\diamond}(x)=\Theta(x)$, therefore $\Theta_{\diamond}(x)=\Theta(x)\cup\displaystyle\bigcup_{y\in\mathcal{D}}\Theta_{\diamond}(y)$.

$(3)$ $(i)$ Let $C_{1},...,C_{m}$ be clusters of ${\mathcal{M}}_{n}$ and $X_{i}\subseteq C_{i}$ where $\displaystyle\bigcup_{x\in X_{i}}\Theta(x)=\Theta_{\diamond}(C_{i})$ for each $i=1,...,m$, and $x\in W=\displaystyle\bigcup_{i=1}^{m}X_{i}$. Let $x\in X_{j}\subseteq C_{j}$ and $\Theta_{\diamond}(x)=\Theta_{\diamond}(y)$ Then $X_{j}\cup\{y\}\subseteq C_{j}$ and $\displaystyle\bigcup_{x\in X_{j}\cup\{y\}}\Theta(x)=\Theta_{\diamond}(C_{j})$. Then $W\cup\{y\}=\displaystyle\bigcup_{j\neq i=1}^{m}X_{i}\cup(X_{j}\cup\{y\})\in Supp_{1}$.

$(ii)$ Let $Y_{i}\subseteq W_{1}\in Supp_{1}$ and $X_{i}\subseteq C_{i}$ where $C_{i}$ is a cluster, $\forall y\in Y_{i}\leavevmode\nobreak\ \leavevmode\nobreak\ \Theta_{\diamond}(y)\subseteq\Theta_{\diamond}(C_{i})$ and $\Theta_{\diamond}(C_{i})=\displaystyle\bigcup_{y\in Y_{i}}\Theta(y)\cup\displaystyle\bigcup_{x\in X_{i}}\Theta(x)$, for $i=1,...,n$, and $x\in W=W_{1}\cup\displaystyle\bigcup_{i=1}^{n}X_{i}$. If $x\in W_{1}$, then by induction hypothesis $W_{1}\cup\{y\}\in Supp_{1}$, so $W\cup\{y\}=(W_{1}\cup\{y\})\cup\displaystyle\bigcup_{i=1}^{n}X_{i}$ by Definition $3.3(2)$. If $x\in X_{j}$, then $X_{j}\cup\{y\}\subseteq C_{j}$ and $\displaystyle\bigcup_{w\in X_{j}}\Theta(w)\subseteq\displaystyle\bigcup_{w\in X_{j}\cup\{y\}}\Theta(w)\subseteq\Theta_{\diamond}(C_{j})$ thus $\Theta_{\diamond}(C_{j})=\displaystyle\bigcup_{w\in Y_{j}}\Theta(w)\cup\displaystyle\bigcup_{w\in X_{j}\cup\{y\}}\Theta(w)$. Therefore $W\cup\{y\}=W_{1}\cup(X_{j}\cup\{y\})\cup\displaystyle\bigcup_{j\neq i=1}^{n}X_{i}\in Supp_{1}$ by Definition $3.3(2)$.

$(5)$ $(\Leftarrow)$ By induction on $W$. Let $W=\displaystyle\bigcup_{i=1}^{n}X_{i}$ where $\emptyset\neq X_{i}\subseteq C_{i}$ and $C_{i}$ is a cluster for $i=1,...,n$. $\displaystyle\bigcup_{x\in X_{i}}\Theta(x)\subseteq\Theta_{\diamond}(C_{i})$ is trivial. If $p_{k}\in\Theta_{\diamond}(C_{i})$ and $\varphi_{j}\in X_{i}\subseteq C_{i}$, since $\varphi_{j}\vDash\varphi_{j}$ and $p_{k}\in\Theta_{\diamond}(\varphi_{j})=\Theta_{\diamond}(C_{i})$ then $\varphi_{j}\vDash\diamond p_{k}$. Let $\varphi_{j}R\varphi_{j^{\prime}}$ and $\varphi_{j^{\prime}}\vDash p_{k}$ then $p_{k}\in\Theta(\varphi_{j^{\prime}})$. If $W^{\prime}=\emptyset$ then $\varphi_{j^{\prime}}\in X_{i}$ thus $p_{k}\in\displaystyle\bigcup_{x\in X_{i}}\Theta(x)$, so $\displaystyle\bigcup_{x\in X_{i}}\Theta(x)=\Theta_{\diamond}(C_{i})$ and therefore $W\in Supp_{1}$ by Definition $2(1)$. If $W^{\prime}\neq\emptyset$ and $X_{i}\subseteq W-W^{\prime}$ and $Y_{i}=\{\varphi_{j^{\prime}}\in W^{\prime}\mid\varphi_{j}\in X_{i}\}$, then $\varphi_{j^{\prime}}\in X_{i}\cup Y_{i}$ and $p_{k}\in\displaystyle\bigcup_{x\in X_{i}}\Theta(x)\cup\displaystyle\bigcup_{y\in Y_{i}}\Theta(y)$, so $\displaystyle\bigcup_{x\in X_{i}}\Theta(x)\cup\displaystyle\bigcup_{y\in Y_{i}}\Theta(y)=\Theta_{\diamond}(C_{i})$. Since the hypothesis ${\mathcal{M}}(\Theta_{n},W^{\prime}),\varphi_{j}\vDash\varphi_{j}$ is true for all $\varphi_{j}\in W^{\prime}$, by induction hypothesis $W^{\prime}\in Supp_{1}$ and therefore $W=W^{\prime}\cup\displaystyle\bigcup_{X_{i}\subseteq W-W^{\prime}}X_{i}\in Supp_{1}$ by Definition $3.3(2)$

$(\Rightarrow)$is trivial.

$(6)$ It is trivial by $(5)$ and Definition $3.3(3).$

$(7)$ By the part $(5)$ and Theorem $3.4$.

$(8)$ By the part $(6)$ and Theorem $3.6$.

$(9)$ If $r=\frac{I,J}{J}$ is inadmissible then by the part $(8)$, there is $W\in Supp_{2}$ such that $W\cap I\neq\emptyset$ and $W\subseteq I\cup J$. Since $Supp_{2}\subseteq Supp_{1}$ then $W\in Supp_{1}$ so $r$ is invalid by the part $(7)$.

$(10)$ If $r=\frac{I,J}{J}$ is invalid then by the part $(7)$, there is $W\in Supp_{1}$ such that $W\cap I\neq\emptyset$ and $W\subseteq I\cup J$. Let $z\in W\cap I$ and $W^{\prime}=\{w\in W\mid zRw\}$ and $\theta(x)=\theta_{\diamond}(x)=\theta_{\diamond}(z)$. By the parts $(2)$ and $(3)$, $W^{\prime}\cup\{x\}\in Supp_{2}$, $(W^{\prime}\cup\{x\})\cap I\neq\emptyset$ and $(W^{\prime}\cup\{x\})\subseteq(I\cup\{x\})\cup J$. Then by the part $(8)$ $r^{\prime}=\frac{x,I,J}{x,J}$ is inadmissible.

$(11)-(13)$ similar to $(7)-(10)$ by Theorems $3.8$ and $3.9$. ∎

4. substitutions for rules in reduced normal forms

Suppose $S:{\mathcal{P}}(\Theta_{n})\rightarrow{\mathcal{P}}(\Sigma)$ with $S(A)=\{\sigma\leavevmode\nobreak\ |\leavevmode\nobreak\ \nvdash_{\sigma}r\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ for\leavevmode\nobreak\ each\leavevmode\nobreak\ r\leavevmode\nobreak\ in\leavevmode\nobreak\ A\}$. The following theorem shows properties of $S$ and the way of composing $S(A)$ to its components in the form of $S(\frac{W}{W-\{i_{1}\}},...,\frac{W}{W-\{i_{n}\}})$ where $W=\{i_{1},...,i_{n}\}$ according to the parts $(7)$ and $(8)$ of the theorem. Some applications of the theorem is given by figures $(4)$, $(5)$ and $(6)$ and example $(4.1)$. The compositions are done using the actions a $+J$ and $-J$ in the parts $(12)$ and $(13)$, decreasing premisses and adding demands $(9)$ and $(11)$, and simplifications $(4)$, $(5)$ and $(6)$. The method introduced in the theorem and the following examples justify the approach of the paper in considering substitutions rejecting a set of rules instead of a single rule as usual and generalizing the results in the previous sections.

Theorem 4.1**.**

*The above function $S$ satisfies the following:

$(1)$* If $A\subseteq B$ then $S(B)\subseteq S(A)$.

$(2)$* $S(A)\cap S(B)=S(A\cup B)$.

$(3)$* $S(A)\cup S(B)\subseteq S(A\cap B)$.

$(4)$*  $S(\{\frac{W_{1}}{J_{1}},\frac{W_{2}}{J_{2}}\}\cup A)=S(\{\frac{W_{2}}{J_{2}}\}\cup A)$ if $J_{1}\subseteq J_{2},W_{2}\subseteq W_{1}.$

$(5)$* If $J_{i}\subseteq J^{\prime}_{i},W_{i}\subseteq W^{\prime}_{i}$ and $\frac{W^{\prime}_{i}}{W_{i}}$ and $\frac{J^{\prime}_{i}}{J_{i}}$ are admissible for $i=1,...,m$ then $S(\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}})=S(\frac{W^{\prime}_{1}}{J^{\prime}_{1}},...,\frac{W^{\prime}_{m}}{J^{\prime}_{m}})$ (special cases of $(10)$ and $(11)$).

$(6)$* If $W=\displaystyle\bigcap_{i=1}^{m}W_{i}$ then $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})=S(\{\frac{W}{J_{1}\cap W},...,\frac{W}{J_{m}\cap W}\})$ then $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})=\varnothing$ if $J_{1}\subseteq W$ or … or $J_{m}\subseteq W$

$(7)$* If $W=\{i_{1},...,i_{n}\}$ and $J_{1},...,J_{m}\subseteq W$ then $S(\frac{W}{W-\{i_{1}\}},...,\frac{W}{W-\{i_{n}\}})\subseteq S(\frac{W}{J_{1}},...,\frac{W}{J_{m}})$

$(8)$* If $W_{1}=\{i_{1},...,i_{n}\}$ and $W_{2}=\{j_{1},...,j_{m}\}$ and $W_{1}\neq W_{2}$ then

$S(\frac{W_{1}}{W_{1}-\{i_{1}\}},...,\frac{W_{1}}{W_{1}-\{i_{n}\}})\cap S(\frac{W_{2}}{W_{2}-\{j_{1}\}},...,\frac{W_{2}}{W_{2}-\{j_{m}\}})=\emptyset$

$(9)$* If $J_{1},...,J_{n}\varsubsetneq W_{1}\varsubsetneq W_{2}\subseteq\Theta(r)$ then

$(i)$* $S(\{\frac{W_{2}}{J_{1}},...,\frac{W_{2}}{J_{m}}\})=S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{1}}{J_{m}}\})\bigcup S(\{\frac{W_{2}}{W_{1}}\})$.

$(ii)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{1}}{J_{m}}\})\cap S(\{\frac{W_{2}}{W_{1}}\})=\emptyset$.

$(iii)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{1}}{J_{m}}\})\subseteq S(\{\frac{W_{2}}{J_{1}},...,\frac{W_{2}}{J_{m}}\})$.

$(10)$* If $J_{i}\varsubsetneq W_{i}\subseteq W^{\prime}_{i}\subseteq\Theta(r)$ for $i=1,...,m$ then

$(i)$* $S(\{\frac{W^{\prime}_{1}}{J_{1}},...,\frac{W^{\prime}_{m}}{J_{m}}\})=S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})\bigotimes S(\{\frac{W^{\prime}_{1}}{W_{1}},...,\frac{W^{\prime}_{m}}{W_{m}}\}):=\displaystyle\bigcup_{\alpha\in 2^{m}}S_{\alpha}$ where

$S_{\alpha}=S(\{r_{1},...,r_{m}\})$* and*

[TABLE]

*for $i=1,...,m$.

$(ii)$* $S(\alpha)\cap S(\beta)=\emptyset$ if $\alpha\neq\beta.$

$(iii)$*  $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})\cap S(\{\frac{W^{\prime}_{1}}{W_{1}},...,\frac{W^{\prime}_{m}}{W_{m}}\})=\emptyset$.

$(iv)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})\subseteq S(\{\frac{W^{\prime}_{1}}{J_{1}},...,\frac{W^{\prime}_{m}}{J_{m}}\})$.

$(11)$* If $J_{i}\subseteq J^{\prime}_{i}\subseteq W_{i}\subseteq\Theta_{n}(r)$ for $i=1,...,m$ then

$(i)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})=S(\{\frac{W_{1}}{J^{\prime}_{1}},...,\frac{W_{m}}{J^{\prime}_{m}}\})\bigotimes S(\{\frac{J^{\prime}_{1}}{J_{1}},...,\frac{J^{\prime}_{m}}{J_{m}}\}):=\displaystyle\bigcup_{\alpha\in 2^{m}}S_{\alpha}$ where $S_{\alpha}=S(\{r_{1},...,r_{m}\})$ and*

[TABLE]

*for $i=1,...,m$.

$(ii)$* $S(\alpha)\cap S(\beta)=\emptyset$ if $\alpha\neq\beta.$

$(iii)$* $S(\{\frac{W_{1}}{J^{\prime}_{1}},...,\frac{W_{m}}{J^{\prime}_{m}}\})\cap S(\{\frac{J^{\prime}_{1}}{J_{1}},...,\frac{J^{\prime}_{m}}{J_{m}}\})$.

$(iv)$* $S(\{\frac{W_{1}}{J^{\prime}_{1}},...,\frac{W_{m}}{J^{\prime}_{m}}\})\subseteq S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})$.

$(12)$* If $J_{i}\subsetneq W_{i}\subseteq\Theta(r)$ for $i=1,...,m$, and $\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}-J=\{\frac{W_{1}-J}{J_{1}-J},...,\frac{W_{m}-J}{J_{m}-J}\}$ then

$(i)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})=S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}-J)\cup\displaystyle\bigcup_{i=1}^{m}(S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})\cap S(\{\frac{W_{i}}{W_{i}-J}\}))$

$=S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}-J)\cup\displaystyle\bigcup_{i=1}^{m}S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}},\frac{W_{i}}{W_{i}-J}\})$*.

$(ii)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}-J)\cap S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}},\frac{W_{i}}{W_{i}-J}\})=\emptyset$.

$(iii)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}-J)\subseteq S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})$.

$(13)$* If $J\subseteq W_{i}\subseteq\Theta(r)$ for $i=1,...,m$, and $\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}+J=\{\frac{W_{1}}{J_{1}\cup J},...,\frac{W_{m}}{J_{m}\cup J}\}$ then

$(i)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})=S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}+J)\bigotimes S(\{\frac{J_{1}\cup J}{J_{1}},...,\frac{J_{m}\cup J}{J_{m}}\}):=\displaystyle\bigcup_{\alpha\in 2^{m}}S_{\alpha}$ where $S_{\alpha}=S(\{r_{1},...,r_{m}\})$ and*

[TABLE]

*for $i=1,...,m$.

$(ii)$* $S(\alpha)\cap S(\beta)=\emptyset$ if $\alpha\neq\beta.$

$(iii)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}+J)\cap S(\{\frac{J_{1}\cup J}{J_{1}},...,\frac{J_{m}\cup J}{J_{m}}\})=\emptyset$.

$(iv)$* $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\}+J)\subseteq S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})$.

Theorem 4.2**.**

If $J_{i}\subseteq W_{i}\subseteq\Theta_{n}(r)$ for $i=1,...,m$, then $S(\{\frac{W_{1}}{J_{1}},...,\frac{W_{m}}{J_{m}}\})\neq\emptyset$ iff there is $W\in Supp_{2}$ such that $J_{i}\subsetneq W\subseteq W_{i}$, for $i=1,...,m$.

Proof.

By Theorem $3.10(12)$. ∎

Remark 4.1**.**

In the items $(9)-(13)$ in the Theorem $4.1$, by Theorem $4.2$, we can replace $\subsetneq$ with $\subseteq$ if for some $W\in Supp_{2}$ we suppose $W_{1}\varsubsetneq W\subseteq W_{2}$ in $(9)$, $W_{i}\subsetneq W\subseteq W^{\prime}_{i}$ for $i=1,...,m$ in $(10)$, $J_{i}\subsetneq W\subseteq J^{\prime}_{i}$ for $i=1,...,m$ in $(11)$, $J_{i}\subsetneq W\subseteq W_{i}$ for $i=1,...,m$ in $(12)$. $J_{i}\varsubsetneq W\subseteq J_{i}\cup J$ for $i=1,...,m$ in $(13)$.

Remark 4.2**.**

Theorem $4.1$ suggests a method to decompose $S(A)$ to its components as in $(7)$ and $(8)$. If for some $\frac{W}{J}\in A$, $i\in W-J$ and $W-J$ has another element, we use the part $(13)$ for $J=\{i\}$ and denote this action by $\stackrel{{\scriptstyle+i}}{{=}}$. If $i\in W\cap J$ for some $\frac{W}{J}\in A$ and the previous action can’t be done, we use $(12)$ for $J=\{i\}$ and denote this action by $\stackrel{{\scriptstyle-i}}{{=}}$. After each of these actions we use $(4)$ and $(5)$ and $(6)$ to simplify or unify the premisses or demands and to remove unnecessary rules. We denote this action by $\stackrel{{\scriptstyle s}}{{=}}$. When the premisses are the same the action $\stackrel{{\scriptstyle-i}}{{=}}$ is simplified to

$S(\{\frac{W}{J_{1}},...,\frac{W}{J_{m}}\})=S(\{\frac{W-\{i\}}{J_{1}-\{i\}},...,\frac{W-\{i\}}{J_{m}-\{i\}}\})\cup S(\{\frac{W}{J_{1}},...,\frac{W}{J_{m}},\frac{W}{W-\{i\}}\})$.

If $i\in J^{\prime}$ for each $\frac{W}{J^{\prime}}\in A$ and $i\in W-J$ then the action $\stackrel{{\scriptstyle+i}}{{=}}$ is simplified to

$S(\{\frac{W}{J}\}\cup A)=S(\{\frac{W}{J\cup\{i\}}\}\cup A)\cup S(\{\frac{J\cup\{i\}}{J}\}\cup A)$.

By Theorem $4.2$ in each step the empty or nonempty sets are determined.

Figure $4$ shows the tree of applications of actions $\pm i$ for an example from Figure $2$. Figure $5$ shows the simplified forms of the rules in figure $4$. Since the tree for $3$ variables is very big, Figure $6$ shows a branch of application $\pm i$ for an example from figure $3$.

Example 4.1**.**

$S(\frac{1,2,3,4,10,12}{1})\stackrel{{\scriptstyle+2}}{{=}}S(\frac{1,2,3,4,10,12}{1,2})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle-2}}{{=}}S(\frac{1,3,4,10,12}{1})\cup S(\frac{1,2,3,4,10,12}{1,2},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle+3}}{{=}}S(\frac{1,3,4,10,12}{1,3})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,3,4,10,12}{1,2,3},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle s}}{{=}}S(\frac{1,3,4,10,12}{1,3})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,3,4,10,12}{1,2,3},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle+4}}{{=}}S(\frac{1,3,4,10,12}{1,3,4})\cup S(\frac{1,3,4}{1,3})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle s}}{{=}}S(\frac{1,3,4,10,12}{1,3,4})\cup S(\frac{1,3,4}{1,3})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3})\cup S(\frac{1,2}{1})$**

$\stackrel{{\scriptstyle-3}}{{=}}S(\frac{1,4,10,12}{1,4})\cup S(\frac{1,3,4,10,12}{1,3,4},\frac{1,3,4,10,12}{1,4,10,12})\cup S(\frac{1,4}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,4,10,12}{1,2,4},\frac{1,2,4,10,12}{1,4,10,12})\cup\\ S(\frac{1,2,3,4,10,12}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2}{1,2},\frac{1,2}{1})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3},\frac{1,2,3}{1,2})\cup S(\frac{1,2}{1})$**

$\stackrel{{\scriptstyle s}}{{=}}S(\frac{1,4,10,12}{1,4})\cup S(\frac{1,3,4,10,12}{1,3,4},\frac{1,3,4,10,12}{1,4,10,12})\cup S(\frac{1,4}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,4,10,12}{1,2,4},\frac{1,2,4,10,12}{1,4,10,12})\cup\\ S(\frac{1,2,3,4,10,12}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3})\cup S(\frac{1,2}{1})$**

$\stackrel{{\scriptstyle-4}}{{=}}S(\frac{1,10,12}{1})\cup S(\frac{1,4,10,12}{1,4},\frac{1,4,10,12}{1,10,12})\cup S(\frac{1,3,10,12}{1,3},\frac{1,3,10,12}{1,10,12})\cup S(\frac{1,3,4,10,12}{1,3,4},\frac{1,3,4,10,12}{1,4,10,12},\frac{1,3,4,10,12}{1,3,10,12})\cup S(\frac{1,4}{1})\cup S(\frac{1,3}{1,3},\frac{1,3}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,10,12}{1,2},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,4,10,12}{1,2,4},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,3,10,12}{1,2,3},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,2}{1,2},\frac{1,2}{1})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4},\frac{1,2,4}{1,2})\cup S(\frac{1,2,3}{1,2,3},\frac{1,2,3}{1,3},\frac{1,2,3}{1,2})\cup\\ S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4},\frac{1,2,3,4}{1,2,3})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle s}}{{=}}S(\frac{1,10,12}{1})\cup S(\frac{1,4,10,12}{1,4},\frac{1,4,10,12}{1,10,12})\cup S(\frac{1,2,3,10,12}{1,3},\frac{1,2,3,10,12}{1,10,12})\cup S(\frac{1,2,3,4,10,12}{1,3,4},\frac{1,2,3,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,4}{1})\cup\\ S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,10,12}{1,2},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,4,10,12}{1,2,4},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,3,10,12}{1,2,3},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2,3}{1,2},\frac{1,2,3}{1,3})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle+10}}{{=}}S(\frac{1,10,12}{1,10})\cup S(\frac{1,10}{1})\cup S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12})\cup S(\frac{1,4,10}{1,4},\frac{1,4,10,12}{1,10,12})\cup S(\frac{1,2,3,10,12}{1,3,10},\frac{1,2,3,10,12}{1,10,12})\cup S(\frac{1,3,10}{1,3},\frac{1,2,3,10,12}{1,10,12})\cup S(\frac{1,2,3,4,10,12}{1,3,4,10},\frac{1,2,3,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,3,4,10}{1,3,4},\frac{1,2,3,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,4}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup\\ S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,10}{1,2},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,4,10}{1,2,4},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup\\ S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,10}{1,2,3},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup$* $S(\frac{1,2,3,4,10}{1,2,3,4},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2,3}{1,2})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle s}}{{=}}S(\frac{1,10,12}{1,10})\cup S(\frac{1,10}{1})\cup S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12})\cup S(\frac{1,4,10}{1,4},\frac{1,4,10}{1,10})\cup S(\frac{1,2,3,10,12}{1,3,10},\frac{1,2,3,10,12}{1,10,12})\cup S(\frac{1,3,10}{1,3},\frac{1,3,10}{1,10})\cup\\ S(\frac{1,2,3,4,10,12}{1,3,4,10},\frac{1,2,3,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,3,4,10}{1,3,4},\frac{1,3,4,10}{1,4,10},\frac{1,3,4,10}{1,3,10})\cup S(\frac{1,4}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,10}{1,2},\frac{1,2,10}{1,10})\cup S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,4,10}{1,2,4},\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,2,10})\cup S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})\cup\\ S(\frac{1,2,3,10}{1,2,3},\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10}{1,2,10})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,2,3,4,10}{1,2,3,4},\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10}{1,2,4,10},\frac{1,2,3,4,10}{1,2,3,10})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2,3}{1,2})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle-10}}{{=}}S(\frac{1,12}{1})\cup S(\frac{1,10,12}{1,10},\frac{1,10,12}{1,12})\cup S(\frac{1,10}{1})\cup S(\frac{1,4,12}{1,4},\frac{1,4,12}{1,12})\cup S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12},\frac{1,4,10,12}{1,4,12})\cup S(\frac{1,4,10}{1,4},\frac{1,4,10}{1,10})\cup\\ S(\frac{1,2,3,12}{1,3},\frac{1,2,3,12}{1,12})\cup S(\frac{1,2,3,10,12}{1,3,10},\frac{1,2,3,10,12}{1,10,12},\frac{1,2,3,10,12}{1,2,3,12})\cup S(\frac{1,3,10}{1,3},\frac{1,3,10}{1,10})\cup S(\frac{1,2,3,4,12}{1,3,4},\frac{1,2,3,4,12}{1,4,12},\frac{1,2,3,4,12}{1,2,3,12})\cup\\ S(\frac{1,2,3,4,10,12}{1,3,4,10},\frac{1,2,3,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})\cup S(\frac{1,3,4,10}{1,3,4},\frac{1,3,4,10}{1,4,10},\frac{1,3,4,10}{1,3,10})\cup S(\frac{1,4}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup\\ S(\frac{1,2,12}{1,2},\frac{1,2,12}{1,12})\cup S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12},\frac{1,2,10,12}{1,2,12})\cup S(\frac{1,2,10}{1,2},\frac{1,2,10}{1,10})\cup S(\frac{1,2,4,12}{1,2,4},\frac{1,2,4,12}{1,4,12},\frac{1,2,4,12}{1,2,12})\cup\\ S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12},\frac{1,2,4,10,12}{1,2,4,12})\cup S(\frac{1,2,4,10}{1,2,4},\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,2,10})\cup S(\frac{1,2,3,12}{1,2,3},\frac{1,2,3,12}{1,3,12},\frac{1,2,3,12}{1,2,12})\cup\\ S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12},\frac{1,2,3,10,12}{1,2,3,12})\cup S(\frac{1,2,3,10}{1,2,3},\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10}{1,2,10})\cup S(\frac{1,2,3,4,12}{1,2,3,4},\frac{1,2,3,4,12}{1,3,4,12},\frac{1,2,3,4,12}{1,2,4,12},\frac{1,2,3,4,12}{1,2,3,12})\cup\\ S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})\cup S(\frac{1,2,3,4,10}{1,2,3,4},\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10}{1,2,4,10},\frac{1,2,3,4,10}{1,2,3,10})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2,3}{1,2})\cup S(\frac{1,2}{1})$

$\stackrel{{\scriptstyle s}}{{=}}S(\frac{1,10,12}{1,10})\cup S(\frac{1,10}{1})\cup S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12})\cup S(\frac{1,4,10}{1,4},\frac{1,4,10}{1,10})\cup S(\frac{1,2,3}{1,3})\cup S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10}{1,2,10},\frac{1,2,3,10}{1,2,3})\cup S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,10}{1,10},\frac{1,2,10}{1,2})\cup S(\frac{1,3,10}{1,3},\frac{1,3,10}{1,10})\cup S(\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,3})\cup$* $S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup\\ S(\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10}{1,2,4,10},\frac{1,2,3,4,10}{1,2,3,10},\frac{1,2,3,4,10}{1,2,3,4})\cup S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,2,10},\frac{1,2,4,10}{1,2,4})\cup\\ S(\frac{1,3,4,10}{1,3,4},\frac{1,3,4,10}{1,4,10},\frac{1,3,4,10}{1,3,10})\cup S(\frac{1,4}{1})\cup S(\frac{1,3,4}{1,3},\frac{1,3,4}{1,4})\cup S(\frac{1,3}{1})\cup S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,10}{1,2},\frac{1,2,10}{1,10})\cup\\ S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,4,10}{1,2,4},\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,2,10})\cup S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,10}{1,2,3},\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10}{1,2,10})\cup\\ S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup S(\frac{1,2,3,4,10}{1,2,3,4},\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10}{1,2,4,10},\frac{1,2,3,4,10}{1,2,3,10})\cup S(\frac{1,2,4}{1,2},\frac{1,2,4}{1,4})\cup\\ S(\frac{1,2,3,4}{1,2,3},\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,4})\cup S(\frac{1,2,3}{1,2})\cup S(\frac{1,2}{1})$

*In the last equation note $\frac{1,12}{1},\frac{1,4,12}{1,4},\frac{1,2,3,12}{1,2,3},\frac{1,2,3,4,12}{1,2,3,4}$ are admissible, otherwise by Theorem $3.6$ and Figure $(2)$ there is $W\subseteq\{1,3,4,12\}$ such that $12\in W$ and ${\mathcal{M}}(\Theta_{2},W),\phi_{12}\vDash\phi_{12}$ then ${\mathcal{M}}(\Theta_{2},W),\phi_{12}\vDash\diamond p_{2}$ which is false. Then by Theorem $4.1$, parts $(4)$ and $(5)$

$S(\frac{1,12}{1})=\varnothing$

$S(\frac{1,4,12}{1,4},\frac{1,4,12}{1,12})=S(\frac{1,4,12}{1,4},\frac{1,4,12}{1})=S(\frac{1,4,12}{1,4})=\varnothing$

$S(\frac{1,2,12}{1,2},\frac{1,2,12}{1,12})=S(\frac{1,2,12}{1,2},\frac{1,2,12}{1})=S(\frac{1,2,12}{1,2})=\varnothing$

$S(\frac{1,2,4,12}{1,2,4},\frac{1,2,4,12}{1,4,12},\frac{1,2,4,12}{1,2,12})=S(\frac{1,2,4,12}{1,2,4},\frac{1,2,4,12}{1,4},\frac{1,2,4,12}{1,2})=S(\frac{1,2,4,12}{1,2,4})=\varnothing$

$S(\frac{1,2,3,12}{1,2,3},\frac{1,2,3,12}{1,3,12},\frac{1,2,3,12}{1,2,12})=S(\frac{1,2,3,12}{1,2,3},\frac{1,2,3,12}{1,3},\frac{1,2,3,12}{1,2})=S(\frac{1,2,3,12}{1,2,3})=\varnothing$**

$S(\frac{1,2,3,4,12}{1,2,3,4},\frac{1,2,3,4,12}{1,3,4,12},\frac{1,2,3,4,12}{1,2,4,12},\frac{1,2,3,4,12}{1,2,3,12})=S(\frac{1,2,3,4,12}{1,2,3,4},\frac{1,2,3,4,12}{1,3,4},\frac{1,2,3,4,12}{1,2,4},\frac{1,2,3,4,12}{1,2,3})=S(\frac{1,2,3,4,12}{1,2,3,4})=\emptyset$**

$S(\frac{1,10,12}{1,10},\frac{1,10,12}{1,12})=S(\frac{1,10,12}{1,10},\frac{1,10,12}{1})=S(\frac{1,10,12}{1,10})$,

$S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12},\frac{1,4,10,12}{1,4,12})=S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12},\frac{1,4,10,12}{1,4})=S(\frac{1,4,10,12}{1,4,10},\frac{1,4,10,12}{1,10,12}),$**

$S(\frac{1,2,3,12}{1,3},\frac{1,2,3,12}{1,12})=S(\frac{1,2,3}{1,3},\frac{1,2,3}{1})=S(\frac{1,2,3}{1,3}),$**

$S(\frac{1,2,3,10,12}{1,3,10},\frac{1,2,3,10,12}{1,10,12},\frac{1,2,3,10,12}{1,2,3,12})\stackrel{{\scriptstyle+2}}{{=}}S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,2,10,12},\frac{1,2,3,10,12}{1,2,3,12})\cup S(\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10,12}{1,2,10,12},\frac{1,2,3,10,12}{1,2,3,12})\cup S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,10,12}{1,10,12},\frac{1,2,3,10,12}{1,2,3,12})\cup S(\frac{1,2,3,10}{1,3,10},\frac{1,2,10,12}{1,10,12},\frac{1,2,3,10,12}{1,2,3,12})=S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,2,10,12},\frac{1,2,3,10,12}{1,2,3})\cup S(\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10}{1,2,10},\frac{1,2,3,10}{1,2,3})\cup S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12},\frac{1,2,10,12}{1,2})\cup S(\frac{1,2,10}{1,10},\frac{1,2,10}{1,10},\frac{1,2,10}{1,2})=S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,2,10,12})\cup S(\frac{1,2,3,10}{1,3,10},\frac{1,2,3,10}{1,2,10},\frac{1,2,3,10}{1,2,3})\cup S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12})\cup S(\frac{1,2,10}{1,10},\frac{1,2,10}{1,2})$**

$S(\frac{1,2,3,4,12}{1,3,4},\frac{1,2,3,4,12}{1,4,12},\frac{1,2,3,4,12}{1,2,3,12})=S(\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,4},\frac{1,2,3,4}{1,2,3})=S(\frac{1,2,3,4}{1,3,4},\frac{1,2,3,4}{1,2,3})$,

$S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12},\frac{1,2,10,12}{1,2,12})=S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12},\frac{1,2,10,12}{1,2})=S(\frac{1,2,10,12}{1,2,10},\frac{1,2,10,12}{1,10,12})$,

$S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12},\frac{1,2,4,10,12}{1,2,4,12})=S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12},\frac{1,2,4,10,12}{1,2,4})=\\ S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})$,

$S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12},\frac{1,2,3,10,12}{1,2,3,12})=S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12},\frac{1,2,3,10,12}{1,2,3})=\\ S(\frac{1,2,3,10,12}{1,2,3,10},\frac{1,2,3,10,12}{1,3,10,12},\frac{1,2,3,10,12}{1,2,10,12})$,

$S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})=\\ S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4})=\\ S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,3,4,10,12},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})$,

$S(\frac{1,2,3,4,10,12}{1,3,4,10},\frac{1,2,3,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})\stackrel{{\scriptstyle+2}}{{=}}S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})\cup S(\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})\cup S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})\cup\\ S(\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4,12})=S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12},\frac{1,2,3,4,10,12}{1,2,3,4})\cup\\ S(\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10}{1,2,4,10},\frac{1,2,3,4,10}{1,2,3,10},\frac{1,2,3,4,10}{1,2,3,4})\cup S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12},\frac{1,2,4,10,12}{1,2,4})\cup\\ S(\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,2,10},\frac{1,2,4,10}{1,2,4})=S(\frac{1,2,3,4,10,12}{1,2,3,4,10},\frac{1,2,3,4,10,12}{1,2,4,10,12},\frac{1,2,3,4,10,12}{1,2,3,10,12})\cup\\ S(\frac{1,2,3,4,10}{1,3,4,10},\frac{1,2,3,4,10}{1,2,4,10},\frac{1,2,3,4,10}{1,2,3,10},\frac{1,2,3,4,10}{1,2,3,4})\cup S(\frac{1,2,4,10,12}{1,2,4,10},\frac{1,2,4,10,12}{1,4,10,12},\frac{1,2,4,10,12}{1,2,10,12})\cup S(\frac{1,2,4,10}{1,4,10},\frac{1,2,4,10}{1,2,10},\frac{1,2,4,10}{1,2,4}).$**

5. Conclusion and Discussion

In this paper, we investigated some relations between the method in [1], [15], based on the reduced normal form rules in Theorem $3.5$, and sets of substitutions which reject them in section $4$. We also generalized the method for one rule to inadmissibility of a set of rules. We did some case studies for the cases $2$ and $3$ variables. The case studies show complexity of the problem. The decomposition of sets of substitutions rejecting sets of rules to its components algorithmically is done in section $4$ and calculating at least one member of each components is leaved. Partially the problem is solved for some cases in the paper.