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A SIMPLIFIED FINITE AXIOMATIZATION FOR THE SAIN

TYPE ALGEBRAIZATION OF THE FIRST ORDER LOGIC

Sandor Csizmazia

Abstract. The problem of finding a finite axiomatization to the algebraic counterpart of the first order logicwas solved by Ildiko Sain. She gave a finite scheme of axioms, however she stated as an open problem thetask of giving amore simple one. In this work we give a small elegant system of axioms, solving the abovementioned problem. I would like to thank Ildiko Sain and Istvan Nemeti to inspire this work.

This paper is a continuation of the work was initiated earlier together with Ildiko Sain and announcedin [I.Sain 87] I. Thm 1. (p.3.).

The problem of finding a finite scheme algebraization of the first order logic goes back to [J.D.Monk70] and [L.Henkin,J.D.Monk 74] and was recalled by [I.Sain 87]:

Devise an algebraic version of predicate logic in which the class ofrepresentable algebras forms a finitely based equational class

An equational class is finitely based if it is axiomatizable by a finite amount of equations. An algebra

is called representable if it is isomorphic to a subdirect product of set algebras. According to [A.Tarski66] we add the requirement that the operations should be logical. An operation is called logical iff it isinvariant under permutation of the base of the algebra. By [I.Sain 93] (p.2.) the logical counterpart ofthis requirement is that isomorphic models satisfy the same formulas. Sain found the following solutionto the above stated problem. We shall use the notation of [HMT I] and [HMT II].

[i, j] is the permutation of i and j, [i/j] is the replacement of i by j (i.e. [i/j](i) = jand [i/j](k) = k if k = i). Let f and i, j , then let f(i/j) according to the definitionf(i/j)(i) = f(j) and f(i/j)(n) = f(n) if n = i. Sb(X) is the class of all subsets of X.

Definition 1.:([Sain 87.] I. Def.1. p.3.)

By a Sgws we understand a subalgebra:

A Sb(U), , , V, , V, Ssuc, Spred, Sij , cii,j

where: S(x) = SV (x) = {q V | q x} for every and x V,

Sij = S[i/j], and suc

is the usual successor, and pred is its inverse with pred(0) = 0,

V U let Gwsunit in the sense of [HMT II.] Def.3.1.1. (5.o.) definition and let Ssuc(V) =Spred(V) = Sij(V) = ci(V) = V

Remark 2.: It is clear from the definition that Sgws algebras do not contain the constants of the cylindricalgebras dij ,i ,j < . The logical counterpart of this is that Sgws algebras are related to the first order

logic without equality.

Typeset by AMS-TEX

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Theorem 3.:([Sain 87.] I. Thm.1. (Sain & student), p.3.)

ISgws is a variety axiomatizable by a finite scheme of equations.

A detailed introduction to the history of the problem solved above and an overview of the results can be

found in [I.Nemeti 91]. During the last years many results were published in this area.The negative results are: in [B.Biro 87] the non-finitizability of logic with equality, the strengthening ofBiros result in [I.Nemeti 91] and [H.Andreka 91]. In [I.Sain, R.J.Thompson 91] there is a non-finitizabilityresult for quasi-polyadic algebras. In [Sagi 95] and [Sagi,Nemeti 95] there are a non-finitizability resultsfor polyadic algebras.The positive results are: in [I.Sain 93] there are results on adding di,j , i , j < constants to Sgws. In[I.Sain, V.Gyuris 95] there are finitizability results for first order logic with equality too. Also in [Sagi 95]it was shown that Sgws can be replaced by Cs-style algebras i.e. by such Sgwss the geatest elements ofwhich are (full) Cartesian spaces. Strong positive results in non-well-founded set theories are in [NemetiTit1],[Nemeti Tit2] and in [Simon-Nemeti 95].

Our main theorem originated in [I.Sain 87] II. p.38. Claim (under Remark 10.1.) as an open problem.The [I.Sain 87] preprint introduced the class of Sgws-like algebras. She gave a finite axiom scheme [I.Sain87] II. p.37. (Remark 10.1) axiomatizing the class of Sgws algebras. However she left the task of findingsimpler and more elegant axiomatization(s) as an open problem. We list the original system of axiomsand we will give a simplified version of it as stated in [S.Csizmazia 92], [S.Csizmazia 95]. The presentlyreported research was conducted with guidance from I.Sain and R.J.Thompson during the period 1987-1992 (though some corrections were made later). During 1993-1995 strongly related and sometimesslightly overlapping results were found by I.Sain and V.Gyuris of [I.Sain, V.Gyuris 95].

We will denote the variable symbols of the language of Sgws algebras with x, y.

The function symbols of the type tSgws are the following ones:

0 : , the least element constant symbol

1 : , the greatest element constant symbol- : < d : d >, the symbol of the complementation+ : < d, d : d >, the symbol of the join : < d, d : d >, the symbol of the meetci : < d : d >, the symbols of the cilindrifications (i < )Sij : < d : d >, the symbols of the replacements (i,j < )

Ssuc : < d : d >, the symbol of the successorSpred : < d : d >, the symbol of the predecessor

Let us denote with LSgws the tSgws type first order language.

We will introduce the following defined symbols:

Pij : < d : d >, the symbols of the interchange (where i,j < )

Pijdef= SpredS

0j+1S

j+1i+1S

i+10 Ssuc

First we will detail the AX system of axioms according to [I.Sain 87] II. (in the [I.Sain 87] II. Proof ofTheorem 1. 34-38.o.), after this we will give its simplified version as PAX.

The AX system of axioms contains the following schemes:

With arbitrary i, j, k,l < :

(S0) The usual axioms of the Boolean algebras

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(S1) The Spred, Ssuc, Sij are Boolean endomorphisms

With arbitrary S {Spred, Ssuc, Sij}:

S(x + y) = Sx + Sy

S(x y) = Sx Sy

S0 = 0

S1 = 1

(S2) The connections of Spred, Ssuc with each other and with Sij :

(1.1) SpredSsucx = x

(1.2) SsucSpredx = S01x

(1.3) SsucSijx = Si+1j+1Ssucx if i = j

(1.4) SijSpredx = SpredSi+1j+1x if i = 0 and i = j

(1.5) S0jSpredx = SpredS0j+1S

1j+1x if j = 0

(1.6) Jonssons seven schemes about the connections of the Sij-s and the Pij-s (see [HMT II.] (p.68.)):

(1.6.1) Pijx = Pjix

(1.6.2) PijPjix = x

(1.6.3) PijPikx = P

jkP

ijx if j = k, i = k, i = j

(1.6.4) PijSki x = S

kj P

ijx if j = k, i = k

(1.6.5) Pij

Sj

ix = Si

jx if i = j

(1.6.6) SijSkl x = S

kl S

ijx if k = j, i = k, l

(1.6.7) SijSikx = S

ikx if i = k

(S3) The q1 q9 axiom schemes of Pinter [73]:

(q1) Sij(x) = S

ijx

(q2) Sij(x + y) = S

ijx + S

ijy

(q3) Siix = x

(q4

) Sij

Sk

ix = Si

jSk

jx

(q5) ci(x + y) = cix + ciy

(q6) x cix

(q7) Sijcix = cix

(q8) ciSijx = S

ijx if i = j

(q9) Sijckx = ckS

ijx if k = i, j

We give the simplified version as PAX:

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Let PAX be the following finite amount of axiom schemes with arbitrary i, j, k,l < :

(S0) The usual axioms of the Boolean algebras

(S1) The Spred, Ssuc, Sij are Boolean endomorphisms:

with arbitrary S {Spred, Ssuc, Sij}:

(0.1) S(x + y) = Sx + Sy

(0.2) S( -x ) = - Sx

(S2) The connections of Spred and Ssuc with each other and with Sij:

(1.1) SpredSsucx = x

(1.2) SsucSpredx = S01x

(1.3) SsucSijx = Si+1j+1Ssucx if i = j

(S3) The connections of ci-s and Sij-s:

(p1) Sijcix = cix

(p2) ciSijx = S

ijx if i = j

(p3) Sijckx = ckS

ijx if k = i, j

The properties of the Sij-s:

(p4) SijS

ki x = S

ijS

kj x

(p5) Siix = x

The properties of the ci-s:

(p6) cix x

(p7) ci(x + y) = cix + ciy

PAX fully describes Sgws:

Theorem 4.: ISgws = Mod(PAX)

The proof uses significantly [I.Sain 87] I. Lemma 1. (p.6.): ISgws = Mod(AX). As it will be provedin Lemma 5.: AX and PAX are equivalent, thus Mod(PAX) = Mod(AX). Out of this we concludeISgws = Mod(PAX).

Lemma 5.: AX |= PAX and PAX |= AX

Proof 5.:1. The part AX |= PAX is obvious, because every axiom and axiom scheme of PAX except of (S1)(0.2)are part of AX. The axiom (S1)(0.2) is a consequence of AX(S0), (S1):1 = S(1) = S(x + x) = S(x) + S(x)

0 = S(0) = S(x x) = S(x) S(x)

and because of the existence of the unique complementer in the Boolean algebras: S(x) = S(x).

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2. We have to prepare some lemmas as evidence to the other part: PAX |= AX

As a proof we will use the results of [C.C.Pinter 73]. Because PAX contains every axiom and scheme of[C.C.Pinter 73] we will not repeat the proofs of these results. The results are the following ones, wherei,j,k,l < are arbitrary:

(i) [C.C.Pinter 73] Lemma 2.2 (i) p.363.:

PAX |=ci0 = 0

(ii) [C.C.Pinter 73] Lemma 2.2 (ii) p.363.:

PAX |=ci(x ciy) = cix ciy

(iii) [C.C.Pinter 73] Lemma 2.2 (iii) p.363.:

PAX |=cicjx = cjcix

(iv) [C.C.Pinter 73] Lemma 2.2 (iv) p.363.:

PAX |=SijSikx = S

ikx if i = k

(v) [C.C.Pinter 73] Lemma 2.2 (v) p.363.:

PAX |=SijSkl x = S

kl S

ijx if i = k, l and k = j

(vi) [C.C.Pinter 73] Lemma 5 Proof (4) p.363.:

PAX |=cix = min{y | y x and y = Sijz for some z } if i = j

Lemma 6.: PAX |= SijSki x = S

kj x, if cix = x

Proof 6.:

Let us assume that cix = x

If i = j then

SiiS

ki x =

Ski x

PAX(S3)/(p5)

If i = k then

SijS

kkx =

Sijx

PAX(S3)/(p5)

We can assume that i = j, k

x = cix =

Sijcix =

Sijx(3)

PAX(S3)/(p1) cix = x

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Then if j = k

SijS

ki x =

SijS

kj x =

Skj S

ijx =

Skj x

PAX(S3)/(p4) (v) (3)

because i = j, k and j = k

If j = k

SijS

ki x =

SijS

jjx =

Sjjx =

x

PAX(S3)/(p4) PAX(S3)/(p5) (3)

Lemma 7.:

Let S {Spred, Ssuc, Sij} be arbitrary. Then the following statements are consequences of PAX:

S(x y) = S(x) S(y)

S(0) = 0

S(1) = 1

Proof 7.:

S(x y) =

S((x y)) =

S(x + y) =

S(x) + S(y) =

S(x) + S(y)

PAX(S1)(0.2) PAX(S0) PAX(S1)(0.1) PAX(S1)(0.2)

=

(S(x) S(y))

PAX(S0)

Then according to (S0):S(x y) = S(x) S(y) follows.

S(1) = S(x + x) =

S(x) + S(x) =

S(x) + S(x) =

1(4)

PAX(S1)(0.1) PAX(S1)(0.2) PAX(S0)

S(0) =

S(1) =

S(1) =

1 =

0

PAX(S0) PAX(S1)(0.2) (4) PAX(S0)

Lemma 8.:

PAX |= ciSjix = cjS

ijx

Proof 8.:

Sijx cix PAX(S3)/(p6), (S3)/(p1), (S1)(0.1)

cjSijx cjcix PAX(S3)/(p7)(5)

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a. (i = j) case:

ciSjix =

cicjS

jix =

cjciS

jix

cjSijS

jix =

cjS

ijx

PAX(S3)/p2

(iii) (5)-ben PAX(S3)/(p4

), (p5

)

i = j i = j with x.

= Sjix

The direction is verifiable interchanging the role of the i and j.

b. (i = j) case: obvious.

Lemma 9.:

PAX |=c0Ssucx = Ssucx

Proof 9.:

Ssucx =

SsucSpredSsucx =

S01Ssucx =

c0S

01Ssucx =

c0Ssucx

PAX(S2)(1.1) PAX(S2)/(1.2) PAX(S3)/(p2) PAX(S2)(1.2), (1.1)

Lemma 10.:

PAX |= if x = Sijy for some y and i = j, then Sijx = x

Proof 10.:

Sijx = S

ijS

ijy =

Sijy =

x

(iv)

(because i = j) (because x = Sijy)

The following lemma have been already published in [I. Sain, V. Gyuris 94] (p.17.) from the author.

Lemma 11.:

PAX |=c1Ssucx = Ssucc0x

Proof 11.:

We remark that the minimums exist in this proof is due to the celebrated Jonsson-Tarski theorem:every Mod(PAX) algebra is embeddable into a complete algebra ([HMT I.] Thm. 10.5.(i) (p.412.)),where the positive equations (our system of axioms essentially looks like) are preserved. Because of thecompletness the minimums exist in the extended algebra, as well as in its subalgebra, in Mod(PAX).

because of (vi): c1Ssucx = min {y | y Ssucx and y = S12z for some z }

B1

(1)

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because of (vi): Ssucc0x = Ssucmin {y | y x and y = S01z for some z }

B2

(2)

In the next step we will verify that (1) = (2). Starting with (1) we will get (2) applying sometransformations.

a.) The following identity is valid:

min {y | y Ssucx and y = S12z for some z }

B1

= min {y | y Ssucx and

(3) y = S12z1 for some z1 and y = S

01z2 for some z2 }

A1

The identity above is true because of A1 B1 and for every y B1 there is less y

A1, especially

y def

= c0(y) will be available as we shall see later.Let y B1.a.1.)

Because of PAX(S3)/(p6): c0(y) y c0(y) y y

ya.2.)

y

Ssucx because if y Ssucx, then

y = c0(y)

c0(Ssucx) =

Ssuc(x) =

Ssucx

PAX(S3)/(p7) Lemma 9. PAX(S1)(0.2)

PAX(S1)(0.2)

a.3.)

S12y

= S12(c0(y)) =

c0(S12y) =

c0(y) = y

PAX(S3)/(p3), (S1)(0.2) (3) and Lemma 10.

and

S01y

= S01(c0(y)) =

c0(y) = y

PAX(S1)(0.2), (S3)/(p1)

Thus because of a.2.) and a.3.) it is true that y

A1b.) It is valid that

min {y | y Ssucx and y = S12z1 for some z1 and

(4) y = S01z2 for some z2 }

A1

=

min {Ssucy | y x and

(5) y = S01z for some z } A2

because A1 = A2.

b.1.) In the identity above first we shall proof the direction A1 A2:Let y A1.

y =

S01y =

SsucSpredy

Lemma 10. PAX(S2)(1.2)

y Ssucx Spredy

SpredSsucx =

x

PAX(S1)(0.1) PAX(S2)(1.1)

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y =

S01y =

SsucSpredy and y Ssucx Spredy

SpredSsucx =

x

Lemma 10 PAX(S2)(1.2) PAX(S1)(0.1) PAX(S2)(1.1)

and

S0

1

Spredy =

SpredSsucS0

1

Spredy =

SpredS1

2

SsucSpredx =

SpredS1

2

S0

1

y =

Spredy

PAX(S2)(1.1) PAX(S2)(1.3) PAX(S2)(1.2) y A1

Thus it is true that y A2.b.2.) After this we shall proof the direction A2 A1:

Let y A2. Then y = Ssucy

for some y

x, and for which (5) is valid too.

Because of PAX(S1) if y

x y = Ssucy

Ssucx.

S01y = S

01Ssucy

=

SsucSpredSsucy

=

Ssucy

= y

PAX(S2)(1.2) PAX(S2)(1.1)

and

S12y = S

12Ssucy

=

SsucS01y

=

Ssucy

= y

PAX(S2)(1.3) (5) and Lemma 10

verifies that y A1.c.) The following identity is valid:

min {Ssucy | y x and y = S01z for some z }

A2

= Ssucmin {y | y x and y = S01z for some z }

B2

Because y B2 Ssucy A2 thus according to PAX(S1)(0.1) the identity above is true and this verifiesthe lemma.

Remark 12.:

Similarly with arbitrary i < : PAX |=ci+1Ssucx = Ssuccix.

Lemma 13.:

PAX |=Ssucc0x = c0S10Ssucx

Proof 13.:

Spredc0S10Ssucx =

SpredS01c0S

10Ssucx =

SpredS01c1S

01Ssuc =

(1)

PAX(S3)/(p1) Lemma 8 PAX(S2)(1.2)

SpredSsucSpredc1SsucSpredSsucx =

Spredc1Ssucx =

SpredSsucc0x =

c0x

PAX(S2)(1.1) Lemma 10. PAX(S2)(1.1)

from (1):

Spredc0S10Ssucx = c0x applying Ssuc

SsucSpredc0S10Ssucx = Ssucc0x because of PAX(S2)/(1.2) and PAX(S3)/(p1)

c0S10Ssucx = Ssucc0x

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We will give the deduction of the system of axioms AX from the system of axioms PAX

AX(S0) exactly PAX(S0).

AX(S1) is provable from PAX(S1) because of the Lemma 7.

AX(S2)(1.1)-(1.3) exactly PAX(S2)(1.1)-(1.3).

AX(S2)(1.4): PAX |=SijSpredx = SpredSi+1j+1x if i = 0 and i = j proof:

Let i = j and i = 0, then because of PAX(S2)(1.3):

SsucSijSpredx = S

i+1j+1SsucSpredx PAX(S2)(1.2)

SsucSijSpredx = S

i+1j+1S

01x

SpredSsucSijSpredx = SpredS

i+1j+1S

01x PAX(S2)(1.1)

SijSpredx = SpredSi+1j+1S

01x (v)(because i = 0)

SijSpredx = SpredS

01S

i+1j+1x PAX(S2)(1.2)

SijSpredx = SpredSsucSpredS

i+1j+1x PAX(S2)(1.1)

SijSpredx = SpredS

i+1j+1x

AX(S2)(1.5): PAX |= S0jSpredx = SpredS0j+1S

1j+1x if j = 0 proof:

Let j = 0. Because of PAX(S2)(1.1):

S0jx = SpredSsucS

0jx PAX(S2)(1.3) (because j = 0)

S0jx = SpredS

1j+1Ssucx Lemma 6 (by Lemma 9.: 0 (Ssucx))

S0jx = SpredS

0j+1S

10Ssucx PAX(S2)(1.2)

S0jSpredx = SpredS

0j+1S

10S

01x PAX(S3)/(p4)

S0jSpredx = SpredS

0j+1S

10S

00x PAX(S3)/(p5)

S0jSpredx = SpredS

0j+1S

10x PAX(S3)/(p4)

S0jSpredx = SpredS

0j+1S

1j+1x

We will not proof Jonsson schemes in their original sequence from PAX, because some of the schemeswill be used later to prove the other ones.

AX(S2)(1.6.6): PAX |=SijSkl x = S

kl S

ijx if (k = j, i = k, l) proof:

Similar as (v).

AX(S2)(1.6.7): PAX |=SijSikx = S

ikx if (i = k) proof:

Similar as (iv).

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AX(S2)(1.6.5): PAX |=PijSjix = S

ijx if (i = j) proof:

Let i = j. Then:

PijSjix =

SpredS0j+1Sj+1i+1Si+10 SsucSjix =

SpredS0j+1Sj+1i+1 Si+10 Sj+1i+1Ssucx =

Pij definition PAX(S2)(1.3.) PAX

(because i = j) (S3)/(p4)

SpredS0j+1S

j+1i+1 S

i+10 S

j+10 Ssucx =

SpredS

0j+1S

j+1i+1S

j+10 S

i+10 Ssucx =

(v) PAX(S3)/(p4)

(because i = j)

SpredS0j+1S

j+1i+1S

j+10 S

i+1j+1Ssucx =

SpredS

0j+1S

j+1i+1S

j+10 SsucS

ijx =

PAX(S2)(1.3) (iv)

(because i = j)SpredS

0j+1S

j+10 SsucS

ijx =

SpredS

j+1j+1SsucS

ijx =

Lemma 6. PAX(S3)/p5

(because 0 (SsucSijx):Lemma 9.)

SpredSsucSijx =

Sijx

PAX(S2)(1.1)

AX(S2)(1.6.4): PAX |=PijSki x = S

kj P

ijx if (j = k, i = k) proof:

a. Let k = 0 and k = i, j

PijS

ki x =

SpredS

0j+1S

j+1i+1S

i+10 SsucS

ki x =

SpredS

0j+1S

j+1i+1S

i+10 S

k+1i+1 Ssucx =

Pijdefinition PAX(S2)(1.3.) PAX

(because i = k) (S3)/(p4)

SpredS0j+1S

j+1i+1S

i+10 S

k+10 Ssucx =

SpredS

0j+1S

j+1i+1S

k+10 S

i+10 Ssucx =

(v) (v)

(because i = k) (because j, i = k)

SpredS0j+1S

k+10 S

j+1i+1 S

i+10 Ssucx =

SpredS

0j+1S

k+1j+1S

j+1i+1S

i+10 Ssucx =

PAX(S3)/(p4) (v)

(because k = j)

SpredSk+1j+1S

0j+1S

j+1i+1 S

i+10 Ssucx =

Skj SpredS

0j+1S

j+1i+1 S

i+10 Ssucx =

Skj P

ij

AX(S2)(1.4.) Pij definition

(because k = j, k = 0)

b. Let k = 0 and k = i, j

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Because k = j, i, then j, i = 0

PijS

0i x =

SpredS

0j+1S

j+1i+1S

i+10 SsucS

0i x =

SpredS

0j+1S

j+1i+1S

i+10 S

1i+1Ssucx =

Pijdefinition PAX(S2)(1.3.) PAX

(because i = 0) (S3)/(p4)

SpredS0j+1Sj+1i+1S

i+10 S

10Ssucx =

SpredS0j+1S

j+1i+1S

10S

i+10 Ssucx =

(v) (v)

(because i = 0) (because j, i = 0)

SpredS0j+1S

10S

j+1i+1 S

i+10 Ssucx =

SpredS

0j+1S

1j+1S

0j+1S

j+1i+1S

i+10 Ssucx =

(iv), PAX(S3)/(p4), (v) AX(S2)(1.5)

(because j = 0) (because 0 = j)

S0jSpredS

0j+1S

j+1i+1 S

i+10 Ssucx =

S0jP

ij

Pij definition

AX(S2)(1.6.2): PAX |=PijPji x = x proof:

a. Let i = j

PijP

ji x =

Pijdefinition

SpredS

0j+1S

j+1i+1 S

i+10 SsucSpredS

0i+1S

i+1j+1S

j+10 Ssucx =

PAX(S2)(1.2)

SpredS0j+1S

j+1i+1 S

i+10 S

01S

0i+1S

i+1j+1S

j+10 Ssucx =

(iv)

SpredS0j+1S

j+1i+1S

i+10 S

0i+1S

i+1j+1S

j+10 Ssucx =

PAX(S3)/(p4)

SpredS0j+1S

j+1i+1 S

i+10 S

00S

i+1j+1S

j+10 Ssucx =

PAX(S3)/(p5), (iv)

(because i = j)

SpredS0j+1S

j+1i+1 S

i+1j+1S

j+10 Ssucx =

SpredS

0j+1S

j+1i+1 S

j+10 Ssucx =

PAX(S3)/(p4), (p5) (iv)

SpredS0j+1S

j+10 Ssucx =

SpredS

j+1j+1Ssucx =

SpredSsucx =

x

Lemma 6. PAX(S3)/(p5) PAX

(because 0 (Ssucx):Lemma 9. ) (S2)(1.1)

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b. Let i = j

PiiP

iix =

Pi

idefinition

SpredS0i+1S

i+1i+1S

i+10 SsucSpredS

0i+1S

i+1i+1S

i+10 Ssucx =

PAX(S2)(1.2)

SpredS0i+1S

i+1i+1S

i+10 S

01S

0i+1S

i+1i+1S

i+10 Ssucx =

(iv)

SpredS0i+1S

i+1i+1S

i+10 S

0i+1S

i+1i+1S

i+10 Ssucx =

Lemma 6.

because i + 1 (Si+1i+1Si+10 Ssucx)

SpredS0i+1S

i+1i+1S

00S

i+1i+1S

i+10 Ssucx =

PAX(S3)/(p5)

SpredS0i+1S

i+10 Ssucx =

SpredS

i+1i+1Ssucx =

SpredSsucx =

x

Lemma 6. PAX(S3)/(p5) PAX

(because 0 (Ssucx):Lemma 9. ) (S2)(1.1)

Lemma 14.:[I.Sain,R.J.Thompson 88] (p.553.)

PAX |=Ski SijS

jkS

kj S

ji S

ikx = S

ki x if i = j, k

Proof 14.:

Let i = j, k. Then:

Sk

iSi

jSj

kSk

jSj

iSi

kx =

Sk

iSi

jSj

kSj

iSi

kx =

Sk

iSi

jSj

iSi

kx =

PAX(S3)/(p4), (p5) (iv) PAX(S3)/(p4), (p5)

(because i = j)

Ski S

ijS

ikx =

Ski S

ikx =

Ski x

(iv) PAX(S3)/(p4), (p5)

(because i = k)

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Lemma 15.:

PAX |=cicix = cix if i = j, k

Proof 15.:

Let j = i. Then:

cicix =

ciSijcix =

Sijcix =

cix

PAX(S3)/(p1) PAX(S3)/(p2) PAX(S3)/(p1)

(because i = j)

Lemma 16.:

PAX |=Ski cix = Pikcix if k = i

Proof 16.:

a. Let k = 0 and k = i

Ski cix =

Ski SpredSsuccix =

SpredS

k+1i+1 c0Ssuccix =

PAX(S2)(1.1) Lemma 9. PAX

AX(S2)(1.4)(because k = 0, k = i)(S3)/(p3)

Spredc0Sk+1i+1 Ssuccix =

SpredS

0k+1c0S

k+1i+1 Ssuccix =

SpredS

0k+1S

k+1i+1 Ssuccix =

PAX(S3)/(p1) Lemma 9. Remark

PAX(S3)/(p3) 11.

SpredS0k+1S

k+1i+1 ci+1Ssucx =

SpredS

0k+1S

k+1i+1 S

i+10 ci+1Ssucx =

PAX(S3)/(p1) Remark 12.

SpredS0k+1S

k+1i+1 S

i+10 Ssuccix =

Pikcix

Pikdefinition

b. Let k = 0 and k = i

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Because k = i, then i = 0

S0i cix =

S0i SpredSsuccix =

SpredS

0i+1S

1i+1c0Ssuccix =

PAX(S2)(1.1) Lemma 9. (v)

AX(S2)(1.5)(because 0 = i) (because i = 0)SpredS

1i+1S

0i+1c0Ssuccix =

SpredS

1i+1c0Ssuccix =

SpredS

1i+1ci+1Ssuccix =

PAX(S3)/(p1) Lemma 9., 15. PAX

Remark 12. (S3)/(p1)

SpredS1i+1S

i+10 ci+1Ssuccix =

SpredS

1i+1S

i+10 Ssuccix =

Remark 12,15 PAX(S2)(1.1)

SpredSsucSpredS1i+1S

i+10 Ssuccix =

SpredS

01S

1i+1S

i+10 Ssuccix =

Pi0cix

PAX(S2)(1.2) Pi0definition

Lemma 17.:[I.Sain,R.J.Thompson 88] (p.553.) :

PAX |=Skj Pijx = S

kj S

ji S

ikx if i,j,k are distinct.

Proof 17.:

Let i,j,k be distinct. Then:

Skj P

ij(x S

ikx)

Skj P

ijcj(x S

ikx) =

PAX(S3)/(p6) PAX(S3)/(p1)

Skj P

ijS

ji cj(x S

ikx) =

Skj S

ijcj(x S

ikx) =

Skj S

ikcj(x S

ikx) =

AX(1.6.5.) PAX(S3)/(p4) PAX(S3)/(p3)

(because i = j) (because j = i, k)

Skj cjS

ik(x S

ikx) =

Skj cj(S

ikx S

ik(S

ikx)) =

Skj cj(S

ikx S

ikS

ikx) =

PAX(S1) PAX(S1) PAX(S3)/(p1), (p2)

(because i = k)

Skj cj(S

ikx S

ikx) =

Skj cj(0) =

Skj (0) =

0

PAX(S0) (i) PAX(S1)

Thus Skj Pijx S

kj P

ijS

ikx = 0 because of PAX(S1).

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Then:

Skj P

ijx

Skj P

ijS

ikx =

Skj P

ijciS

ikx =

Skj S

ji ciS

ikx =

Skj S

ji S

ikx(1)

PAX(S0) PAX(S3)/(p2) Lemma 16. PAX(S3)/(p2)

(because i = k) (because i = j) (because i = k)

Ski x =

Ski SijS

jkS

kj S

ji S

ikx

Ski SijS

jkS

kj P

ijx =

Ski S

ijS

jkP

ijx(2)

Lemma 14. (1) (because i = k, j) and PAX(S3)/(p4), (p5)

(because i = k, j) PAX(S1)(0.1)

Thus:

Ski P

jix

Ski S

ijS

jkP

ijP

ji x =

Ski S

ijS

jkx

(2) bol AX(1.6.2.)

Interchanging the role of i and j we will get: (3) Skj Pijx S

kj S

jiS

ik

The statement follows from (1) and (3).

Lemma 18.:

PAX |=SsucPijx = Pi+1j+1Ssucx if i = j

Proof 18.:

Let i = j

SsucPijx =

SsucSpredS

0j+1S

j+1i+1S

i+10 Ssucx =

S01S

0j+1S

j+1i+1S

i+10 Ssucx =

Pijdefinition PAX(S2)/(1.2) (iv)

S0j+1S

j+1i+1 S

i+10 Ssucx =

S0j+1P

i+1j+1Ssucx =

Pi+1j+1S

0i+1Ssucx =

Lemma 17. AX(1.6.4) Lemma 9.

(because i = j)

Pi+1j+1S

0i+1c0Ssucx =

Pi+1j+1c0Ssucx =

Pi+1j+1Ssucx

PAX(S3)/(p1) Lemma 9.

Lemma 19.:

PAX |=Ski ciSi0c0x = PikP0i c0x if i = 0, k = i

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Proof 19.:

Let i = 0, k = i. Then:

S

ki ciS

i0c0x = P

ikciS

i0c0x = P

ikS

i0c0x = P

ikP

0i c0x

Lemma 16. PAX(S3)/(p2) Lemma 16.

(because k = i) (because 0 = i) (because i = 0)

AX(S2)(1.6.3): PAX |=PijPikx = P

jkP

ijx if j = k, i = k, i = j proof:

We will prove it in two steps. In the step a. we will prove the statement for x where c0x = x. In thestep b. the case c0x = x will be attributed to the application of step a.

Let i,j,k be distinct.

a. Let c0x = x

Starting from PAX(S3)/(p4) we can deduce:

SjkS

ijx = S

jkS

ikx

PjkS

kj P

ijS

jix= S

jkP

ikS

ki x AX(S3)(1.6.5) (because i = j, j = k, i = k)

PjkP

ijS

ki S

jix= P

ikS

ji S

ki x AX(S3)(1.6.4) (because i = j, j = k, i = k)

PjkP

ijS

ji S

ki x= P

ikS

ji S

ki x (v) (because i = j, j = k, i = k)(3)

Starting from (iv) we conclude the following, because i = k:

SijS

ikx = S

ikx

PijS

ji P

ikS

ki x = P

ikS

ki x AX(S3)(1.6.5) (because i = j, i = k)

PijS

ji P

ikS

ki S

jix= P

ikS

ki S

jix replacing x by S

ji (x)

PijS

ji P

ikS

ji S

ki x= P

ikS

jiS

ki x (v) (because i = j, j = k, i = k)(4)

According to (3) and (4) starting from the following identity:

PjkP

ijS

ji S

ki x = P

ijS

ji P

ikS

ji S

ki x

PjkP

ijS

ji S

ki x = P

ijS

ji S

jkP

ikS

ki x AX(S3)(1.6.4) (because i = j, i = k)

PjkP

ijS

ji S

ki x = P

ijS

jkP

ikS

ki x (iv) (because j = k)

PjkP

ijS

ji S

ki x = P

ijP

ikS

ji S

ki x AX(S3)(1.6.4) (because i = j, j = k)(5)

PjkP

ijS

ji S

ki ciS

i0c0x= P

ijP

ikS

ji S

ki ciS

i0c0x replacing x by ciS

i0c0x

PjkPijSji PikP0i c0x = PijPikSji PikP0i c0x because of Lemma 19.

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(6) PjkPijS

jiP

ikP

0i x = P

ijP

ikS

ji P

ikP

0i x because c0x = x

We can transform the identity (6) with help of AX(S2)(1.6.2) for the following form, when the x isreplaced by Pi0x and P

ki x:

(7)Pj

kPi

jSj

ix =Pi

jPi

kSj

ix

The method eliminating the Ski -s from (5) is called Method of the elimination of replacements

Let us repeat the Method of the elemination of replacements in the identity (7) for the Sji . We shallget the following identity.

(8) PjkPijx = P

ijP

ikx, thus we have proved (1.6.3), if c0x = x.

b. Let c0x = x

PjkP

ijx =

SpredSsucP

jkP

ijx =

SpredP

j+1k+1P

i+1j+1Ssucx =

PAX(S2)(1.1) Lemma 18. step a. of the deduction

(because j = k, i = j) (c0Ssuxx =Ssucx:Lemma 9.)

SpredPi+1j+1P

i+1k+1Ssucx =

SpredSsucP

ijP

ikx =

PijP

ikx


(because i = k, i = j

Thus we have completed the proof of the identity (1.6.3).

AX(S2)(1.6.1): PAX |=Pijx = Pjix proof:

If i = j then Pii x = Pii x is obvious. Thus we can assume that i = j.

Let i = j.

a. Let c0x = x

Let k = i, j. Then:

SjkS

ijx =

SjkS

ikx =

SikS

jkx =

SikS

jix(1)

PAX(S3)/(p4) (v) PAX(S3)/(p4)

(because j = i, k and k = i)

Let us start with the identity (1):

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SjkS

ijx = S

ikS

jix

PjkS

kj P

ijS

jix = P

ikS

ki P

ji S

ijx AX(S2)(1.6.5) (because j = i, k and k = i)

PjkPijS

ki S

jix = P

ikP

ji S

kj S

ijx AX(S2)(1.6.4) (because j = k and k = i)

PijP

ikS

ki S

jix = P

ji P

jkS

kj S

ijx AX(S2)(1.6.3) (because j = i, k and k = i)

PijS

ikS

jix = P

ji S

jkS

ijx AX(S2)(1.6.5) (because j = k and k = i)

PijS

ikS

jkx = P

ji S

jkS

ikx PAX(S3)/(p4)

PijS

ikS

jkx = P

ji S

ikS

jkx (v) (because j = i, k and k = i)

In the last identity we can apply the Method of the elimination of replacements to the Sjk and Sik. We

will get the following identity:

Pijx = P

ji x (if j = i and c0x = x) b. Let c0x = x

Pji x =

SpredSsucP

ji x =

SpredP

j+1i+1Ssucx =

PAX(S2)(1.1) Lemma 16. step a. in the deduction

(because i = j) (c0Ssuxx = Ssucx:Lemma 9.)

SpredPi+1j+1Ssucx =

SpredSsucP

ijx =

Pijx


(because i = j)

Thus we have completed the proof of the identity (1.6.1).

AX(S3)/(q1) exactly PAX(S1)(0.2.) if we shall choose as S the Sij.

AX(S3)/(q2) exactly PAX(S1)(0.1.) if we shall choose as S the Sij.

AX(S3)/(q3) exactly PAX(S3)/(p5).







Thus we have completed the proof of the Lemma 5. as well as the Theorem 4.

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