distinct models, still insists on a function always returning a particular value, given a particular list of arguments. In the case of nondeterministi

Transcription

1 On Specialization of Derivations in Axiomatic Equality Theories A. Pliuskevicien_e, R. Pliuskevicius Institute of Mathematics and Informatics Akademijos 4, Vilnius 2600, LITHUANIA M. Walicki, S. Meldal University of Bergen Department of Informatics HiB, N-020 Bergen, NORWAY Abstract Walicki and Meldal have dened a calculus DEQ (\Disjunctive EQuational calculus") for reasoning about nondeterministic operators when specifying nondeterministic systems in an equation-oriented style. A variant of DEQ, the calculus DEQ for axiomatic equality theories with cut-like rules introducing as cut formulas only the negative equalities of a specic axiom, is constructed. For pure positive specic axioms (i.e. with empty antecedents) and for so called non-contrary equality theories DEQ does not contain cut-like rules at all. The variant of the calculus DEQ without structural rules of contraction and exchange is constructed. A simple cut-elimination procedure for axiomatic equality theories is presented. 1. Introduction The motivation for introducing DEQ was the need for the disjunctive formulae when specifying and reasoning about nondeterministic operations [9]. The notion of nondeterminism arises naturally in describing concurrent systems. Nondeterminism is also a natural concept in describing sequential programs, either as a means of indicating a don't care attitude as to which among a number of computational paths will actually be utilized in a particular computation or as a means of increasing the level of abstraction [3]. In terms of satisability, allowing nondeterminism is quite distinct from underspecication. The latter, though admitting 1

2 distinct models, still insists on a function always returning a particular value, given a particular list of arguments. In the case of nondeterministic operators this is no longer true. In [9, 10] we presented a function oriented view of nondeterministic operators. The result of a (non-)deterministic operator is considered a single (rather than set) value, namely the result of a particular application. Distinct occurrences of a particular term each denote a single value, but not necessarily the same one for each. For instance, specifying an operator t:s which chooses nondeterministically an element from the set s, one needs to say that for any application i, t i :f0; 1g = 0or t i :f0; 1g = 1. In [9,10] it is shown how viewing nondeterministic operations as determinisitc functions dependent on an additional index (application) argument, allows one to translate nondeterministic specications into deterministic ones using the language of conditional equations extended with disjunction. Then we presented a natural deduction system, DEQ, as a reasoning tool for transformed algebraic specications of nondeterministic operators, using axiomatic equality theories. Along with the natural aspect and modularity, the system DEQ contains a cut rule which is not amenable to automated reasoning. This paper introduces an equivalent variant DEQ of the DEQ calculus where cut is replaced by cut-like rules making DEQ more amenable to automation. A general theory of specialization of derivations in axiomatic equality theories was proposed by M. Rogava in [] and A. Pliuskevicien_e in [4, ]. In [] specic axioms of axiomatic equality theories were divided into two types: The axioms having a single equality, i.e. only one item-equality, were identied as being of type 1. The other specic axioms were identied as axioms of type 2. For specic axioms of type 1 equality-like rules were introduced in []. For specic axioms of type 2 unspecialized cut-like rules were introduced in []. In [4, ] we introduced specialized cut-like rules for specic axioms of type 2, introducing positive equalities of specic axioms as cut formulas. There we also introduced the complex notion of variants of equality formulas by specic axioms with a view to eliminate the introduction of the negative equalities of specic axioms as cut formulas. The main purpose of this paper is to construct the calculus DEQ which is equivalent to DEQ. DEQ contains a cut-like rule introducing as cut formulas only the negative equalities of specic axioms. Therefore in DEQ the complex notion 2

3 of variant of [4, ] is not necessary. For pure positive specic axioms (i.e. with empty antecedents) and for so-called non-contrary equality theories one can construct the calculus DEQ not containing cut-like rules at all. In this process no variant notion is used. The alternate version of the calculus DEQ not containing structural rules of contraction and exchange is also constructed. In the paper a simple cut-elimination procedure for axiomatic equality theories is presented. 2. The natural deduction-like calculi DEQ and DEQ 0 Denition 2.1 (sequent) A sequent is an expression?! ; where?; are nite sets of equalities r = s; where r; s are arbitrary terms. Denition 2.2 (calculus DEQ). A calculus DEQ is dened by the following postulates (see e.g. [9, 10]): (1) specic axioms (specications) (Sp. AX k ) r 1 = s 1 ; : : : ; r m = s m! u 1 = v 1 ; : : : ; u n = v n ; where k; m; n are some natural numbers; the index k identies the specic axioms; (2) equality postulates (where e represents a single equation, t; t 1 ; t 2 are arbitrary terms) R1.! t = t R2.? x t 1! x t 1 ;? 0! t 1 = t 2? x t 2 ;? 0! x t 2 R3.?!? x t! x t R4. e! e (3) Structural rules R.?!?! ; e?! e;?! (weakening) R6.?! ; e; e;? 0! 0?;? 0! ; 0 (cut) Remark 2.1. (i) DEQ is sound and complete w.r.t. standard semantics (see [9] with reference to [1]). 3

4 (ii) DEQ represents a variant of the natural deduction system for the applied axiomatic theory of equations. (iii) In DEQ the rule? x t 1! x t 1 ;? 0! t 1 = t 2 ; 0? x t 2 ;? 0! x t 2 ; 0 is derivable (using R2, R4 and R6). (iv) From denition 2.2 it follows that DEQ implicitly contains the structural rules \exchange" and \contraction". In the following assume given a set of specic axioms (Sp: AX k ). Denition 2.3 (the calculus DEQ 0 ). A calculus DEQ 0 is constructed from DEQ by removing rule R3 and changing (Sp: AX k ) of DEQ in the following manner: let S be (Sp: AX k ) of DEQ; then S x 1;:::;xn t 1 ;:::;tn is the set of specic axioms (Sp: AXk 0 ) of DEQ0 ; where t 1 ; : : : ; t n are arbitrary terms. Denition 2.4 (inclusion and equivalence of calculi). Let I 1 ; I 2 be arbitrary calculi then the notation I 1 ) I 2 indicates that if I 1 ` S then I 2 ` S (where S is a sequent) and the notation I 1 I 2 means I 1 ) I 2 and I 2 ) I 1 : Lemma 2.1. DEQ DEQ 0 : Proof: Follows from the fact that DEQ 0 `?! ) DEQ 0 `? x t! x t : 3. Kanger-like calculus DEQ 1 Denition 3.1 (calculus DEQ 1 ). A calculus DEQ 1 is obtained from DEQ 0 by replacing axiom R4 and rule R2 with the rules (see e.g. [2]): r = s;? x s! x s r = s;? x r! x r (= 1 ) s = r;? x s! x s s = r;? x r! x r (= 2 ) Lemma 3.1. DEQ 1 DEQ 0 : Proof: Follows from the following facts: (1) rules (= 1 ); (= 2 ) are derivable in DEQ 0 ; (2) R4 and R2 are derivable in DEQ 1 : 4. Rogava-like calculi DEQ 2 ; DEQ 3 4

5 Denition 4.1 (calculus DEQ 2 ). A calculus DEQ 2 is obtained from DEQ 1 by removing (cut) and replacing every specic axiom (Sp: AXk 0 ) by the following (Sp: cut k ) rule (see e.g. [, 4]): S 11 ; : : : ; S 1m ; S 21 ; : : : ; S 2n?! (Sp: cut k ); where?! :=? 11 ; : : : ;? 1m ;? 21 ; : : : ;? 2n! 11 ; : : : ; 1m ; 21 ; : : : ; 2n S 11 :=? 11! 11 ; r 1 = s 1 : : : : : : : : : : : : : : : : : : : : : S 1m :=? 1m! 1m ; r m = s m S 21 := u 1 = v 1 ;? 21! 21 : : : : : : : : : : : : : : : : : : : : : S 2n := u n = v n ;? 2n! 2n Denition 4.2 (calculus I + cut). Let I be some calculus not containing (cut), then I + cut denotes the calculus obtained from I by adding (cut). Lemma 4.1. (Sp: AXk 0 ) are derivable in DEQ 2: Proof: Using rules (Sp: cut k ): Lemma 4.2. The rules (Sp: cut k ) are derivable in DEQ 1 : Proof: Using the (cut) rule. Theorem 4.1. DEQ 1 DEQ 2 + (cut): Proof: From Lemmas 4.1 and 4.2. Denition 4.3 (calculus DEQ 3 ). A calculus DEQ 3 is obtained from the calculus DEQ 2 by replacing rules (= 1 ); (= 2 ) by the following rules (see e.g. [7]): Remark 4.1. r = s;?! x s r = s;?! x r (= 1s ) s = r;?! x s s = r;?! x r (= 2s ) Sometimes the notation (= s ) will be used instead of the notation (= is ) (i = 1; 2). Lemma 4.3. DEQ 3 ` S ) DEQ 2 ` S:

6 Proof. : Obvious. Denition 4.4. The notation I `D S will represent the fact that one can construct a derivation D of the sequent S in calculus I: Lemma 4.4. DEQ 2 `D S ) DEQ 3 `D S: Proof. The proof is carried out by induction on n! + h(d); where n is the number of antecedent applications of (= 1 ); (= 2 ) and h(d) is the height of the given derivation D: Consider the upper application of (= i ) (i = 1; 2) in D: We consider only the following case: D + >< >: u = v; r = s(u);? x s(u)! x s(v) (= 2s ) u = v; r = s(u);? x s(u)! x s(u) u = v; r = s(u);? x r! x r We construct the following derivation D 0 in DEQ 2 : D 0 >< >: D 1 n u = v; r = s(u);? x s(u)! x s(v) u = v; r = s(u);? x r! x s(v) u = v; r = s(u);? x r! x s(u) u = v; r = s(u);? x r! x r (= 1 ) (= 1 ) (= 2s ) (= 1s ) As h(d 1 ) < h(d + ); then DEQ 2 `D1 S 1 ) DEQ 3 `D 1 S1 ; where S 1 := u = v; r = s(u);? x r! x s(v) and DEQ 2 `D0 S 2 ) DEQ 3 `D0 S 2 ; where S 2 := u = v; r = s(u);? x r! x r : Theorem 4.2. DEQ 2 DEQ 3 : Proof: Follows from Lemmas 4.3 and 4.4. Remark 4.1. From the proof of Lemma 4.4 it follows that h(d ) > h(d); where D is the derivation of a sequent S in DEQ 3 ; D is the derivation of the same sequent S in DEQ 2 : Lemma 4.. DEQ 3 ` r = r;?! ) DEQ 3 `?! : Proof: By induction on the height of the given derivation. Lemma 4.6. Consider the derivation D : D 1?! ; e; D2 e;! (cut);?;! ; where D 1 ; D 2 are the derivations in DEQ 3 ; then DEQ 3 `?;! ; : 6

7 Proof. The proof is carried out by induction on h(d 1 ): Let h(d 1 ) = 0; then?! ; e :=! r = r: Consider the derivation D 2 in DEQ 3 : By Lemma 4. we get DEQ 3 `D0 2! and D 0 2 is the desired derivation. Let h(d 1 ) > 0; then we consider only the following case: D1 0 u = v;?! ; r = s(v) (= u = v;?! ; r = s(u) s ) D 2 r = s(u);! (cut) u = v;?;! ; Applying (weakening) and (=) to D 2 weget DEQ 2 `D0 2 S 0 2 := r = s(v); u = v;! : Using Lemma 4.4 we get DEQ 3 `D0 2 S2: 0 Applying (cut) to D1; 0 D 0 2 we get DEQ 3 + cut `D0 S := u = v;?;! ; : As h(d1) 0 < h(d 1 ) we get DEQ 3 ` S: Theorem 4.3 (elemination theorem for DEQ 2 ). DEQ 2 + cut ) DEQ 2 : Proof: Follows from Lemma 4.6 and Theorem 4.2. Theorem 4.4. DEQ 1 DEQ 2 : Proof: From Theorems 4.1 and Specialization of (Sp: cut k ) rules Denition.1 (calculus DEQ 4 ). A calculus DEQ 4 is obtained from DEQ 2 by replacing rules (Sp: cut k ) by the rules S 11 ; : : : ; S 1m ; S 21 ; : : : ; S 2n?! (Sp: = k ); where?! :=? 11 ; : : : ;? 1m ;? 21 ; : : : ;? 2n! 11 ; : : : ; 1m ; 21 x 1 11 ; : : : ; 2n xn n1 S 11 :=? 11! 11 ; r 1 = s 1 : : : : : : : : : : : : : : : : : : : : : S 1m :=? 1m! 1m ; r m = s m S 21 :=? 21! 21 x 1 12 : : : : : : : : : : : : : : : : : : : : : S 2n :=? 2n! 2n x n n2 7

8 where j1 ; j2 2 fu j ; v j g; j 2 f1; : : : ; ng and j1 6= j2 : Remark.1. Special cases of (Sp: = k ) rules, corresponding to specic axioms of the form! u = v; were presented in [, 4, ]. In [4, ] the complex notion of variant of formula by specic axioms was used instead of the introduction of negative equalities r j = s j (j = 1; : : : ; m) of specic axioms as cut formulas in the premises s 11 ; : : : ; s 1m ; equalities u j = v j (j = 1; : : : ; n) in the premises s 21 ; : : : ; s 2n were introduced as cut formulas, besides. In rules (Sp: = k ) only the negative equalities of specic axioms are introduced as cut formulas. Therefore the complex notion of the variant of a formula is unnecessary. If m = 0 then the (Sp: = k) rules are purely cut-free. Analogous rules (corresponding to the case when m = 0) were presented in [6] for arithmetical robinson system and at the same time no cut-like rule was not introduced for the specic axiom containing a negative equality. Example.1. Let DEQ 0 have the following three specic axioms (Sp: AX k ) (k 2 f1; 2; 3g) : (1) r = s! u 1 (c 1 ) = v 1 (c 2 ); u 2 (c 3 ) = v 2 (c 4 ) (2) u 1 (c ) = v 1 (c 6 )! (3) u 2 (c 7 ) = v 2 (c )! Then the calculus DEQ 4 has the following three (Sp: = k ) (k 2 f1; 2; 3g) rules: x? 1! 1 ; r = s;? 2! 1 x 2 12 ;? 3! x? 1 ;? 2 ;? 3! 1 ; 1 x 2 11 ; 2 (Sp: = 1 ) 3 21 where?! ; u 1 (c ) = v 1 (c 6 )?!?! ; u 2 (c 7 ) = v 2 (c )?! (Sp: = 2 ) (Sp: = 3 ); 11 ; 12 2 fu 1 (c 1 ); v 1 (c 2 )g 21 ; 22 2 fu 2 (c 3 ); v 2 (c 4 )g Let S :=?!; where? := s = r; c 1 = c ; c 2 = c 6 ; c 3 = c 7 ; c 4 = c ; then applying (Sp: = 3 ); (Sp: = 2 ); (= s ); (Sp: = 1 ) bottom-up we get DEQ 4 ` S:

9 Lemma.1. DEQ 4 ) DEQ 3 : Proof: Follows from the fact that rules (Sp: = k ) are derivable in DEQ 3 : Example.2. Let DEQ 4 be as in Example.1. Consider the rule (Sp: = 1 ): Let us show that (Sp: = 1 ) is derivable in DEQ 3. Let S 1 :=? 1! 1 ; r = s S 2 :=? 2! 2 x 1 12 S 3 :=? 3! 3 x 2 22 Applying (W ); (= is ) (i = 1; 2) to S 2 ; S 3 we get S 0 2 := u 1 (c 1 ) = v 1 (c 2 );? 2! 2 x 1 11 S 0 3 := u 2 (c 3 ) = v 2 (c 4 );? 3! 3 x 2 21 Applying (Sp. cut 1 ) to S 1 ; S 0 2; S 0 3 we get the conclusion S of (Sp.= 1 ). Lemma.2. Consider the derivation D.... S 11 ; : : : ; S 1m ; S 21 ; : : : ;?! S 2n (Sp: cut k ) and let D 11 ; : : : ; D 1m ; D 21 ; : : : ; D 2n be the derivations of the sequents S 11 ; : : : ; S 1m ; S 21 ; : : : ; S 2n ; respectively, in the calculus DEQ 4 ; then DEQ 4 `?! : Proof: By induction on h(d 21 ) + + h(d 2n ): Example.3. Let DEQ 4 be the same as in Example.1 and let S 1 ; S 0 2; S 0 3 be the same as in Example.2. Then (Sp. cut 1 ) has the following form: D 1 fs 1 ; D 2 fs 0 2; D 3 fs 0 3 (Sp:cut 1 ) S Let us consider only the case when D 2 ; D 3 have the following forms: D 2 D 3 >< >: >< >: D 0 2 D 0 3 n u 1 (c 1 ) = v 1 (c 2 );? 2! 2 x 1 12 (= is ) u 1 (c 1 ) = v 1 (c 2 );? 2! 2 x 1 11 n u 2 (c 3 ) = v 2 (c 4 );? 3! 3 x 2 22 (= is ) u 2 (c 3 ) = v 2 (c 4 );? 3! 3 x

10 x Having set? :=? 1 ;? 2 ;? 3 and S 2 := u 1 (c 1 ) = v 1 (c 2 );? 2! ; x S 3 := u 2 (c 3 ) = v 2 (c 4 );? 3! , we construct the following derivation D : D >< >: S 1 ; D 0 2fS 2 ; D 3 fs 0 3 S 1 ; D 2 fs 0 2; D 0 3fS 3 x 1 x 2 x 1 x 2 S 1 ;?! 1 ; 2 ; 3?! 1 ; 2 ; (Sp: = 1 ) x 1 x 2?! 1 ; 2 ; As h(d 0 2) < h(d 2 ) and h(d 0 3) < h(d 3 ), The (Sp. cut 1 ) steps (unlabeled above) can be eliminated from the derivation D : Theorem.1. DEQ 4 DEQ 3 : Proof: Follows from Lemmas.1 and.2. Denition.2 (calculus DEQ ). A calculus DEQ is obtained from calculus DEQ 4 by replacing rules (Sp: = k ) by the rules (Sp: = k ) : where S 11 ; : : : ; S 1m ; S 21 ; : : : ; S 2n?! (Sp: = k);?! :=?! x 1 : : : x n 11 : : : n1 S 11 :=?! x 1 : : : x n 11 : : : n1 ; r 1 = s 1 : : : : : : : : : : : : : : : : : : : : : S 1m :=?! x 1 : : : x n 11 : : : n1 ; r m = s m S 21 :=?! x 1x 2 : : : x n : : : n1 : : : : : : : : : : : : : : : : : : : : : S 2n :=?! x 1 : : : x n?1 x n 11 : : : n?11 n2 Lemma.2. DEQ 4 DEQ : Proof: Using structural rule weakening. Denition.3 (calculus DEQ ). A calculus DEQ is obtained from the calculus DEQ by dropping structural rule weakening and replacing the axiom! r = r by the axiom?! ; r = r: 10

11 Lemma.3. Structural rule weakening is admissible in DEQ : Proof: By induction on the height of a given derivation. Theorem.2. DEQ DEQ: Proof: Follows from Lemmas 2.1, 3.1, Theorems 4.2, 4.4,.1 and Lemmas.2, Some specializations of (Sp: = k ) rules Denition 6.1 (contrary calculus DEQ). A calculus DEQ will be called contrary if there exists some substitution x=t (where x := x 1 ; : : : ; x n ; t := t 1 ; : : : ; t n ) and some specic axiom k containing a negative occurrence of an equality ri k = s k i ; where i 2 f1; : : : ; m k g; such that using only equality rules (= is ) it is possible to construct a derivation of the sequent? t x! (ri k = sk i )x t where? consists of positive occurrences of equality in specic axioms of DEQ and maybe some equalities constituted only from symbols of constants entering in axioms (Sp. AX k ) In the opposite case DEQ will be called non-contrary. Example 6.1. Let DEQ have the following specic axioms (Sp. AX k ) (k 2 f1; 2; 3g) : (1) f(u) = r(a)! t 1 = t 2 (c 1 ) (2)! f(c 1 ) = g(v) (3)! r(b) = g(c 2 ) where a; b; c 1 ; c 2 are some constants. DEQ is contrary because it is possible to construct the derivation D of the sequent? t x! (rk i = sk i )x t, where? := f(c 1) = g(v); r(b) = g(c 2 ); a = b; (ri k = sk i ) := f(u) = r(a) and the substitution x= t; where x := u; v; t := c 1 ; c 2. The derivation D has the form: f(c 1 ) = g(c 2 ); r(b) = g(c 2 ); a = b! r(a) = r(a) (= s ) f(c 1 ) = g(c 2 ); r(b) = g(c 2 ); a = b! r(b) = r(a) (= s ) f(c 1 ) = g(c 2 ); r(b) = g(c 2 ); a = b! g(c 2 ) = r(a) (=s ) f(c 1 ) = g(c 2 ); r(b) = g(c 2 ); a = b! f(c 1 ) = r(a) Example 6.2. Let DEQ has the following (Sp. AX k ) (k 2 f1; 2g) : (1) f(u) = g(v)! t 1 = t 2 (c 1 ) (2)! g(c 1 ) = f(c 2 ) 11

12 where c 1 ; c 2 are some constants. It is easy to see that DEQ is contrary under the substitution x=t, where x := u; v; t := c 2 ; c 1. The following derivation of the sequent y = c 1 ; u = c 2 ; v = c 1! t 1 = t 2 (y) shows the necessity of introducing as cut formula the negative occurrence of equality f(u) = g(v) :?! f(c 2 ) = f(c 2 ) (Sp: = 2 )?! f(c 2 ) = g(c 1 ) (=s ); (= s )?! f(u) = g(v)?! t 1 = t 1 (Sp: = 1 )?! t 1 = t 2 (c 1 ) (= s )?! t 1 = t 2 (y) where? := y = c 1 ; u = c 2 ; v = c 1 : Denition 6.2 (calculus DEQ ). The calculus DEQ is obtained from the calculus DEQ (1) replacing rules (Sp: = k ) by following ones: S 0 21; : : : ; S 0 2n ;?! x 1:::xn 11 :::n1 (Sp: = k ); where := r 0 1 = s 0 1; : : : ; r 0 m = s 0 m; r 0 i 2 fr i ; s i g; s 0 i 2 fr i ; s i g; i 2 f1; : : : ; mg; r 0 i 6= s0 i ; S0 2j is obtained from S 2j of rules (Sp: = k ) changing? by ;?; (2) replacing rules (= is ) by (= i ) (i = 1; 2): Theorem 6.1. Let DEQ be a non-contrary calculus, then DEQ DEQ : Proof: Applying the denition of contrary calculus. Example 6.3. Let DEQ be the following (Sp. AX k ) (k 2 f1; 2g) : (1) g(v) = f(u)! t 1 = t 2 (c 1 ) (2)! g(c 1 ) = g(c 2 ) where c 1 ; c 2 are some constants. It is easy to see that DEQ is non-contrary. We construct the following derivations of the sequent y = c 1 ; v = s; u = r; f(u) = g(s)! t 1 = t 2 (y), which shows how the rule (Sp. = )works :?! t 1 = t 1 (Sp: = )?! t 1 = t 2 (c 1 ) (= s )?! t 1 = t 2 (y) (=)?! t 1 = t 2 (y) where? := y = c 1 ; v = s; u = r; f(u) = g(v);? := y = c 1 ; v = s; u = r; f(u) = g(s): 12

13 Remark 6.1. To prove that DEQ DEQ in case DEQ is contrary it is necessary to change the rules (Sp: = k ) in such a way that they involve the notion of variant of formula by specic axioms (see [4,]). 7. Elimination of contraction and exchange Denition 7.1 (calculus DEQ 6 ). A calculus DEQ 6 is obtained from the calculus DEQ in the following way: (i) the?; in sequent?! are considered arbitrary lists rather than sets; (ii) adding the structural rules of contraction and exchange:?; e 1 ; e 2 ;!?; e 2 ; e 1 ;! (EX!)?! ; e 1 ; e 2 ; (! EX)?! ; e 2 ; e 1 ; e; e;?! e;?! (C!)?! ; e; e (! C)?! ; e Lemma 7.1. DEQ 6 DEQ : Proof: Applying denition 7.1. Denition 7.2. (calculus DEQ 7 ). The calculus DEQ 7 is obtained from the calculus DEQ 6 by dropping structural rules (EX!); (! EX): Lemma 7.2. DEQ 6 DEQ 7 : Proof:d Aanalogously to the proof in [7]. Denition 7.3 (Calculus DEQ ). A calculus DEQ is obtained from the calculus deq 7 by dropping the structural rules (C!); (! C) and changingthe rules (=) and (Sp: = k ) in such a way that the main formulas are duplicated in the premises of the rule. For example, instead of the rule (= 1s ) the following rule: r = s;?! x s; x r r = s;?! x r (= 1s) is in the calculus DEQ : Denition 7.4 (calculus DEQ 0 7). A calculus DEQ 0 7 is obtained from the calculus DEQ 7 by dropping the structural rule (C!): Lemma 7.3. DEQ 7 `D S ) DEQ 0 7 `D S: 13

14 Proof: By induction on n! + h(d); where n is the number of applications of (C!) in D: Lemma 7.4. DEQ 0 7 `D) DEQ `D S: Proof. The proof is carried out by n! + h(d); where n is the number of applications of (! C) in D: We consider only the following case D 0 >< >: D 0 1 n?; r = s! ; p(s) = q; p(r) = q; p(r) = q (= s )?; r = s! ; p(r) = q; p(r) = q (! C)?; r = s! ; p(r) = q where D 0 1 is the derivation in DEQ : Let us construct the following derivation D 0 : D 0 < : n?; r = s! ; p(s) = q; p(r) = q; p(r) = q D 0 1?; r = s! ; p(s) = q; p(r) = q?; r = s! ; p(r) = q (! C) (= s) As h(d 0 1 ) < h(d1) 0 then DEQ 0 7 `D0 S ) DEQ `D0 S ; where S :=?; r = s! ; p(r) = q: Theorem 7.2. DEQ 7 DEQ : Proof: Follows from Lemmas 7.3 and 7.4. Remark 7.1. Let p = q be any explicitly indicated term in (Sp. AX k ) of DEQ 0 ; then p; q will be called axiom terms. We introduce an ordering of terms: Let r(t) stand for the rank of a term t which is dened in the following way: (1) r(t) = 0; if t is any axiom term; (2) r(t 1 ) < r(t 2 ); if t 1 ; t 2 are not axiom terms and t 1 is less then t 2 in a lexicograpc ordering of terms. Then using the ordering of terms introduced above we can drop duplication of the main formulas in the equality rules and leave this duplication in the rules (Sp: = k ) only. References 1. S. Feferman, Lectures on Proof Theory, LNM, 70, Springer, S. Kanger, A simplied proof method for elementary logic. Comput. Progr. and Formal Systems, North-Holland, Amsterdam, 7{94,

15 3. S. Meldal, An abstract axiomatization of pointer types, in Proc. of the 22nd Annual Hawaii International Conference on System Sciences, B.D. Shriver (ed.), IEEE Computer Society Press, vol. 2, 129{134, A. Pliuskevicien_e, Elimination of cut-type rules in axiomatic theories with equality. Seminars in mathematics V.A. Steklov Mathem. Institute, Leningrad, 16, 90{94, A. Pliuskevicien_e, Specialization of the use of axioms for deduction search in axiomatic theories with equality, Seminars in mathematics V.A. Steklov Mathem. Institute, Leningrad, J. Soviet Math. 1, 110{ 116, A. Pliuskevicien_e, A sequential variant of R.M.Robinson's arithmetic system not containing cut rules. Proc. Steklov Inst. Math., Leningrad, 121, 121{10, R. Pliuskevicius, Sequential variant of the calculus of constructive logic for normal formulas. Proc. Steklov Inst. Math., 9, 17{229, M. Rogava, Sequential variants of applied predicate calculi without structural deductive rules. Proc. Steklov Inst. Math. 121, 136{164, M. Walicki, Algebraic Specications of nondeterminism, Ph. D thesis, University of Bergen, M. Walicki, S. Meldal, A complete calculus for the multialgebraic and functional semantics of nondeterminism, (submitted for publ.)