( ) ! ) Dr. Frederick A. Johnson Department of Philosophy Colorado State University Fort Collins, Colorado 80523 -123A Three-Valued Interpretation for a Relevance Logic * In thi.s paper an entailment relation which holds between certain propositions of ,the propositional calculus will be defined both syntactically and semantically. Some theorems about this relation will show why one could not follow Lewis to prove that a contradiction entails, for the notion of entailment discussed below, every proposition. We will develop a system RC in which the primitive symbols are the five symbols v c ) and the propositional variables The formation rules of RC arei 1) A variable standing alone is a well-formed formula (wff). 2) If A and B are well-formed (wf) then (A v B) is wf. 3) If A and B are wf then (A * B) is wf. 4) If A is wf then -A is wf. 5) If A is wf it is so in virtue of 1) ~ 4). We will let capital letters with or without subscripts be variables which range over occurre~ces of wffs. Following Church (see I'ntroduction to Mathematical Logi~, pp. 135-6) we will say that X is the full disjunctive normal form of A (the FDNF of A), where pi , ... , 1 pi are all and the only n -124propositional variable constituents of A, and ij < ik if j < k, if and only if i) X is a disjunction of at most 2n conjuncts such that each conjunction < < n is identical to (C. • (C .•. C ) ..• ) (1 m 2 ) , where ml m2 mn C (1 ~ j ~ n) is either p. of -p. and ii) '((A* X) v (-A* -X))' is a m. 1. i. J J J tautology. By the full disjunctive normal form of A relative to B (the FDNF of A/B) we will mean the FDNF of A if every p. in B is a p. in A. If 1 1 p. 1 ••• , pi are all and the only propositional variables which occur in 1 1 n B but not in A then by the FDNF of A/B we will mean the FDNF of (A * ( (p. v -p. ) * ( (p. v -p. ) . .. (p. v -p. ) ... ) 11 11 12 12 in ln Let us now define a relation which we will call syntactic-relevance-entailment and denote by ' + ' A + B if and only if i) Every disjunct in the FDNF of A/B is a disjunct in the FDNF of B/ A;. ii) There is at least one disjunct in the FDNF of A/B; and iii) Every p. in B is a p. in A. 1 1 To define semantic-relevance-entailment, denoted by '~ ', we will use the notion of a valuation of a wff of RC. Let V be a valuation of A if V is a function which i) assigns 0, 1 or 2 to each p. in A, ii) assigns 1 the same value to different occurrences, if any, of the same p. in A and 1 iii) assigns O, 1 or 2 to A as directed by the following tables: v 0 1 2 0 1 2 0 0 0 2 0 0 1 2 0 1 1 0 1 2 1 1 1 2 1 0 2 2 2 . 2 2 2 2 2 2 2 ÃB if and only if i) Every valuation that assigns 0 to A assigns 0 to B; and ii) There is at least one valuation which assigns 0 to A. ( ) -125The above notions of syntactic-relevance-entailment and semantic-relevance-entailment are extensionally equivalent. To show this (Theorem 1, below) we.will make use of the following lemmas. Lemma 1. If for every p. in X V (p.) = 0 or 1 then V (X) = 0 or 1. Proof: 1. 1. By strong induction on the number of symbols in X. Lemma 2. If for every pi in X (V(pi) = 0 or 11 then V(X) = 0 if and only if V(the FDNF of X) = 0 and V(X) = 1 if and only if V(the FDNF of X) = 1. Proof: Standard result. Lemma 3. If V(X) = 0 or 1 and pi is a wf part of X then V(pi) = 0 or 1. Proof: By using strong induction on the number of symbols in X we will prove that if p. is a wf part of X and V(p.) = 2 then V(X) = 2. i) Suppose that 1. . 1. there is one symbol in X. Then X = pi. If V(pi) = 2 then V(X) = 2. ii) By the induction hypothesis if there are m symbols in X, where m < n, then if pi is a wf part of X and V(p1) = 2 then V(X) = 2. Consider a formula Yin which there are n symbols where m < n. We need to show that if there is a pi in Y such that V(pi) = 2 then V(Y) = 2. There are three cases to consider. a) y = (Y l v Y2). Suppose there is a p. in Y it must be either Y1 or Y2. 1 If there is a pi in Y1 such that V(p.) 1 = 2 then by the induction hypothesis V(Y 1) ::: 2. But if V(Y 1) = 2 then V(Y) ::: 2. By similar reasoning if there is a p. in y2 such that V(p.) = 2 then V(Y) :: 2. b) y = (Y 1 . Y zl. Similar to 1 1 case a). c) Y = -Y1 . Similar to case a). Lemma 4. If V(the FDNF of A/B) = 0 then V(A) = 0. Proof: i) Suppose that every p. in B occurs in A. Then the FDNF of A/B is identical to the FDNF of 1 A. But if V(the.FDNF of A) = 0 then by lemmas 2 and 3 V(A) = 0. ii) Suppose -126that pi , . . . . , pi are all and the only variables that occur in B but not 1 n in A. Then the FDNF of A/B is identical to the FDNF of (A * ((p. v -p. ) 11 11 (p. v -p. ) 1 1 n n V( (A • ( (p. 11 V(A)= 0. ... ). By lemmas 2 and 3 if V(the FDNF of A/B) = 0 then v -p. ) 11 (p. 1 n v -p. ) ln . .. ) = 0 and thus (by the tables) Lemma S. If V(the FDNF of A/B) = 1 then V(A) = 1. Proof: Similar to the proof for Lemma 4. (Note that if V((A * ((p. v -p. ) 11 11 (p. v -p . ) ... ) = 1 l 1 n n then V (A) .. 1. ) Lemma 6. If the FDNF of A/B is non-empty then there is a valuation which ( ) assigns 0 to A. Proof: If the FDNF of A/B is non-empty let (C 1 * (C1 * ... * c1 ) ... ) be the left-most disjunct of the FDNF of A/B, 1 2 n .. *. where p. , ... , p. are the variables which occur in A or B. If c1 = p. 11 1 1 n r r let V(p. ) = 0 •. if c = -p. let V(p. ) = 1. Then V((C1 . (Cl cl 1 , 1 1 1 r r r r n 1 2 n . . . ) = o . Since c. = p. or -p. (1 ~ j 5. 2 ; l 5. r 5. n) V(any disjunct in Jr 1 1 r r the FDNF of A/B) . = 0 or 1 . So V(the FDNF of A/B) = 0. By Lemma 4 V(A) = 0. Lemma 7. If there is a valuation V such that V(A) = 0 then the FDNF of A/B is non-empty. Proof: Suppose V(A) = 0. Then by lemmas 2 and 3 V(the FDNF of A) = 0. Since there is nothing assigned to the empty symbol by V, the FDNF of A .is non-empty. If the FDNF of A is non-empty then the* FDNF of (A * ((p. v -p. ) ... (p. v -p. ) ... ) is non-empty, where 11 11 1n 1n p. (1 .s. j :5.. n) does not occur in A. So the FDNF of A/B is non-empty if 1. J V(A) = 0. Theorem 1. A + B if and only if A=-~ B. Proof: Consider the three conditions: a) every disjunct in the FDNF of A/B is a disjunct in the FDNF of B/A, • ) -127b) there is at least one disjunct in the FDNF of A/B and c) every p. in B 1 is a p. in A. We must show that these conditions are met if and only if 1 the following two conditions are met: d) every valuation which assigns 0 to A assigns 0 to B and e) there is a valuation which assigns 0 to A. i). (If a),b) and c) then d) .) Assume that a), b) and c) are true and that V(A) = 0. By lemmas 2 and 3 V(the FDNF of A) = 0. Since c) is true the FDNF of A is identical to the FDNF of A/B. So V(the FDNF of A/B) = 0. So V assigns 0 to at least one of the disjuncts of the FDNF of A/B. By a) V assigns 0 to at least one of the disjuncts of the FDNF of B/A. But then V assigns 0 or 1 to each of the disjuncts of the FDNF of B/A and thus V assigns 0 to the FDNF of B/A. By Lemma 4 if V(the FDNF of B/A) = 0 then V(B) = 0. ii) (If a), b) and c) then e) ,) Follows from Lemma 6. iii) (If d) and e) then a).) We will show that if a) is false then d) is false. Suppose there is a disjunct of the FDNF of A/B which is not a disjunct of the FDNF of B/A. Let V assign 0 to this disjunct. Then V assigns 1 to every other disjunct in the FDNF of A/B and 1 to every disjunct in the FDNF of B/A. So V(the FDNF of A/B) = 0 and V(the FDNF of B/A) == 1. By lemmas 4 and S V(A) = 0 and V(B) ~ O. iv) (If d) and e) then b).) Follows from Lemma 7. v) (If d) and e) then c).) We will show that if c) is false then d) is false. Assume that p. is in B but not in A. 1 V(p1) = 2. By.Lemma 3 V(B) = 2. Let V(A) = 0 and Theorem 2. (Simplification) If there is a valuation which assigns 0 to (A * B) then (A * B) ~A .. Proof: By examination of tables. Theorem 3. (Commutation) If there is a valuation which assigns 0 to (A * B) then (A * B) ~ (B * A) . Proof: By examination of tables. -128Theorem 4. (Addition) If there is a valuation which assigns p to A and if every p. in B is a p. *in A then A ~Av B. Proof: Assume that V(A) = 0 l 1 and that every p. in B is a p. in A. By lemmas 1 and 3 V(B) = 0 or 1. So 1 1 V(A v B) = 0. Theorem 5. (Adj unction) If A.~ B and Ã C then A-::;.. (B * C) . Proof: By examination of tables. Theorem 6. (Disjunctive Syllogism) If there is a valuation which assigns 0 to (-A * (A v B)) then (-A * (Av B)) ~ B. Proof: By examination of tables. Theorem 7. (Transitivity of Entailment) If A ~B and B ::~C then A =>-C. Proof: i) if A :::;:;.B then there is a valuation which assigns 0 to A. ii) If V(A) = O then if ÃB V(BJ = 0. But if V(B) = 0 and B~C then V(C) = 0. So if V(A) = 0 and the antecedent of Theorem 7 is true V(C) = 0.