2.4.5 Metatheorems about Deductions

Transcription

1 96 CHAPTER 2. FIRST-ORDER LOGIC Metatheorems about Deductions We have been using the word theorem in two di erent ways. When ` we say that is a theorem of. We have also stated and proven theorems in English about deductions in afirst-orderlanguage. TheseEnglishtheoremsareoftencalledmetatheorems to emphasize that they are statements in English about deductions and first-order logic. How to show that a deduction exists without actually giving one? Answer: One must first prove metatheorems about deductions. Induction on Deductions Let be a set of w s. Since there is a procedure for building all the theorems of by first starting with the formulas in [ and then, applying modus ponens, we have the following induction principle. Induction Principle: Let S be a set of w s such that 1. [ S 2. for all w s and,if 2 S and (! ) 2 S, then 2 S. Then S contains the all the theorems of. Proof Strategy: Let Stmt(') beastatementaboutaw '. Inordertoproveanassertion for all w s ', if ` ', thenstmt(') by induction, use the following diagram: Base Step: Prove Stmt( )forallformulas 2 [. Inductive Step: Let and be w s. Assume Stmt( ) and Stmt(! ). Prove Stmt( ). Remark. The proof strategy, follows from the above induction principle applied to the set S = {' is a w : Stmt(')}. WenowapplythisstrategywhereStmt(') is `8x'. Theorem (Generalization Theorem). Suppose that x does not occur free in any formula in. For all w s ', if ` ', then `8x'. Proof. Assume that x does not occur free in any formula in. We shall prove, using the induction principle, the following: For all w s ', if ` ', then `8x'. Base Step: Let be in [. If 2, then x does not occur free in.thus,(!8x ) is in axiom group 4 of the Logical Axioms Since ` and ` (!8x ), we conclude (by modus ponens) that `8x. If 2, then is a logical axiom. Thus, the generalization 8x is also a logical axiom. Therefore, `8x. Inductive Step: Let and be w s. Assume the induction hypothesis `8x and `8x(! ). (IH) We must prove that `8x. Note that 8x(! )! (8x!8x )isinaxiomgroup3 of the Logical Axioms By (IH) we have that `8x(! )and `8x. Thus, by applying modus ponens twice, we have that `8x. Thiscompletestheproofofthe theorem.

2 2.4. A DEDUCTIVE CALCULUS 97 A converse of the Generalization Theorem holds. Theorem For all w s ', if `8x', then ` '. Proof. Assume that `8x'. Bylogicalaxiom2, wehavethat ` (8x'! '). Thus, by modus ponens, we conclude that ` '. Theorem (Rule T). If ` 1,..., ` n and { 1,..., n } tautologically implies, then `. Proof. Assume that ` 1,..., ` n and that { 1,..., n } tautologically implies. Thus, 1! 2!! n! is a tautology and hence, it is a logical axiom (1). Because ` 1,..., ` n,weconcludethat ` by applying modus ponens n times. For example, since (! ) tautologicallyimplies and, and{, } tautologically implies (! ), we have the following corollary of Theorem Corollary ` (! ) ifandonlyif ` and `. Recall that the notation ; is an abbreviation for [{ }. Theorem (Deduction Theorem). Let be a set of w s and let,' be w s. Then ; ` ' if and only if ` (! '). Proof. Let be a set of w s and let,' be w s. Then ; ` ' i ; [ tautologicallyimplies' by Theorem The proof is complete. i [ tautologically implies (! ') by Problem 8 on page 52 of notes i ` (! ') by Theorem Corollary Let be a set of w s. If and are tautologically equivalent, then (1) ` i `. (2) ; ` ' i ; ` ', foranyw '. Proof. Let be a set of w s and let and be tautologically equivalent. Theorem implies that ` i `,establishing(1). Toprove(2),observethatsince and are tautologically equivalent, we have that! ' and! ' are tautologically equivalent, for any w '. Wecannowprovethat ; ` ' i ; ` ' as follows: The proof is complete. ; ` ' i ` (! ') by Theorem i ` (! ') by (1) i ; ` ' by Theorem

3 98 CHAPTER 2. FIRST-ORDER LOGIC Corollary (Contraposition). ; ' ` i ; ` '. Proof. Let be a set of w s and let ', be w s. Observe that ('! ) and (! ') are tautologically equivalent. We now prove that ; ' ` i ; ` ' as follows: The proof is complete. ; ' ` i ` ('! ) by Theorem i ` (! ') by Corollary (1) i ; ` ' by Theorem Definition Aset offormulasisconsistent if there is no formula such that ` and `. Moreover, a set of w s is inconsistent when it is not consistent; that is, when there is a formula such that ` and `. Remark Suppose that is inconsistent. Then for any w, wehavethat `. The reason for this follows: Let such that ` and `.Then,because! (! ) is a tautology, it follows that `. Corollary (Reductio ad Absurdum). If ; ' is inconsistent, then ` '. Proof. Assume that ; ' is inconsistent. So, there is a formula such that ; ' ` and ; ' `. Thus, ` ('! )and ` ('! )bythedeductiontheorem. Since {'!,'! } tautologically implies ', weconcludethat ` ' by Rule T. Corollary implies that if 0 ', then ; ' is consistent. Since (! )! and (! )! are tautologies, we have the following Strategies for Showing that Deductions Exist To show that a deduction exists, one can use the following deduction strategies: (S1) To show ` (! ), it is su ; `, bythedeductiontheorem. (S2) To show that ; `, itissu cienttoshowthat ; `, by contraposition. Also, ; ` 8x i ; 8x `. (S3) To show that `8x, itissu cienttoshowthat `, ifx does not occur free in, by the generalization theorem. (S4) To show that ` (! ), it is su ` and `, byrulet and the fact that {, } tautologically implies (! ). (S5) To show that `, itissu cienttoshowthat `, byruletandthefactthat { } tautologically implies. (S6) To show that `, itissu cienttoshowthat ; is inconsistent, by reductio ad absurdum. (S7) To show that ` 8x, itissu cienttoshowthat ` t x for some term t. Since ` ( t x! 8x ) bylogicalaxiom2andcorollary2.4.27(1). Thus, ` 8x would follow by modus ponens. (If this is not useful, try strategy (S6).)

4 2.4. A DEDUCTIVE CALCULUS 99 Problem 1. Using the above strategies, show that 1. `8xP (x)!9xp (x). 2. 8x8yP(x, y) `8y8zP(z,y). 3. 8xP (x) `8xQ(x)!8x (P (x)! Q(x)). Solution. (1) We will show that ` 8xP (x)! 9xP (x). By the deduction theorem (see (S1)), it is su 8xP (x) ` 9xP (x). By Theorem , we see that 8xP (x) ` P (x). By Example 31 on page 93 and the deduction theorem, we have that P (x) `9xP (x). Since 8xP (x) ` P (x) andp (x) `9xP (x), Lemma implies that 8xP (x) `9xP (x). Therefore, `8xP (x)!9xp (x). (2) We shall show that 8x8yP(x, y) `8y8zP(z,y). Two applications of logical axiom 2 and modus ponens, shows that 8x8yP(x, y) ` P (z,y). Since z is not free in 8x8yP(x, y), we see that 8x8yP(x, y) `8zP(z,y), by applying (S3). Similarly, as y is not free in 8x8yP(x, y), we conclude that 8x8yP(x, y) `8y8zP(z,y). (3) We shall show that 8xP (x) ` 8xQ(x)! 8x (P (x)! Q(x)). By the deduction theorem, it is su {8xP (x), 8xQ(x)} `8x (P (x)! Q(x)). So by strategy (S3), it is su {8xP (x), 8xQ(x)} ` (P (x)! Q(x)). By logical axiom 2 and modus ponens, we see that {8xP (x), 8xQ(x)} `P (x) and{8xp (x), 8xQ(x)} `Q(x). Thus, {8xP (x), 8xQ(x)} `P (x), and {8xP (x), 8xQ(x)} ` Q(x) bycorollary2.4.27(1). Applying strategy (S4), we conclude that {8xP (x), 8xQ(x)} ` (P (x)! Q(x)). Hence, 8xP (x) `8xQ(x)!8x (P (x)! Q(x)). Problem 2. Show that `9x8y'!8y9x'. Solution. We show that a deduction of 9x8y'!8y9x' exists. By the deduction theorem it is su 9x8y' `8y9x'. Hence, by the Generalization Theorem , it is su 9x8y' `9x'. Removing the abbreviations, we need to show that 8x 8y' ` 8x '. By contraposition, this reduces to showing that 8x ' `8x 8y' (see Corollary ). So, by the generalization theorem, we must show that 8x ' ` 8y' and thus, by reductio ad absurdum, it is now su = {8x ', 8y'} is inconsistent. Note that ` ' and ` ' by logical axiom 2 and modus ponens. Hence, is inconsistent. Therefore, it follows that there is a deduction of 9x8y'!8y9x'. Proposition (Q2A). If x does not occur free in, then ` (!8x ) $8x(! ). Proof. By Rule T, it is su ` (!8x )!8x(! ), (?) `8x(! )! (!8x ). (??) For (?): By the deduction theorem we need to show that (!8x ) `8x(! ). By assumption, x does not occur free in. Hence, x does not occur free in (!8x ). Thus,

5 100 CHAPTER 2. FIRST-ORDER LOGIC by the Generalization Theorem , it is now enough to show that (!8x ) ` (! ). Again, by the deduction theorem, it will be su {(!8x ), }`.Let ={(!8x ), }. Thus, (1) `!8x Premise. (2) ` Premise. (3) `8x By (1), (2) and modus ponens. (4) `8x! Logical axiom 2. (5) ` By (3), (4) and modus ponens. This completes the proof of (?). 6 For (??): By the deduction theorem we need to show that 8x(! ) ` (!8x ). So, again by the deduction theorem, we must show that {8x(! ), }`8x. Since x does not occur free in 8x(! )orin, thegeneralizationtheoremimpliesthatwejustneed to show {8x(! ), }`.Let ={8x(! ), }. Thus, (1) `8x(! ) Premise. (2) ` Premise. (3) `8x(! )! (! ) Logical axiom 2. (4) `! By (1), (3) and modus ponens. (5) ` By (2), (4) and modus ponens. This completes the proof of (??). Propositions and , below, are presented in Exercise 8 on page 130 of text. Proposition (Q2B). If x does not occur free in, then` (!9x ) $9x(! ). Proposition (Q3A). If x does not occur free in, then` (8x! ) $9x(! ). Proposition (Q3B). If x does not occur free in, then` (9x! ) $8x(! ). Proof of Proposition By Rule T, it is su ` (9x! )!8x(! ), (?) `8x(! )! (9x! ). (??) We shall first prove (?). It is su (9x! ) `8x(! ), by the deduction theorem. By assumption, x does not occur free in. Hence, x does not occur free in (9x! ). So, by the Generalization Theorem , it is now enough to show that (9x! ) ` (! ). Again, by the deduction theorem, it will be su 6 The list (1) (5) is not a deduction. It is just part of a proof showing that a deduction of (?) exists. In this list we have implicitly applied Lemmas on page 94 of these notes.

6 2.4. A DEDUCTIVE CALCULUS 101 {(9x! ), }`. Let ={(9x! ), }. Thus, 7 (1) ` 8x! Premise. (2) ` Premise. (3) `8x! Logical axiom 2. (4) `! 8x By (3) and Cororollay (1). (5) ` 8x By (2), (4) and modus ponens. (6) ` By (1), (3) and modus ponens. In (4) we used that fact that (8x! ) and(! 8x ) are tautologically equivalent. We shall now prove (??). By the deduction theorem we need to show that 8x(! ) ` (9x! ). So, by the deduction theorem and contraposition, it is su {8x(! ), } ` 9x. As 9x and 8x are tautologically equivalent, by Corollary (1), we only need to show that {8x(! ), } `8x. By assumption, x does not occur free in. Hence, x does not occur free in any formula in {8x(! ), }. Thus,bythegeneralizationtheorem,itisnowenoughtoshowthat {8x(! ), } `. Furthermore, by contraposition, we just need to show that Let = {8x(! ), }. Thus, This completes the proof. {8x(! ), }`. (1) `8x(! ) Premise. (2) ` Premise. (3) `8x(! )! (! ) Logical axiom 2. (4) `! By (1), (3) and modus ponens. (5) ` By (2), (4) and modus ponens. The following theorem can be useful for showing that certain deductions exist. Theorem Let and be w s. If `!,then`8x!8x Proof. This is assigned Exercise #6(a) on page 130 of text. 7 Recall that 9x is an abbreviation for 8x.

7 102 CHAPTER 2. FIRST-ORDER LOGIC Homework: Exercise #6(a), 7, 8, 10 on page 130 of text. Hint for 7(a): This is a strange, but valid, sentence. Use reductio ad absurdum by showing that 8x (P (x)!8xp (x)) `8xP (x) (vialogicalaxiom2, atautology, andgeneralization) and showing, via logical axiom 2 and a tautology, that 8x (P (x)!8xp (x)) ` 8xP (x). Hint for 8: For the first part, by Rule T, it is su Hint for 10: Use Problem 1(2) on page 99. ` (!9x )!9x(! ) (?) `9x(! )! (!9x ). (??)