Semantic Metatheory of SL: Mathematical Induction

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Amanda Townsend
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1 Semantic Metatheory of SL: Mathematical Induction Preliminary matters: why and when do we need Mathematical Induction? We need it when we want prove that a certain claim (n) holds for all n N (0, 1, 2, ) (or for all n k, where k N). o E.g.: For all n 4, n! > 2 n. 1

2 Weak Mathematical Induction: o A valid argument (proof) schema: (k) For every m k: if (m), then (m + 1) (n) for all n k Yuri Balashov, PHIL/LING 4510/6510 (Basis clause) (Inductive step) (m) is the inductive hypothesis. Example: prove by WMI that for all n 4, n! > 2 n. Proof: (i) Basis clause: 4! = 24 > 16 = 2 4. (ii) Inductive step: Suppose m 4 and assume m! > 2 m (inductive hypothesis). Then prove that (m + 1)! > 2 m+1. (m + 1)! = m! (m + 1) (Def. of!) > 2 m (m + 1) (Ind. Hyp.) > 2 m 2 (m 4) = 2 m+1. 2

3 Strong Mathematical Induction: o A valid argument (proof) schema: Yuri Balashov, PHIL/LING 4510/6510 (k) (Basis clause) For every m k, if (l) holds for all l such that k l m, then (m + 1) (Inductive step) (n) for all n k (l) holds for all l such that k l m is the inductive hypothesis. Example: prove by SMI that for all Fibonacci numbers F n (n N), defined recursively as follows: F 0 = 0; F 1 = 1; F n+2 = F n + F n+1, n F i = Fn + 2 i= 0 1 Proof: (i) Basis clause: 0 F i= 0 i = F 0 = 0 = ( F0 + 1) 1 = ( F0 + F1 ) 1 = F2 (ii) Inductive step: Assume inductive hypothesis: For every l i= 0 i l+ 2 m+ 1 F i = Fm + 3 i= 0 0 l m, F = F 1, and prove that 1. (Left as an exercise.) 1. 3

4 Weak vs. Strong Mathematical Induction: o Strong Mathematical Induction is logically equivalent to Weak Mathematical Induction: everything that can be proven by Weak Mathematical Induction is provable by Strong Mathematical Induction and vice versa. o But Strong Mathematical Induction is more convenient for some purposes. 4

5 Why do we need Strong Mathematical Induction in the Metatheory of SL? o Because we need to be able to prove something about all sentences of SL, i.e. all sentences P S, where S is the complete set of sentences of SL: The smallest set S containing every sentence letter (i.e. atomic sentence) of SL and such that: 1. If P S then ~P S; 2. If P, Q S then (P&Q), (P Q), (P Q), (P Q) S. o S is an infinite countable set. o We need a way of going over all the infinite number of sentences of SL (P S) without missing any. o And if we want to use Mathematical Induction we need a way of grouping all the sentences of SL according to the sequence of natural numbers n N (0, 1, 2, ). Such groups may themselves be infinite. 5

6 o One useful way of such grouping is by the number of occurrences of connectives (i.e. the number of connective tokens) in them: Group 0: all the sentences of SL with zero occurrences of connectives = all the atomic sentences (sentence letters) of SL: A, A 1, A 2, B, B 1, B 2, Z, Z 1, Z 2, an infinite countable number of them. Group 1: all the sentences of SL with one occurrence of connectives: ~A, ~A 1, ~Z, ~Z 1, (A&A 1 ), (A&A 2 ) (A&Z) an infinite countable number of them. Important: they all have one of the following forms: ~P, (P&Q), (P Q), (P Q), (P Q), where P and Q have zero occurrences of connectives. Group 2: all the sentences of SL with two occurrences of connectives: ~ ~A, ~ ~A 1, (~A&Z) an infinite countable number of them. Important: they all have one of the following forms: ~P, (P&Q), (P Q), (P Q), (P Q), where P and Q have one or fewer occurrences of connectives. Group n: all the sentences of SL with n occurrences of connectives an infinite countable number of them. Important: they all have one of the following forms: ~P, (P&Q), (P Q), (P Q), (P Q), where P and Q have n 1 or fewer occurrences of connectives. 6

7 o Suppose: we (i) demonstrate that all the atomic sentences of SL have some property ; and (ii) show that if every sentence of SL with k or fewer occurrences of connectives has, then every sentence of SL with k+1 occurrences of connectives has. Then, by SMI, every sentence of SL has. o More generally, arguments by SMI work as follows: (TLB, p. 230): First, we group the items [e.g. sentences of SL] about which we wish to prove some claim into a series of cases, each associated with a nonnegative integer k. 7

8 Some additional definitions and lemmas Def: A sentence P of SL is a TWA-sentence iff the only connectives of P are ~, &, and. (TLB: 231) Def: Sentence P is the dual of a TWA-sentence P of SL iff P results from P by: (i) replacing each occurrence of in P with & ; (ii) replacing each occurrence of & in P with ; (iii) adding ~ in front of each atomic component of P. (TLB: 231) Def: A sentence P of SL is a -sentence iff every connective occurring in P is an instance of. (Cross, F2015) Def: A sentence P of SL is an &-sentence iff every connective occurring in P is an instance of &. (Cross, F2015) Def: (Replacement products) (TLB: Ex. 6.1E:1e; Cross, F2015): Where Q is a sentential component of P, let [P] (R//Q) be defined as the result of replacing at least one occurrence of Q in P with R. 8

9 CL1: If (P 1 P 2 ) [(P 1 & P 2 )] is an -sentence (&-sentence) of SL, then so are P 1 and P 2. If Q is an atomic component of (P 1 P 2 ) [(P 1 & P 2 )], then Q is an atomic component of P 1 or an atomic component of P 2. (Cross, F2015) CL2: If P is a sentence of SL with no binary connectives, then no sentential component of P has binary connectives. (Cross, F2015) CL3: If P is a negation-free sentence of SL, then every sentential component of P is negation-free. (Cross, F2015) 6.1.1: If Q Q 1 and R R 1, then ~Q ~Q 1, (Q & R) (Q 1 & R 1 ), (Q R) (Q 1 R 1 ), (Q R) (Q 1 R 1 ), and (Q R) (Q 1 R 1 ). (TLB: solution to Ex. 6.1E:1e) 9

10 Some metatheoretic facts about SL established by SMI 6.1.2: In every sentence of SL the number of left parentheses equals the number of right parentheses. (TLB: 227ff) 6.1.3: Every TWA-sentence P is such that P and its dual have opposite truth-values on each TVA. (TLB: 231ff) 6.1.4: No negation-free sentence of SL is truth-functionally false. (TLB: Ex. 6.1E: 1a) 6.1.5: Any negation-free sentence P of SL is true on any TVA on which all the atomic components of P are true. (Cross, F2015) 6.1.6: Every sentence of SL that contains no binary connectives is truth-functionally indeterminate. (TLB: Ex. 6.1E:1b) 6.1.7: If A 1 and A 2 assign the same truth-values to all the atomic components of P, then P has the same truth-value on A 1 and A 2. (TLB: Ex. 6.1E:1c) 10

11 6.1.8: An iterated conjunction ( (P 1 & P 2 ) & P n ) of sentences of SL is true on a TVA iff P 1, P 2, P n are all true on that TVA. (TLB: Ex. 6.1E: 1d) 6.1.9: An iterated disjunction ( (P 1 P 2 ) P n ) of sentences of SL is true on a TVA iff at least one of P 1, P 2, P n is true on that TVA. (C. Cross, F2015) : If Q is a sentential component of P and Q 1 Q, then P [P] (Q 1 //Q). (TLB: Ex. 6.1E: 1e) : If P is an &-sentence and Q is the only atomic component of P, then {Q} P. (Cross 2015) : If P is a -sentence and Q is an atomic component of P, then {Q} P. (Cross 2015) 11

12 Exercises (a) Prove : If P is an &-sentence and Q is the only atomic component of P, then {Q} P. 12

13 (b) Prove 6.1.9: An iterated disjunction ( (P 1 P 2 ) P n ) of sentences of SL is true on a TVA iff at least one of P 1, P 2, P n is true on that TVA. (C. Cross, handout, F2015) 13

14 (c) Prove : Yuri Balashov, PHIL/LING 4510/6510 If Q is a sentential component of P and Q 1 Q, then P [P] (Q 1 //Q). (TLB: Ex. 6.1E: 1e) 14

Preliminary Matters. Semantics vs syntax (informally):

Preliminary Matters. Semantics vs syntax (informally): Preliminary Matters Semantics vs syntax (informally): Yuri Balashov, PHIL/LING 4510/6510 o Syntax is a (complete and systematic) description of the expressions of a language (e.g. French, PYTHON, or SL),