Overview. I Review of natural deduction. I Soundness and completeness. I Semantics of propositional formulas. I Soundness proof. I Completeness proof.

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1 Overview I Review of natural deduction. I Soundness and completeness. I Semantics of propositional formulas. I Soundness proof. I Completeness proof.

2 Propositional formulas Grammar: ::= p j (:) j ( ^ ) j ( _ ) j (! ) Precedence rules: : > _; ^ >! Example: p ^ (q _ :r! q) (p ^ ((q _ (:r))! q))

3 Natural deduction 1 ; : : : ; n ` Formula can be proved from formulas 1 ; : : : ; n. Rules for reasoning about formulas. I Introduction rules: Connective in the conclusion. Construct a complex formula..!!-i I Elimination rules: Connective in a premise. Extract information from proved formulas. xxx!!-e

4 Basic rules ^ Introduction xxx ^ ^-i ^ Elimination ^-e 1 ^ ^-e2 _-i 1 -i 2 _ xxx... xxx... _-e!...!-i xxx!!-e! :...? : :-i xxx:? :-e? no rule??-e :: :: ::-e

5 Proof p! q; r! a ` (p _ r)! (q _ s) 1: p _ r assumption 2: p assumption 3: p! q premise 4: q!-e : 2; 3 5: q _ s _-i 1 : 4 6: r assumption 7: r! s premise 8: s!-e : 6; 7 9: q _ s _-i 2 : 8 10: q _ s _-e : 2 9

6 Proof p! q; r! a ` (p _ r)! (q _ s) 1: p _ r assumption 2: p assumption 3: p! q premise 4: q!-e : 2; 3 5: q _ s _-i 1 : 4 6: r assumption 7: r! s premise 8: s!-e : 6; 7 9: q _ s _-i 2 : 8 10: q _ s _-e : 2 9

7 Reasoning vs. reality Reasoning: argument based on observations and derivational rules. Reality: reality. How does reasoning relate to reality? I Soundness: reasoning derives only true statements. Trivial solution: reasoning that derives nothing. I Completeness: reasoning derives all true statements. Trivial solution: reasoning that derives everything.

8 Semantics What is a true formula? A formula is either true T or false F. The meaning depends on the meaning of the subterms. Examples: I [[p]] = T and [[q]] = F implies [[p ^ q]] = F I [[p]] = T implies [[:p]] = F I [[p]] = T and [[q]] = F implies [[p ^ :q]] = T

9 Truth tables The value of a formula for all possible inputs. Basic connectives: > T xxx? F xxx : F T T F xxx ^ T T T T F F F T F F F F _ T T T T F T F T T F F F xxx xxx! T T T T F F F T T F F T Other tables dene other functions.

10 Meaning of a formula Begin with the meanings of atoms. Compute the value bottom-up. Example: (p! q) ^ (p! r) p q r p! q p! r (p! q) ^ (p! r) T T T T T F T F T T F F F T T F T F F F T F F F I Tautology: Always true. I Contradiction: always false.

11 Semantic entailment true when 1 ; : : : n are true. 1 ; : : : n j= 1. Compute truth tables of the 1 ; : : : n. 2. Collect lines where 1 ; : : : n are all true. 3. Evaluate in these cases.

12 Example q ^ r j= (p! q) ^ (p! r) p q r q ^ r p! q p! r (p! q) ^ (p! r) T T T T T T F F T F T F T F F F F T T T F T F F F F T F F F F F Thus, q ^ r j= (p! q) ^ (p! r)

13 Example q ^ r j= (p! q) ^ (p! r) p q r q ^ r p! q p! r (p! q) ^ (p! r) T T T T T T T T T F F T F T F T F F F F T T T T T T F T F F F F T F F F F F Thus, q ^ r j= (p! q) ^ (p! r)

14 Soundness and completeness Relate provability to semantic entailment. I Soundness: 1 ; : : : ; n ` implies that 1 ; : : : ; n j=. I Completeness: 1 ; : : : ; n j= implies that 1 ; : : : ; n `.

15 Soundness Theorem: Let 1 ; : : : ; n and be propositional logic formulas. Then, if 1 ; : : : ; n `, then 1 ; : : : ; n j=. Proof idea: each proof step is justied by truth tables.

16 Proof: inductive denition I : premise is a proof of `, where 2. I Let: i prove ` i j prove ; j ` j For any rule R: 1 : : : m xxx 1. 1 : : :. n n Then 1 ; : : : ; m ; 1 ; : : : ; n ; R proves `.

17 Base case Given: ` Show: j= Proof of ` : : premise Show: j= Evaluate for truth table lines where are all T. is a premise, so 2. Thus is T.

18 Induction case Given: ` Show: j= ` must have a proof 1 ; : : : ; m ; 1 ; : : : ; n ; R, for R: 1 : : : m xxx 1. 1 : : :. n n Induction hypothesis: I i proves ` i implies j= i. I j proves ; j ` j implies ; j j= j : Proceed by cases on the possible rules.

19 Rule: ^-i Given: ` ^, show: j= ^. Proof of ` ^ : 1 ; 2 ; ^-i xxx ^ ^-i 1 proves `, 2 proves ` By induction: j=, j= Evaluate ^ when [[]] = [[ ]] = T: Thus, j= ^. ^ T T T

20 Rule: ^-i Given: ` ^, show: j= ^. Proof of ` ^ : 1 ; 2 ; ^-i xxx ^ ^-i 1 proves `, 2 proves ` By induction: j=, j= Evaluate ^ when [[]] = [[ ]] = T: Thus, j= ^. ^ T T T

21 Rule: ^-i Given: ` ^, show: j= ^. Proof of ` ^ : 1 ; 2 ; ^-i xxx ^ ^-i 1 proves `, 2 proves ` By induction: j=, j= Evaluate ^ when [[]] = [[ ]] = T: Thus, j= ^. ^ T T T

22 Rule: :-e Given: `?, show: j=?. Proof of `?: 1 ; 2 ; :-e xxx:? :-e 1 proves `, 2 proves ` : By induction: j=, j= : Truth table for : : F T T F never all T. Thus, trivially: j=? NB: Truth table for?:? F

23 Rule: :-e Given: `?, show: j=?. Proof of `?: 1 ; 2 ; :-e xxx:? :-e 1 proves `, 2 proves ` : By induction: j=, j= : Truth table for : : F T T F never all T. Thus, trivially: j=? NB: Truth table for?:? F

24 Rule: :-e Given: `?, show: j=?. Proof of `?: 1 ; 2 ; :-e xxx:? :-e 1 proves `, 2 proves ` : By induction: j=, j= : Truth table for : : F T T F never all T. Thus, trivially: j=? NB: Truth table for?:? F

25 Rule:?-e Given: `, show: j=. Proof of ` : ; :-e??-e proves `?. By induction: j=?. Truth table for?: never all T.? F Thus, trivially: j=

26 Rule:?-e Given: `, show: j=. Proof of ` : ; :-e??-e proves `?. By induction: j=?. Truth table for?: never all T.? F Thus, trivially: j=

27 Rule:?-e Given: `, show: j=. Proof of ` : ; :-e??-e proves `?. By induction: j=?. Truth table for?: never all T.? F Thus, trivially: j=

28 Rule!-i Given: `!, show: j=!. Proof of `! : ;!-i xxx.!-i proves ; `! By induction: ; j=. Show: when all true, so is!.! T T T T F F xxx! F T T F F T Potential problem when [[]] = T and [[ ]] = F. But, by induction, if all true and [[]] = T, then [[ ]] = T. Thus, j=!.

29 Rule!-i Given: `!, show: j=!. Proof of `! : ;!-i xxx.!-i proves ; `! By induction: ; j=. Show: when all true, so is!.! T T T T F F xxx! F T T F F T Potential problem when [[]] = T and [[ ]] = F. But, by induction, if all true and [[]] = T, then [[ ]] = T. Thus, j=!.

30 Rule!-i Given: `!, show: j=!. Proof of `! : ;!-i xxx.!-i proves ; `! By induction: ; j=. Show: when all true, so is!.! T T T T F F xxx! F T T F F T Potential problem when [[]] = T and [[ ]] = F. But, by induction, if all true and [[]] = T, then [[ ]] = T. Thus, j=!.

31 Completeness Theorem: Let 1 ; : : : ; n and be propositional logic formulas. Then, if 1 ; : : : ; n j=, then 1 ; : : : ; n `. Proof idea: construct a proof from a truth table.

32 Completeness proof structure 1. Eliminate premises: 1 ; 2 ; : : : ; n j= implies j= 1! ( 2! : : :! ( n! )). 2. Show provability: j= 1! ( 2! : : :! ( n! )) implies ` 1! ( 2! : : :! ( n! )). 3. Reintroduce premises: ` 1! ( 2! : : :! ( n! )) implies 1 ; 2 ; : : : ; n `.

33 Eliminating premises Theorem: If 1 ; 2 ; : : : ; n j=, then j= 1! ( 2! : : :! ( n! )). Proof: By induction on n. Base case: n = 0. Clearly j= implies j=. Induction case: We showed ; j= implies j=!. Thus, 1 ; 2 ; : : : ; n j= implies 1 ; : : : n 1 j= n!. By induction, j= 1! ( 2! : : :! ( n! ))

34 Showing provability Theorem: I Let be a formula such that p1; p2; : : : ; p n are its only propositional atoms. I Let ` be any line in 's truth table. I For any atom or formula, let `() be if the truth table entry in line ` for is T, and : if the truth table entry for is F. Then, `(p1); `(p2); : : : ; `(p n ) ` `() is provable.

35 Example (p ^ r)! (q ^ s) A truth table line: p q r s p ^ r q ^ s (p ^ r)! (q ^ s) T F T T T F F Constructed sequent: p; :q; r ; s ` :((p ^ r)! (q ^ s)) Another truth table line: p q r s p ^ r q ^ s (p ^ r)! (q ^ s) F T F T F T T Constructed sequent: :p; q; :r ; s ` (p ^ r)! (q ^ s)

36 Proof Induction on the structure of. Base case: p. Truth table: p T F p T F `(p) ` `(p) is p ` p, or :p ` :p: I Proof of p ` p: 1: p premise 2: p 1 I Proof of :p ` :p: 1: :p premise 2: :p 1 Thus, `(p) ` `(p).

37 Negation : Possible truth table lines: F T : T F If [[ ]] = F and [[: ]] = T: xxxshow `(p1); `(p2); : : : ; `(p n ) ` : By induction, since [[ ]] = F: `(p1); `(p2); : : : ; `(p n ) ` :

38 Negation, continued If [[ ]] = T and [[: ]] = F: xxxshow `(p1); `(p2); : : : ; `(p n ) ` :: By induction, since [[ ]] = T: `(p1); `(p2); : : : ; `(p n ) ` To prove `(p1); `(p2); : : : ; `(p n ) ` :: : First prove `(p1); `(p2); : : : ; `(p n ) `. Use :-i.

39 Implication 1! 2 Proof strategy: I Consider possible truth values of 1! 2. I Use induction to nd proofs of `( 1 ) and `( 2 ). I Construct a proof of `( 1! 2 ). Four lines, so four cases ! 2 T T T T F F F T T F F T

40 Use of the induction hypothesis Consider 1, 2 : 1 2 1! 2 T T T T F F F T T F F T I Propositional atoms of 1 : a1; : : : ; a x. I Propositional atoms of 2 : b1; : : : ; b y. Atoms of 1! 2 = fa1; : : : ; a x g [ fb1; : : : ; b y g. Thus, a truth table line for 1! 2 also gives meaning to 1 and 2.

41 Using the induction hypothesis For a truth table line ` for 1! 2, by induction: `(a1); : : : ; `(a x ) ` `( 1 ) `(b1); : : : ; `(b y ) ` `( 2 ) Our goal: `(a1); : : : ; `(a x ); `(b1); : : : ; `(b y ) ` `( 1! 2 ) Proof structure: I Proof of `(a1); : : : ; `(a x ) ` `( 1 ). I Proof of `(b1); : : : ; `(b y ) ` `( 2 ). I `( 1 ) ^ `( 2 ) : ^-i I Derive `( 1! 2 ) from `( 1 ) ^ `( 2 ).

42 Cases I 1 is T, 2 is T, and is T: Show 1 ^ 2 implies 1! 2. I 1 is T, 2 is F, and is F: Show 1 ^ : 2 implies :( 1! 2 ). I 1 is F, 2 is T, and is T: Show : 1 ^ 2 implies 1! 2. I 1 is F, 2 is F, and is T: Show : 1 ^ : 2 implies 1! 2. Conjunction and disjunction similar.

43 Combining sequents We have a \complete" collection of sequents: `(p1); `(p2); : : : ; `(p n ) ` `() Note: j= implies 8`:`() =. We want to prove: ` Proof idea: Use the Law of the Excluded Middle.

44 Example (p ^ q)! q Constructed sequents: p; q ` (p ^ q)! q :p; q ` (p ^ q)! q p; :q ` (p ^ q)! q :p; :q ` (p ^ q)! q Consider p; q ` (p ^ q)! q and p; :q ` (p ^ q)! q. Proof of p ` (p ^ q)! q: 1: q _ :q LEM 2: q assumption 3 proof of p; q ` (p ^ q)! q 4: :q assumption 5 proof of p; :q ` (p ^ q)! q 6: (p ^ q)! q _-e : 1 5

45 Proof idea Goal: Reduce `(p1); `(p2); : : : ; `(p n ) ` to `. Proof by induction on n. Find pairs of sequents that only dier in `(p n ). Use LEM to prove a single sequent without `(p n ) in the premise.

46 Reintroducing premises Theorem: If ` 1! ( 2! : : :! ( n! )), then 1 ; 2 ; : : : ; n `. Proof idea: Convert ` 1! ( 2! : : :! ( n! )) to 1 ` 2! : : :! ( n! ), and proceed by induction. Problem: Theorem not general enough. Restatement: If ` 1! ( 2! : : :! ( n! )), then ; 1 ; 2 ; : : : ; n `.

47 Proof Given a proof of: ` 1! ( 2! : : :! ( n! )) Prove: ; 1 ` 2! : : :! ( n! ) I Proof of ` 1! ( 2! : : :! ( n! )) I 1 : premise I 2! : : :! ( n! ) :!-e By induction, ; 1 ; 2 ; : : : ; n `. Take = ; to prove the original theorem.

48 Summary Soundness: Show that the deduction rules are justied by truth tables. Completeness: 1. Eliminate premises: 1 ; 2 ; : : : ; n j= implies j= 1! ( 2! : : :! ( n! )). 2. Show provability: j= 1! ( 2! : : :! ( n! )) implies ` 1! ( 2! : : :! ( n! )). 3. Reintroduce premises: ` 1! ( 2! : : :! ( n! )) implies 1 ; 2 ; : : : ; n `. Conclusion: can freely use either natural deduction or truth tables.

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