COMP2411 Lecture 6: Soundness and Completeness. Reading: Huth and Ryan, Sections 1.4

Transcription

1 COMP2411 Lecture 6: Soundness and Completeness Reading: Huth and Ryan, Sections 14 Arithmetic is useful in the world because it is an example of the diagram: symbols - symbolic manipulation -> symbols relationship relationship v v world physical operation/relation > world 1

2 Notation: We write φ 1, φ k ψ when there exists a natural deduction proof with premises φ 1, φ k and conclusion ψ Note: this relation talks only about symbolic manipulation, not the meaning of the formulas But there is a connection to the relation φ 1, φ k = ψ We say is sound if for all formulas φ 1, φ k and conclusion ψ, if φ 1, φ k ψ then φ 1, φ k = ψ (Only valid conclusions can be derived) We say is complete if for all formulas φ 1, φ k and conclusion ψ, if φ 1, φ k = ψ then φ 1, φ k ψ (All valid conclusions can be derived) Theorem: is both sound and complete 2

3 This result can be understood as an example of a commutative diagram: φ 1,, φ n - natural deduction -> ψ truth truth v set of worlds satisfying φ 1,, φ n is a subset of > v set of worlds satisfying ψ The proof of this result uses mathematical induction: Let P be some property of the natural numbers 0, 1, 2, If 1 Base case: it can be shown that 0 has property P 2 Inductive Case: Assuming P is true for n, it can be shown that P is true for n + 1 (for any natural number n) then all natural numbers satisfy P 3

4 Proof of Soundness To prove soundness, we take P(K) = For all formulas φ 1,, φ n, ψ, for all natural deduction proofs for φ 1, φ n ψ containing K or fewer steps, we have φ 1, φ n = ψ The base case is trivial there are no proofs of 0 steps! Proof of Soundness: Inductive Step For the inductive step, we have to show that if P(K) then P(K+1), ie: Assuming (Inductive Hypothesis): For all formulas φ 1,, φ n, ψ, for all natural deduction proofs for φ 1, φ n ψ containing K or fewer steps, we have φ 1, φ n = ψ We have to prove For all formulas φ 1,, φ n, ψ, for all natural deduction proofs for φ 1, φ n ψ containing K + 1 or fewer steps, we have φ 1, φ n = ψ 4

5 So consider a proof for φ 1, φ n ψ of length K+1 (or less) For all but the last step, if α is the conlusion reached at that step, we get φ 1, φ n = α direct from the induction hypothesis Consider the last step Suppose the rule applied is where A 1 and A 2 are formulae A 1 A 2 It is enough to show that for all formulas φ 1,, φ n, if (1) φ 1,, φ n = A 1, and (2) φ 1,, φ n = A 2, then φ 1,, φ n = A (We know (1) and (2) by the induction hypothesis) We have to show this for every proof rule A Example: i A 1 A 2 A 1 A 2 Suppose (1) φ 1,, φ n = A 1, and (2) φ 1,, φ n = A 2 Then (1) On every line of the truth table where φ 1,, φ n are true, A 1 is true, and (2) On every line of the truth table where φ 1,, φ n are true, A 2 is true SO (by the truth table for ), on every line of the truth table where φ 1,, φ n are true, A 1 A 2 is true 5

6 Rules like e and e are a bit more complicated, we skip the details Completeness Proof For completeness, we need to show that if φ 1,, φ n = ψ then φ 1,, φ n ψ The proof has three steps: 1 If φ 1,, φ n = ψ then = φ 1 (φ 2 (φ n ψ)) 2 If = φ 1 (φ 2 (φ n ψ)) then φ 1 (φ 2 (φ n ψ)) 3 If φ 1 (φ 2 (φ n ψ)) then φ 1,, φ n ψ Step 1 is trivial from the truth table for 6

7 Completeness: Step 2 For step 2, we prove something slightly more general: Proposition 1: If = φ then φ To show this, we first prove: Proposition 2: Let p 1,, p n be the propositional atoms (constants) in the formula φ Consider a line l of the truth table of φ Define the formulae ˆp i by ˆp i is p i if p i is true on l, ˆp i is p i if p i is false on l Then If φ is true on l then ˆp 1,, ˆp n φ If φ is false on l then ˆp 1,, ˆp n φ The proof of this is another induction Example Consider φ = (p q) p: p q p q (p q) p F F F T F T T F T F T T T T T T We consider lines 1 and 2 7

8 On line 1, p and q are both false, so ˆp = p and ˆq = q We show p, q (p q), then use this to show p, q (p q) p On line 2, p is false, so ˆp = p and q is true, so ˆq = q We show p, q p q, then use this to show p, q ((p q) p) 1 : p Premise 2 : q Premise 3 : (p q) Assumption 4 : p Assumption 5 : 1, 4, e 6 : q Assumption 7 : 2, 6, e 8 : 3, 4 5, 6 7, e 9 : (p q) 3 8, i 10 : (p q) Assumption 11 : 9, 10, e 12 : p e 13 : (p q) p i 8

9 For line 2: 1 : p Premise 2 : q Premise 3 : p q 2, i 2 4 : (p q) p Assumption 5 : p 3, 4, e 6 : 1, 5, e 7 : ((p q) p) 4 5, i Suppose now that = φ, ie, φ is true on every line of the truth table Suppose φ has just two propositional constants p 1 and p 2 Then we can construct the following natural deduction proof: (Inside the inner boxes, we use Proposition 2 and the fact that φ is true on every line of the truth table Eg, the leftmost inner box corresponds to the fact that φ holds on the line of the truth table where p 1 and p 2 are both true, and we put the proof for p 1, p 2 φ there) 9

10 1 : p 1 p 1 LEM 2 : p 1 Asn 3 : p 2 p 2 LEM 4 : p 2 Asn x : φ x + 1 : φ x + 2 : φ 4 : p 2 Asn x : φ e 2 : p 1 Asn 3 : p 2 p 2 LEM 2 : p 2 Asn x : φ x + 1 : φ 2 : p 2 Asn x : φ e e Completeness: Step 3 Suppose φ 1 (φ 2 (φ n ψ)) with proof PP A proof of the following form establishes φ 1,, φ n ψ 1 : φ 1 Premise n : φ n Premise x : P P φ 1 (φ 2 (φ n ψ)) x + 1 : φ 2 (φ n ψ) 1, x, e x + n : ψ n, x + n 1, e 10