Logical Entailment A set of premises Δ logically entails a conclusion ϕ (Δ = ϕ) if and only if every interpretation that satisfies Δ also satisfies ϕ.
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1 Relational Proofs
2 Logical Entailment A set of premises Δ logically entails a conclusion ϕ (Δ = ϕ) if and only if every interpretation that satisfies Δ also satisfies ϕ.
3 Propositional Truth Assignments p q r With n constants, there are 2 n truth assignments.
4 Truth Assignments p(a) p(f(a)) p(f(f(a))) Infinitely many interpretations!
5 Good News Given a few restrictions (described later), we have good news. Good News: If Δ logically entails ϕ, then there is a finite proof of ϕ from Δ. And vice versa. More Good News: If Δ logically entails ϕ, it is possible to find such a proof in finite time.
6 Programme Proof System Examples, Examples, Examples
7 Logical Rules of Inference Negations ϕ ψ ϕ ϕ ψ ϕ ϕ Implications ϕ ψ ϕ ψ ϕ ϕ ψ ψ Biconditionals ϕ ψ ϕ ψ ψ ϕ ϕ ψ ϕ ψ ψ ϕ ϕ 1... ϕ n Conjunctions ϕ 1... ϕ n ϕ i Disjunctions ϕ 1... ϕ 1 ϕ 1... ϕ n ϕ 1... ϕ n ϕ 1... ϕ n ϕ 1 ψ... ϕ n ψ ψ
8 New Rules of Inference Universal Introduction Universal Elimination Existential Introduction Existential Elimination
9 Universal Introduction (UI) ϕ ν.ϕ where ν does not occur free in ϕ and an active assumption
10 Examples Premise: hates(jane,x) Conclusion: x.hates(jane,x) Premise: hates(jane,jill) Conclusion: x.hates(jane,jill) Premise: hates(jane,y) Conclusion: x.hates(jane,y)
11 Disallowed case p(x) q(x) x.q(x) Assumption UI
12 Allowed Case p(x) q(x) p(x) q(x) y.(p(x) q(x)) Assumption II UI
13 Universal Elimination (UE) ν.ϕ[ν] ϕ[τ ] where τ is substitutable for ν in ϕ
14 Premise: x.hates(jane,x) Examples Conclusions: hates(jane,jill) hates(jane,jane) x jill x jane hates(jane,x) x x hates(jane,y) x y
15 Premise: x. y.hates(x,y) More Examples Conclusions: y.hates(jane,y) x jane y.hates(y,y) x y Wrong!!
16 Capture An occurrence of a variable ν in ϕ captures a term τ if and only if τ contains a variable µ and the occurrence of ν lies in the scope of a quantifier of µ. The occurrence of y in x.hates(x,y) captures x. Why? The occurrence of y lies in scope of quantifier of x.
17 Substitutability A term τ is substitutable for ν in ϕ if and only if no free occurrence of ν captures τ. Some texts say τ is free for ν in ϕ instead of τ is substitutable for ν in ϕ. mother(jane) is free for y in hates(jane,y). mother(x) is free for y in hates(jane,y). mother(x) is free for y in z.hates(z,y). mother(x) is not free for y in x.hates(x,y). mother(x) is free for y in ( x. y.l(x,y) z.h(z,y)).
18 Universal Elimination (UE) ν.ϕ[ν] ϕ[τ ] where τ is substitutable for ν in ϕ
19 Existential Introduction (EI) ϕ[τ] ν.ϕ[ν] where τ does not occur in ϕ
20 Examples Premise: hates(jill,jill) Conclusions: x.hates(x,x) x.hates(jill,x) x.hates(x,jill) x. y.hates(x,y)
21 Existential Elimination (EE) ν.ϕ[ν] ν.(ϕ[ν] ψ) ψ
22 Example Premises: x.hates(jane,x) x.(hates(jane,x) nice(jane)) Conclusion: nice(jane)
23 Example - Lovers
24 Lovers Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill.
25 Proof Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill. 1. y. z.loves(y,z) Premise
26 Proof Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise
27 Proof Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise 3. z.loves(jill,z) UE:1
28 Proof Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise 3. z.loves(jill,z) UE:1 4. y. z.(loves(y,z) loves(jack,y)) UE:2
29 Proof Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise 3. z.loves(jill,z) UE:1 4. y. z.(loves(y,z) loves(jack,y)) UE:2 5. z.(loves(jill,z) loves(jack,jill)) UE:4
30 Proof Everybody loves somebody. Everybody loves a lover. Show that Jack loves Jill. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise 3. z.loves(jill,z) UE:1 4. y. z.(loves(y,z) loves(jack,y)) UE:2 5. z.(loves(jill,z) loves(jack,jill)) UE:4 6. loves(jack,jill) EE:3, 5
31 General Lovers Everybody loves somebody. Everybody loves a lover. Show that everybody loves everybody.
32 Proof Everybody loves somebody. Everybody loves a lover. Show that everybody loves everybody. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise
33 Proof Everybody loves somebody. Everybody loves a lover. Show that everybody loves everybody. 1. y. z.loves(y,z) Premise 2. x. y. z.(loves(y,z) loves(x,y)) Premise 3. z.loves(y,z) UE:1 4. y. z.(loves(y,z) loves(x,y)) UE:2 5. z.(loves(y,z) loves(x,y)) UE:4 6. loves(x,y) EE: 3, 5 7. y.loves(x,y) UI: 6 8. x. y.loves(x,y) UI: 7
34 Example - Harry and Ralph
35 Harry and Ralph Every horse can outrun every dog. Some greyhounds can outrun every rabbit. Harry is a horse. Ralph is a rabbit. Prove that Harry can outrun Ralph.
36 Harry and Ralph Every horse can outrun every dog. Some greyhound can outrun every rabbit. Harry is a horse. Ralph is a rabbit. Prove that Harry can outrun Ralph. 1. x. y.(h(x) d(x) f(x,y)) Premise 2. y.(g(y) z.(r(z) f(y,z))) Premise 3. x.(g(x) d(x)) Premise 4. x. y. z.(f(y,z) f(y,z) f(x,z)) Premise 5. h(harry) Premise 6. r(ralph) Premise
37 Harry and Ralph (continued) 7. g(y) z.(r(z) f(y,z)) Assumption 8. g(y) AE: 7 9. z.(r(z) f(y,z)) AE: 7
38 Harry and Ralph (continued) 7. g(y) z.(r(z) f(y,z)) Assumption 8. g(y) AE: 7 9. z.(r(z) f(y,z)) AE: r(ralph) f(y,ralph) UE: f(y,ralph) IE: 10, 6
39 Harry and Ralph (continued) 7. g(y) z.(r(z) f(y,z)) Assumption 8. g(y) AE: 7 9. z.(r(z) f(y,z)) AE: r(ralph) f(y,ralph) UE: f(y,ralph) IE: 10, g(y) d(y) UE: d(y) IE: 12, 8
40 Harry and Ralph (continued) 7. g(y) z.(r(z) f(y,z)) Assumption 8. g(y) AE: 7 9. z.(r(z) f(y,z)) AE: r(ralph) f(y,ralph) UE: f(y,ralph) IE: 10, g(y) d(y) UE: d(y) IE: 12, y.(h(harry) d(y) f(harry,y)) UE: h(harry) d(y) f(harry,y) UE: h(harry) d(y) AI: 5, f(harry,y) IE: 15, 16
41 Harry and Ralph (concluded) 18. y. z.(f(harry,y) f(y,z) f(harry,z)) UI: z.(f(harry,y) f(y,z) f(harry,z)) UI: f(harry,y) f(y,z) f(harry,z) UI: 19
42 Harry and Ralph (concluded) 18. y. z.(f(harry,y) f(y,z) f(harry,z)) UI: z.(f(harry,y) f(y,z) f(harry,z)) UI: f(harry,y) f(y,z) f(harry,z) UI: f(harry,y) f(y,ralph) AI: 17, f(harry,ralph) IE: 20, 21
43 Harry and Ralph (concluded) 18. y. z.(f(harry,y) f(y,z) f(harry,z)) UI: z.(f(harry,y) f(y,z) f(harry,z)) UI: f(harry,y) f(y,z) f(harry,z) UI: f(harry,y) f(y,ralph) AI: 17, f(harry,ralph) IE: 20, g(y) z.(r(z) f(y,z)) II: 7, 22 f(harry,ralph)) 24. y.(g(y) z.(r(z) f(y,z)) UI: 23 f(harry,ralph)) 25. f(harry,ralph) EE: 2, 24
44
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