Logic, Resolution and Prolog
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1 Logic, Resolution and Prolog Michael Zakharyaschev Department of Computer Science and Information Systems Birkbeck, University of London homepage: course site:
2 1. First-Order Logic
3 First-order logic: syntax Exercise: represent Goldbach s Conjecture (1742) in first-order logic: Every even integer greater than 2 is the sum of two primes or more verbosely For every even x > 2, there exist primes y and z such that x = y + z Logic, Resolution and Prolog (1) 1
4 First-order logic: syntax Exercise: represent Goldbach s Conjecture (1742) in first-order logic: Every even integer greater than 2 is the sum of two primes or more verbosely For every even x > 2, there exist primes y and z such that x = y + z x ( Even(x) (x > 2) y z (Prime(y) Prime(z) (x = +(y, z)) ) Logic, Resolution and Prolog (1) 1
5 First-order logic: syntax Exercise: represent Goldbach s Conjecture (1742) in first-order logic: Every even integer greater than 2 is the sum of two primes or more verbosely For every even x > 2, there exist primes y and z such that x = y + z x ( Even(x) (x > 2) y z (Prime(y) Prime(z) (x = +(y, z)) ) x ( Prime(x) (x > 1) y z ((y > 1) (z > 1) (x = (y, z))) ) Logic, Resolution and Prolog (1) 1
6 First-order logic: syntax Exercise: represent Goldbach s Conjecture (1742) in first-order logic: Every even integer greater than 2 is the sum of two primes or more verbosely For every even x > 2, there exist primes y and z such that x = y + z x ( Even(x) (x > 2) y z (Prime(y) Prime(z) (x = +(y, z)) ) x ( Prime(x) (x > 1) y z ((y > 1) (z > 1) (x = (y, z))) ) x ( Even(x) y (x = (y, 2)) ) Do these sentences hold true in the standard model of arithmetic? Logic, Resolution and Prolog (1) 1
7 First-order logic: syntax Exercise: represent Goldbach s Conjecture (1742) in first-order logic: Every even integer greater than 2 is the sum of two primes or more verbosely For every even x > 2, there exist primes y and z such that x = y + z x ( Even(x) (x > 2) y z (Prime(y) Prime(z) (x = +(y, z)) ) x ( Prime(x) (x > 1) y z ((y > 1) (z > 1) (x = (y, z))) ) x ( Even(x) y (x = (y, 2)) ) Do these sentences hold true in the standard model of arithmetic? Logic, Resolution and Prolog (1) 1
8 First-order logic: semantics Exercise: is the sentence ϕ = x ( Square(x) White(x) y (Ball(y) Black(y) Under(x, y)) ) true in the following interpretations? Interpretation I 1 : 7 7 Logic, Resolution and Prolog (1) 2
9 First-order logic: semantics Exercise: is the sentence ϕ = x ( Square(x) White(x) y (Ball(y) Black(y) Under(x, y)) ) true in the following interpretations? Interpretation I 1 : 7 7 I 1 = ϕ Logic, Resolution and Prolog (1) 2
10 First-order logic: semantics Exercise: is the sentence ϕ = x ( Square(x) White(x) y (Ball(y) Black(y) Under(x, y)) ) true in the following interpretations? Interpretation I 1 : 7 7 I 1 = ϕ 7 Interpretation I 2 : Logic, Resolution and Prolog (1) 2
11 First-order logic: semantics Exercise: is the sentence ϕ = x ( Square(x) White(x) y (Ball(y) Black(y) Under(x, y)) ) true in the following interpretations? Interpretation I 1 : 7 7 I 1 = ϕ 7 Interpretation I 2 : I 2 = ϕ Logic, Resolution and Prolog (1) 2
12 Satisfiability and validity Exercise: which of the following formulas are satisfiable? 1. P (x 1 ) P (x 2 ) which of them are valid or contradictions? Logic, Resolution and Prolog (1) 3
13 Satisfiability and validity Exercise: which of the following formulas are satisfiable? which of them are valid or contradictions? 1. P (x 1 ) P (x 2 ) 2. xp (x) xp (x) Logic, Resolution and Prolog (1) 3
14 Satisfiability and validity Exercise: which of the following formulas are satisfiable? which of them are valid or contradictions? 1. P (x 1 ) P (x 2 ) 2. xp (x) xp (x) 3. xp (x) xp (x) Logic, Resolution and Prolog (1) 3
15 Satisfiability and validity Exercise: which of the following formulas are satisfiable? which of them are valid or contradictions? 1. P (x 1 ) P (x 2 ) 2. xp (x) xp (x) 3. xp (x) xp (x) 4. x (P (x) P (x)) Logic, Resolution and Prolog (1) 3
16 Satisfiability and validity Exercise: which of the following formulas are satisfiable? which of them are valid or contradictions? 1. P (x 1 ) P (x 2 ) 2. xp (x) xp (x) 3. xp (x) xp (x) 4. x (P (x) P (x)) 5. x (P (x) P (x)) Logic, Resolution and Prolog (1) 3
17 Logical consequence Exercise: Suppose that (1) Dasha likes Sasha; (2) Sasha likes beer; (3) Pasha likes beer and all those who like what Pasha likes Is there anyone who likes Dasha? Logic, Resolution and Prolog (1) 4
18 Logical consequence Exercise: Suppose that (1) Dasha likes Sasha; (2) Sasha likes beer; (3) Pasha likes beer and all those who like what Pasha likes Is there anyone who likes Dasha? First-order formalisation: is it the case that every interpretation validating the sentences ϕ 1 = Likes(d, s) ϕ 2 = Likes(s, b) ϕ 3 = Likes(p, b) x ( y (Likes(p, y) Likes(x, y)) Likes(p, x) ) also validates the sentence ϕ = z Likes(z, d) Is it the case that {ϕ 1, ϕ 2, ϕ 3 } = ϕ or, which is the same, = ϕ 1 ϕ 2 ϕ 3 ϕ Logic, Resolution and Prolog (1) 4
19 Resolution method for checking validity Exercise: is the following sentence valid? ϕ = x y P (x, y) y x P (x, y) Logic, Resolution and Prolog (1) 5
20 = x y P (x, y) y x P (x, y)? Logic, Resolution and Prolog (1) 6
21 Resolution method for checking validity Exercise: is the following sentence valid? ϕ = x( (P (x) ( xp (x) yr(x, y))) yr(x, y) ) Step 1 ϕ ϕ 0 = ϕ we have: ϕ is valid iff ϕ 0 is not satisfiable (contradiction) ϕ 0 = x( (P (x) ( xp (x) yr(x, y))) yr(x, y) ) Step 2 ϕ 0 ϕ 1 = Q 1 x 1... Q n x n (D 1 D N ) prenex normal form (with matrix in CNF) ϕ 1 = x 1 x 2 y 1 y 2 (P (x 1 ) ( P (x 2 ) R(x 1, y 1 )) R(x 1, y 2 ) ) Step 3 ϕ 1 ϕ 2 = x i1... x ik (D 1 D N ) Skolem normal form ϕ 2 = x 1 y 2 (P (x 1 ) ( P (f(x 1 )) R(x 1, g(x 1 ))) R(x 1, y 2 ) ) ϕ 0 is satisfiable iff D 1 = x 1 P (x 1 ), D 2 = x 1 ( P (f(x 1 )) R(x 1, g(x 1 ))), D 3 = x 1 y 2 R(x 1, y 2 ) Is {D 1, D 2, D 3 } satisfiable? are satisfiable (in the same interpretation) Logic, Resolution and Prolog (1) 7
22 = x((p (x) ( xp (x) yr(x, y))) yr(x, y))? If D 1 = x 1 P (x 1 ) is true in an interpretation I, then P (f(a)) is true (for any a) If D 2 = x 1 ( P (f(x 1 )) R(x 1, g(x 1 ))) is true in I, then But then R(a, g(a)) is true in I P (f(a)) R(a, g(a)) is also true in I If D 3 = x 1 y 2 R(x 1, y 2 ) is true in I, then so is R(a, g(a)) which is a contradiction Thus, ϕ 0 is not satisfiable, and so ϕ is valid Logic, Resolution and Prolog (1) 8
Logic, Resolution and Prolog
Logic, Resolution and Prolog Michael Zakharyaschev Department of Computer Science and Information Systems Birkbeck, University of London email: zmishaz@gmail.com homepage: http://www.dcs.bbk.ac.uk/~michael
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