Automatic Reasoning: Resolution for Propositional and First-Order Logic

1. Section 9.1 Automatic Reasoning Recall that a wff W is valid iff ¬ W is unsatisfiable. Resolution is an inference rule used to prove unsatisfiability in a “mechanical” way. Resolution for Propositional Logic The resolution rule for propositions is shown to the right, where A and B are disjunctions of literals. Notation:A – p denotes A with all occurrences of p removed and B – ¬ p denotes B with all occurrences of ¬ p removed. Notation: We denote false by a box []. In applying resolution, we assume A B  B  A and A []  A. If A and B are false, then the rule is simplified as shown. To prove a wff is valid: negate the wff and convert it to CNF; write down the fundamental disjunctions as premises; use resolution to find a false statement. Example. Prove (pq)  (qr)  (pr) is a tautology. Solution. Negate the wff and transform it into CNF: (¬ pq)  (¬ qr) p ¬ r. Now write down the fundamental disjunctions as premises to start the proof. 1. ¬ pq P 2. ¬ qr P 3. p P 4. ¬ r P 5. q 1, 3, R 6. r 2, 5, R 7. []4, 6, R QED.

2. Example/Quiz. Use resolution to prove that the following wff (an application of constructive dilemma) is a tautology. (AB)  (AC  D)  (BE  F)  (C  D)  (E  F). Solution. Negate the wff and transform the result into CNF. (AB)  (¬ AC ) (¬ AD)  (¬ BE) (¬ BF) (¬ C  ¬ D)  (¬ E  ¬ F). 1. AB P 2. ¬ AC P 3. ¬ AD P 4. ¬ BE P 5. ¬ BF P 6. ¬ C  ¬ D P 7. ¬ E  ¬ F P 8. ¬ A ¬ C 3, 6, R 9. ¬ B ¬ F 4, 7, R 10. ¬ B ¬ B 5, 9, R 11. ¬ A ¬ A 2, 8, R 12. B 1, 11, R 13. [] 10, 12, R QED. Question. Are there proofs that use less than 13 steps. Answer: Probably not since each of the 7 premises must be used. Why?

3. Resolution for First-Order Logic Example (to Show the Idea). Suppose some proof contains the following two lines, where we assume that the wffs (called clauses) are universally closed. 1. p(a, x) q(x, y, a) P 2. ¬ p(x, b) r(x, y, b) P To apply resolution to the wffs on lines 1 and 2 several actions need to be taken: Change the names so the clauses have distinct variable names. In this case, we’ll keep line 1 as it is and change the names on line 2. 3. ¬ p(v, b) r(v, w, b) P Unify (i.e., match) p(a, x) on line 1 and p(v, b) on line 3 to get a substitution (i.e., a set of bindings) = {v/a, x/b}. Apply  to clauses on lines 1 and 3. p(a, b) q(b, y, a) ¬ p(a, b) r(a, w, b) Apply resolution to the transformed wffs. 4. q(b, y, a) r(a, w, b)1, 3, R, = {v/a, x/b} So we need to discuss three things for first-order resolution. The clausal form of wffs Unification of atoms The inference rule.

4. Clausal Forms A clause is a disjunction of literals. (Recall a literal is an atom or its negation.) A clausalform is the universal closure of a conjunction of clauses. Notation. A clausal form is also represented as a set of its clauses. Example. The clausal form x y(p(x)  (q(x, y)  ¬ r(x))  (¬ p(x) s(y))) Is represented by the set {p(x), q(x, y)  ¬ r(x), ¬ p(x) s(y)}. Not every wff is equivalent to a clausal form. For example, the wff x p(x) is not equivalent to a clausal form. But we have the following result of Skolem, which is sufficient for resolution. Skolem’s Algorithm Every wff can be associated with a clausal form in which the two are either both satisfiable or both unsatisfiable. Construct the prenex CNF for the wff. Replace all occurrences of each free variable by a new constant. Eliminate the existential quantifiers by Skolem’s Rule: A. If x W(x) is not inside the scope of a universal quantifier, then replace x W(x) by W(c) for a new constant c. B. If x W(x) is inside the scope of x1 ...xn, then replace x W(x) by W(ƒ(x1, …, xn)) for a new function symbol ƒ.

5. Quiz. Find a clausal form for each wff. 1. x A(x). 2. x y B(x, y). 3. x y z C(x, y, z). 4. x y z D(x, y, z). Answers. 1. A(c). 2. xB(x, ƒ(x)). 3. x yC(x, y, g(x, y)). 4. xD(x, ƒ(x), g(x)). Example/Quiz. Find a clausal form for the wff y xp(x, y)  z (q(x)  r(z)). Solution: Put the wff in prenex CNF. y xp(x, y)  z (q(w)  r(z)) (renamed)  ¬ y xp(x, y)  z (q(w)  r(z)) (removed )  y x ¬ p(x, y)  z (q(w)  r(z)) (moved ¬ inside)  y (x ¬ p(x, y)  z (q(w)  r(z))) (moved y outside)  y x z (¬ p(x, y) (q(w)  r(z))) (moved x and z outside)  y x z ((¬ p(x, y)  q(w))  (¬ p(x, y)  r(z))) (constructed CNF) (replace free variable w by constant c) y x z ((¬ p(x, y)  q(c))  (¬ p(x, y)  r(z))) Apply Skolem’s Rule to eliminate x and z. y ((¬ p(ƒ(y), y)  q(c))  (¬ p(ƒ(y), y)  r(g(y))). So there are two clauses in the clausal form, which can be denoted by the set {¬ p(ƒ(y), y)  q(c), ¬ p(ƒ(y), y)  r(g(y)}.

6. Substitutions and Unification A binding of a variable x to a term t is denoted x/t and it means replace x by t. Applying a Binding to an Expression If x/t is a binding and E is an expression, then E(x/t) denotes the expression obtained from E by replacing all free occurrences of x with t. Examples. p(x, y, z)(x/y) = p(y, y, z) p(x, y, z)(y/ƒ(z)) = p(x, ƒ(z), z) A substitution is a finite set of bindings with distinct numerators. (Use lowercase greek letters for substitutions.) Example. q = {x/y, y/ƒ(z)} and s = {y/ƒ(a), z/b} are substitutions. Applying a Substitution to an Expression or a Set of Expressions If q is a substitution and E is an expression, then Eq denotes the expression obtained from E by simultaneously applying the bindings in q to E. Example. If q = {x/y, y/ƒ(z)}, then p(x, y, z)q = p(x, y, z){x/y, y/ƒ(z)} = p(y, ƒ(z), z). If S is a set of expressions, then Sq is the set of expresisons obtained from S by applying q to each expression of S. Example. If q = {x/y, y/ƒ(z)} and S = {p(x, y), q(y, g(z))}, then Sq = {p(x, y)q, q(y, g(z))q}= {p(y, ƒ(z)), q(ƒ(z), g(z))}.

7. Composing Substitutions If q and s are two substitutions, then the compositionqs is applied to an expression E by E(qs) = (Eq)s. We can calculate qsby applying it to an atom that contains all the numerator variables of the two substitutions. Then collect the bindings that result from the application. Example. Let q = {x/y, y/ƒ(z)} and s = {y/ƒ(a), z/b}. Since the numerator variables are x, y, and z, we calculate p(x, y, z)(qs) = (p(x, y, z)q)s = p(y, ƒ(z), z))s = p(ƒ(a), ƒ(b), b). So qs = {x/ƒ(a),y/ƒ(b), z/b}. We can also calculate qs by the following procedure, where q = {xi/ti} and s = {yi/si}. Construct {xi/(tis)}. Delete any xi/xifrom Step 1. Delete yi/si from s if yi is a numerator of q. 4. qs is the union of the substitutions from Steps 2 and 3. Example. Let q = {x/y, y/ƒ(z)} and s = {y/ƒ(a), z/b}. Calculate qs. Step 1. Construct {x/ys, y/ƒ(z)s} = {x/ƒ(a), y/ƒ(b)}. Step 2. No deletions. So no changes from Step 1. Step 3. Delete y/ƒ(a) from s (because y is a numerator of q) to obtain {z/b}. Step 4. The union of sets from Steps 2 and 3 give qs = {x/ƒ(a), y/ƒ(b), z/b}. Properties of Composition q(sg) = (qs)g and qe =eq =q where e ={}.

8. A unifier of a set S of expressions is a substitution q such that Sq is a singleton set. Example. {x/a} is a unifier of {p(x), p(a)} because {p(x), p(a)}{x/a} = {p(a)}. Example/Quiz. What are some unifiers of S = {p(x, a), p(y, z)}? Answer: Since a is a constant, any unifier must include the binding z/a. A unifier might also include x/y or y/x. So two unifiers of S are {x/y, z/a} and {y/x, z/a} because S{x/y, z/a} = {p(y, a)} and S{y/x, z/a} = {p(x, a)}. Notice also that {x/t, y/t, z/a} is a unifier of S for any term t because S{x/t, y/t, z/a} = {p(t, a)}. A most general unifier (mgu) of a set S of expressions is a unifier of q of S such that any other unifier s of S can be written as s =qa for some substitution a. Example. Let S = {p(x, a), p(y, z)} from the preceding example. The unifiers of S are {x/y, z/a} and {y/x, z/a} and {x/t, y/t, z/a} for any term t. The unifier {x/y, z/a} is an mgu for S because {y/x, z/a} = {x/y, z/a}{y/x} and {x/t, y/t, z/a} = {x/y, z/a}{y/t}. Quiz. Continue the example and show that, {y/x, z/a} is an mgu of S. Proof: {x/y, z/a} = {y/x, z/a}{x/y} and {x/t, y/t, z/a} = {y/x, z/a}{x/t}. QED. Quiz. Continue the example and show for a constant c that {x/c, y/c, z/a} is not an mgu of S. Proof: Assume, BWOC, that {x/c, y/c, z/a} is an mgu. Then {x/y, z/a} = {x/c, y/c, z/a}a for some substitution a. But {x/c, y/c, z/a}a contains the subset {x/c, y/c, z/a}, so it can’t equal {x/y, z/a}. This contradiction tells us that {x/c, y/c, z/a} is not an mgu. QED.

9. Unification Algorithms Robinson: For a finite set S of atoms find whether S has an mgu. 1. k := 0; q0:= e; go to Step 2. 2. If Sqk is a singleton then stop with mgu qk Otherwise construct Dk (set of terms in leftmost position of disagreement); go to Step 3. 3. If Dk has a variable v and a term t such that v does not occur in t then qk+1 := qk{v/t}; k := k + 1; go to Step 2. Otherwise stop (S is not unifiable). Example. Trace the algorithm for S = {p(x, h(x, g(y)), y), p(x, h(a, z), b)}. 1. k := 0; q0:= e. 2. Sq0 = Se = {p(x, h(x, g(y)), y), p(x, h(a, z), b)} is not a singleton; D0 = {x, a}. 3. q1 := q0{x/a} = {x/a}; k := 1. 2. Sq1 = {p(a, h(a, g(y)), y), p(a, h(a, z), b)} is not a singleton; D1 = {g(y), z}. 3. q2 := q1{z/g(y)} = {x/a, z/g(y)}; k := 2. 2. Sq2 = {p(a, h(a, g(y)), y), p(a, h(a, g(y)), b)} is not a singleton; D2 = {y, b}. 3. q3 := q2{y/b} = {x/a, z/g(b), y/b}; k := 3. 2. Sq3 = {p(a, h(a, g(b)), b)} is a singleton; stop with mgu q3 = {x/a, z/g(b), y/b}. Quiz. Apply the algorithm to S = {p(x), p(ƒ(x))}. Answer. S is not unifiable because in the set D0 = {x, ƒ(x))} xoccurs in ƒ(x).

10. Martelli-Montanari: To see whether two atoms A and B have an mgu, form the set {A = B} and apply the following rules repeatedly in any order. (ƒ and g are either function or predicate symbols.) 1.Replaceƒ(s1, …, sn) = ƒ(t1, …, tn)with equations s1 = t1, …, sn = tn. 2. If ƒ(s1, …, sn) = g(t1, …, tm)with ƒ ≠ g or m ≠ n, then stop with failure. 3. Delete x = x. 4. Replace t = x, where t is not a variable, with x = t. 5. If x = t and x does not occur in t, apply {x/t} to the other equations that contain x. 6. If x = t and t is not a variable and x occurs in t, then stop with failure. If there is no failure and the rules cannot be applied, then transform {xi = ti} to mgu {xi/ti}. Example. Trace the algorithm for A = p(x, h(x, g(y)), y) and B = p(x, h(a, z), b). {p(x, h(x, g(y)), y) = p(x, h(a, z), b)} ({A = B}) {x = x, h(x, g(y)) = h(a, z), y = b} (Rule 1) {h(x, g(y)) = h(a, z), y = b} (Rule 3) {x = a, g(y) = z, y = b} (Rule 1) {x = a, z = g(y), y = b} (Rule 4) {x = a, z = g(b), y = b} (Rule 5) Done. The mgu is {x/a, z/g(b), y/b}. Quiz. Apply the algorithm to the two atoms p(x, x) and p(y, ƒ(y)). Solution: {p(x, x) = p(y, ƒ(y))} and Rule 1 gives {x = y, x = ƒ(y)}. Apply Rule 5 to get {x = y, y = ƒ(y)}. Apply Rule 6 to get failure.

11. Resolution Rule (R) Given the following two clauses. L1 … Lk  C and ¬ M1 …  ¬ Mn  D, where Li and Mi are atoms and C and D are disjunctions of other literals. Assume also that 1. The clauses have distinct sets of variable names (rename if necessary). 2. q is the mgu of {L1, …, Lk, M1, …, Mn}. 3. N = L1q, where {L1, …, Lk, M1, …, Mn}q = {N}. Then we have: Example. Given the two clauses in the following proof segment k. p(a, y)p(a, z)q(x, y, z)P k + 1. ¬ p(w, ƒ(b))r(w, v, g(w))  ¬ p(a, ƒ(b)) P These two clauses have the form L1L2 C and¬ M1 D. The two clauses have distinct sets of variables. The set of atoms {p(a, y), p(a, z), p(w, ƒ(b))} has mgu q = {w/a, y/ƒ(b), z/ƒ(b)}. Notice that {p(a, y), p(a, z), p(w, ƒ(b))}q = {p(a, ƒ(b))}. So the resolution rule can be applied to the two clauses to obtain the resolvant: k + 2. q(x, ƒ(b), ƒ(b)) r(a, v, g(a))k, k + 1, R, {w/a, y/ƒ(b), z/ƒ(b)}.

12. To prove a wff is valid: negate the wff and convert it to clausal form; write down the clauses as premises; use resolution to find a false statement. Example/Quiz. Use resolution to prove the following wff is valid. x (y p(x, y)  z q(x, z) y (p(x, y) q(x, y))). Solution: Negate the wff: ¬ x (y p(x, y)  z q(x, z) y (p(x, y) q(x, y))). Convert it to clausal form: ¬ x (y p(x, y)  z q(x, z) w (p(x, w) q(x, w))) (renamed variables) x ((y p(x, y)  z q(x, z)) w (¬ p(x, w)  ¬ q(x, w))) (removed  and moved ¬ right) xw ((y p(x, y)  z q(x, z)) ¬ p(x, w)  ¬ q(x, w)) (moved w left) xw y z ((p(x, y)  q(x, z)) ¬ p(x, w)  ¬ q(x, w)) (moved y andz left) xy z ((p(x, y)  q(x, z)) ¬ p(x, ƒ(x))  ¬ q(x, ƒ(x))) (removed w by Skolem) So the set of clauses is {p(x, y)  q(x, z), ¬ p(x, ƒ(x)), ¬ q(x, ƒ(x)))}. Now do a resolution proof by making each the three clauses a premise (rename to get distinct variable names). 1. p(x, y)  q(x, z) P 2. ¬ p(u, ƒ(u)) P 3. ¬ q(w, ƒ(w)) P 4. q(u, z) 1, 2, R, {x/u, y/ƒ(u)} 5. [] 3, 4, R, {w/u, z/ ƒ(u)} QED.