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. 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.pP 4.¬ rP 5.q1, 3, R 6.r2, 5, R 7. [] 4, 6, R QED. 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.

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  ¬ FP 8.¬ A  ¬ C3, 6, R 9.¬ B  ¬ F4, 7, R 10.¬ B  ¬ B5, 9, R 11.¬ A  ¬ A2, 8, R 12.B1, 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 clausal form 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. 1. Construct the prenex CNF for the wff. 2. Replace all occurrences of each free variable by a new constant. 3. 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  x 1...  x n, then replace  x W(x) by W(ƒ(x 1, …, x n )) 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.  x B(x, ƒ(x)). 3.  x  y C(x, y, g(x, y)). 4.  x D(x, ƒ(x), g(x)). Example/Quiz. Find a clausal form for the wff  y  x p(x, y)   z (q(x)  r(z)). Solution: Put the wff in prenex CNF.  y  x p(x, y)   z (q(w)  r(z))(renamed)  ¬  y  x p(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.  = {x/y, y/ƒ(z)} and  = {y/ƒ(a), z/b} are substitutions. Applying a Substitution to an Expression or a Set of Expressions If  is a substitution and E is an expression, then E  denotes the expression obtained from E by simultaneously applying the bindings in  to E. Example. If  = {x/y, y/ƒ(z)}, then p(x, y, z)  = p(x, y, z){x/y, y/ƒ(z)} = p(y, ƒ(z), z). If S is a set of expressions, then S  is the set of expresisons obtained from S by applying  to each expression of S. Example. If  = {x/y, y/ƒ(z)} and S = {p(x, y), q(y, g(z))}, then S  = {p(x, y) , q(y, g(z))  }= {p(y, ƒ(z)), q(ƒ(z), g(z))}.

7 Composing Substitutions If  and  are two substitutions, then the composition  is applied to an expression E by E(  ) = (E  ) . We can calculate  by 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  = {x/y, y/ƒ(z)} and  = {y/ƒ(a), z/b}. Since the numerator variables are x, y, and z, we calculate p(x, y, z)(  ) = (p(x, y, z)  )  = p(y, ƒ(z), z))  = p(ƒ(a), ƒ(b), b). So  = {x/ƒ(a), y/ƒ(b), z/b}. We can also calculate  by the following procedure, where  = {x i /t i } and  = {y i /s i }. 1. Construct {x i /(t i  }. 2. Delete any x i /x i from Step Delete y i /s i from  if y i is a numerator of .  is the union of the substitutions from Steps 2 and 3. Example. Let  = {x/y, y/ƒ(z)} and  = {y/ƒ(a), z/b}. Calculate . Step 1. Construct {x/y , y/ƒ(z)  } = {x/ƒ(a), y/ƒ(b)}. Step 2. No deletions. So no changes from Step 1. Step 3. Delete y/ƒ(a) from  (because y is a numerator of  ) to obtain {z/b}. Step 4. The union of sets from Steps 2 and 3 give  = {x/ƒ(a), y/ƒ(b), z/b}. Properties of Composition  =  and  =  =  where  =  {}.

8 A unifier of a set S of expressions is a substitution  such that S  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  of S such that any other unifier  of S can be written as  =  for some substitution . 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}  for some substitution . But {x/c, y/c, z/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;  0  :=  ; go to Step If S  k is a singleton then stop with mgu  k Otherwise construct D k (set of terms in leftmost position of disagreement); go to Step If D k has a variable v and a term t such that v does not occur in t then  k+1 :=  k {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;  0  := . 2. S  0 = S  = {p(x, h(x, g(y)), y), p(x, h(a, z), b)} is not a singleton; D 0 = {x, a}. 3.  1 :=  0 {x/a} = {x/a}; k := S  1 = {p(a, h(a, g(y)), y), p(a, h(a, z), b)} is not a singleton; D 1 = {g(y), z}. 3.  2 :=  1 {z/g(y)} = {x/a, z/g(y)}; k := S  2 = {p(a, h(a, g(y)), y), p(a, h(a, g(y)), b)} is not a singleton; D 2 = {y, b}. 3.  3 :=  2 {y/b} = {x/a, z/g(b), y/b}; k := S  3 = {p(a, h(a, g(b)), b)} is a singleton; stop with mgu  3 = {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 D 0 = {x, ƒ(x))} x occurs 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. Replace ƒ(s 1, …, s n ) = ƒ(t 1, …, t n ) with equations s 1 = t 1, …, s n = t n. 2. If ƒ(s 1, …, s n ) = g(t 1, …, t m ) 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 {x i = t i } to mgu {x i /t i }. 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. L 1  …  L k  C and ¬ M 1  …  ¬ M n  D, where L i and M i 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.  is the mgu of {L 1, …, L k, M 1, …, M n }. 3. N = L 1 , where {L 1, …, L k, M 1, …, M n }  = {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 L 1  L 2  C and ¬ M 1  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  = {w/a, y/ƒ(b), z/ƒ(b)}. Notice that {p(a, y), p(a, z), p(w, ƒ(b))}  = {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.