333 Answers and Solutions to Selected Exercises Now we transform the formulas ∀x(P (x) → B (x)), ∃x(P (x) ∧ W (x)), ∀x(P (x) → E (x)), and ¬∃x(W (x) ∧ B (x) ∧ E (x)) to clausal form: C1 = {¬P (x), B (x)} C2 = {P (c)} C3 = {W (c)} C4 = {¬P (x), E (x)} C5 = {¬W (x), ¬B (x), ¬E (x)} for some Skolem constant c Now, applying the Resolution rule successively, we get C6 = Res(C3 , C5 ) = {¬B (c), ¬E (c)} MGU[c/x] C7 = Res(C4 , C6 ) = {¬P (c), ¬B (c)} MGU[c/x] C8 = Res(C1 , C7 ) = {¬P (c)} MGU[c/x] C9 = Res(C2 , C8 ) = {} The empty clause is derived, hence the argument is valid (f) Using predicates Y (x) for “x is yellow,” P (x) for “x is a plonk,” and Q(x) for “x is a qlink,” we can formalize the argument as follows: ¬∃x(Y (x) ∧ P (x) ∧ Q(x)), ∃x(P (x) ∨ ¬Q(x)) ∃x¬(Y (x) ∧ Q(x)) We next transform the formulas ¬∃x(Y (x) ∧ P (x) ∧ Q(x)), ∃x(P (x) ∨ ¬Q(x)), and ¬∃x¬(Y (x) ∧ Q(x)) to clausal form: C1 = {¬Y (x), ¬P (x), ¬Q(x)} C2 = {P (c), ¬Q(c)} C3 = {Y (x)} C4 = {Q(x)} for some Skolem constant c Now, applying the Resolution rule successively, we get C5 = Res(C1 , C2 ) = {¬Y (c), ¬Q(c)} MGU[c/x] C6 = Res(C3 , C5 ) = {¬Q(c)} MGU[c/x] C7 = Res(C4 , C6 ) = {} MGU[c/x] The empty clause is derived, hence the argument is valid 4.5.6 We formalize A and B in the domain of all men, using P (x) for “x is happy” and Q(x) for “x is drunk” as follows: A := ¬∃x(P (x) → Q(x)) B := ∃yP (y ) ∧ ¬∃zQ(z ) (a) First we check if A implies B Clausification of A and ¬B : C1 := {P (x), ¬Q(x)}, C2 := {¬P (y )}, C3 := {Q(s1 )}, where s1 is a new Skolem constant