335 Answers and Solutions to Selected Exercises (c) We can formalize the argument as follows: ∀x(T (x) → ((E (x) ∧ ¬G(x)) ∨ (G(x) ∧ ¬E (x)))) ∀y (T (y ) ∧ F l(y ) → E (y )) ∀x(D(x) → F l(x)) ∀x(G(x) ∧ U (x) → T (x)) ∃x(G(x) ∧ U (x)) ∃x(U (x) ∧ ¬D(x)) From all premises and the negated conclusion we obtain the following clauses: C1 = {¬T (x), E (x), G(x)} C2 = {¬T (x), ¬E (x), ¬G(x)} C3 = {¬T (y ), ¬F l(y ), E (y )} C4 = {¬D (u), F l(u)} C5 = {¬G(v), ¬U (v), T (v )} C6 = {G(c)} C7 = {U (c)} C8 = {¬U (w), D (w)} for some Skolem constant c Now, applying the Resolution rule successively, we get C9 = Res(C3 , C4 ) = {¬T (y ), E (y ), ¬D (y )} MGU[y/u] C10 = Res(C8 , C9 ) = {¬T (y ), E (y ), ¬U (y )} MGU[y/w] C11 = Res(C2 , C10 ) = {¬T (y ), ¬U (y ), ¬G(y )} MGU[y/x] C12 = Res(C5 , C11 ) = {¬G(y ), ¬U (y )} MGU[y/v ] C13 = Res(C6 , C12 ) = {¬U (c)} MGU[c/y ] C14 = Res(C7 , C13 ) = {} The empty clause is derived, hence the argument is valid 4.5.8 (a) Using predicates B (x) for “x is a bachelor,” M (x) for “x is a man,” W (x) for “x is a woman,” K (x, y ) for “x knows y ,” and L(x, y ) for “x loves y ,” we can formalize the argument as follows: ∃x(B (x) ∧ ∀y (W (y ) ∧ K (x, y ) → L(x, y ))) ∀x(B (x) → M (x)) W (Eve) ∧ ∀x(M (x) → K (x, Eve)) ∃x(B (x) ∧ ∃y (W (y ) ∧ L(x, y ))) We then obtain the following clausal form: C1 = {B (c)} C2 = {¬W (y ), ¬K (c, y ), L(c, y )} C3 = {¬B (z ), M (z )} C4 = {W (Eve)} C5 = {¬M (w), K (w, Eve)} C6 = {¬B (u), ¬W (v ), ¬L(u, v )}