Answers and Solutions to Selected Exercises 289 Now, transform the result to clausal form: C1 = {p, q } and C2 = {¬q } Applying Resolution, we get C3 = Res(C1 , C2 ) = {p} No more clauses are derivable Therefore, the empty clause cannot be derived, so the formula is not a tautology (c) Transform ¬(((p → q ) ∧ (p → ¬q )) → ¬p) into a CNF: ¬(((p → q ) ∧ (p → ¬q )) → ¬p) ≡ ((p → q ) ∧ (p → ¬q )) ∧ p ≡ (¬p ∨ q ) ∧ (¬p ∨ ¬q ) ∧ p Now, transform the result to clausal form: C1 = {¬p, q }, C2 = {¬p, ¬q }, and C3 = {p} Finally, applying Resolution successively, we get C4 = Res(C1 , C2 ) = {¬p} C5 = Res(C3 , C4 ) = {} The empty clause has been derived, therefore the formula is a tautology (e) Transform ¬(((p → q ) ∧ (p → r)) → (p → (q ∧ r))) into a CNF: ¬(((p → q ) ∧ (p → r)) → (p → (q ∧ r))) ≡ ((p → q ) ∧ (p → r)) ∧ ¬(p → (q ∧ r )) ≡ (¬p ∨ q ) ∧ (¬p ∨ r) ∧ p ∧ (¬q ∨ ¬r) Now, transform the result to clausal form: C1 = {¬p, q }, C2 = {¬p, r}, C3 = {p}, and C4 = {¬q, ¬r} Finally, applying Resolution successively, we get C5 = Res(C2 , C4 ) = {¬p, ¬q } C6 = Res(C1 , C5 ) = {¬p} C7 = Res(C3 , C6 ) = {} The empty clause has been derived, therefore the formula is a tautology (g) First, transform ¬(((p → r) ∨ (q → r)) → ((p ∨ q ) → r)) into a CNF: ¬(((p → r ) ∨ (q → r )) → ((p ∨ q ) → r )) ≡ ((p → r ) ∨ (q → r )) ∧ ¬((p ∨ q ) → r) ≡ (¬p ∨ r ∨ ¬q ∨ r) ∧ (p ∨ q ) ∧ ¬r ≡ (¬p ∨ r ∨ ¬q ) ∧ ¬r ∧ (p ∨ q )