((p∧q)⇒(r∨q))∧((p∧r)⇒q) We know, p⇒q is equivalent to ‌∼p∨q ‌(∼(p∧q)∨(r∨q))∧(∼(p∧r))∨q)) ‌⇒(∼p∨∼q∨r∨q)∧(∼p∨∼r∨q) ‌⇒(∼p∨r∨t)∧(∼p∨∼r∨q) ‌⇒(t)∧(∼p∨∼r∨q) For this to be tautology, (∼p∨∼r∨q) must be always true which follows for r=∼ p or r=q.