MAT1830 - Discrete Mathematics for Computer ScienceAssignment #2 Solutions1.pqrp∨qp∨r¬(p∨r)(p∨q)→ ¬(p∨r)¬p((p∨q)→ ¬(p∨r))Y¬pTTTTTFFFFTTFTTFFFFTFTTTFFFFTFFTTFFFFFTTTTFFTTFTFTFTTTFFFTFTFTTFFFFFFTTTF[4]Since the last column contains F and T, the statement is neither a tautology nor a contradiction.[1]2.¬p→(¬p∨q)[1]≡ ¬¬p∨(¬p∨q)(by the implication law)[1]≡p∨(¬p∨q)(by the double negation law)[1]≡(p∨ ¬p)∨q)(by the associative law)[1]≡T∨q(by the inverse law)[1]≡q∨T(by the commutative law)[1]≡T(by the annihilation law)So¬p→(¬p∨q) is logically equivalent to T and so is a tautology.[1](You don’t have to follow these exact steps. However, you must identify the law that you use ateach step, and you must only use one!)3. Contrapositive: “If X is neither a fruit nor a device, then X is not an apple.”[1]Negation: “X is an apple and X is not a fruit and X is not a device.”[1]