Assignment #2 Solutions.pdf - MAT1830 - Discrete...

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MAT1830 - Discrete Mathematics for Computer ScienceAssignment #2 Solutions1.pqrpqpr¬(pr)(pq)→ ¬(pr)¬p((pq)→ ¬(pr))Y¬pTTTTTFFFFTTFTTFFFFTFTTTFFFFTFFTTFFFFFTTTTFFTTFTFTFTTTFFFTFTFTTFFFFFFTTTF[4]Since the last column contains F and T, the statement is neither a tautology nor a contradiction.[1]2.¬p(¬pq)[1]≡ ¬¬p(¬pq)(by the implication law)[1]p(¬pq)(by the double negation law)[1](p∨ ¬p)q)(by the associative law)[1]Tq(by the inverse law)[1]qT(by the commutative law)[1]T(by the annihilation law)So¬p(¬pq) 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]

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