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Q: Show using truth tables that $\lnot(p \to q)$ and $(p \land q)$ are logically equivalent.

So I thought that the negation of $(p \to q)$ was $(p \land \lnot q)$ so not sure if "logically equivalent" means their truths tables have to be identical or if they only need to have the same number of True and same number of false?

Henrik supports the community
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Sofia Zapata
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    You're right, "equivalent" means that the truth tables must be identical in every row. Is it possible that you misread the question? – MJD Jun 11 '16 at 18:23
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    They aren't equivalent. You're right that $\neg (p\to q) \equiv (p\land \neg q)$. – BrianO Jun 12 '16 at 00:07

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\begin{matrix} P & Q & \sim Q & P\to Q & \sim (P\to Q) & P\wedge \sim Q & P\wedge Q \\ T & T & F & T & F & F & T \\ T & F & T & F & T & T & F \\ F & T & F & T & F & F & F \\ F & F & T & T & F & F & F \\ \end{matrix}

Behrouz Maleki
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