Explain how $(p \land \lnot q) \iff (\lnot p \lor q)$

Question

Is this DeMorgan's law?

I know DeMorgan's law in the form of

$\lnot(p \land q) \iff \lnot p \lor \lnot q$

and

$\lnot(p \lor q) \iff \lnot p \land \lnot q$

But I cannot conceptualize it in the form of

$p \land \lnot q) \iff (\lnot p \lor q)$

Answer 1

This is false. If you construct the truth table for these, you get: $$\begin{array}{cc|cc|cc} p & q & \neg p & \neg q & p \wedge \neg q & \neg p \vee q\\ \hline 0 & 0 & 1 & 1 & 0& 1\\ 1 & 0 & 0 & 1 & 1& 0\\ 0 & 1 & 1 & 0 & 0& 1\\ 1 & 1 & 0 & 0 & 0& 1\\ \end{array}$$ So they are not equivalent (the columns should be identical), but in fact they are opposites: $p \wedge \neg q \iff \neg (\neg p \vee q)$.

If you want to prove this last thing, the truth table is pretty clear about that, but you can also use the second De Morgan's law you posted: $$\neg (p \vee q) \iff \neg p \wedge \neg q$$ And now replace $p$ by $\neg p$: $$\neg (\neg p \vee q) \iff \neg (\neg p) \wedge \neg q$$ And since $\neg(\neg p) \iff p$, you obtain $$\neg (\neg p \vee q) \iff p \wedge \neg q$$