Is $\neg(\neg p \lor \neg q) \iff (p \land q)$ a tautology?

Question

This was an example from my textbook and answer is not provided.

Basically I'm asking because if it is, then $\neg(\neg p \lor \neg q)$ is equivalent to $p \land q$, meaning that $\neg \neg p$ is $p$, $\neg \neg q$ is $q$, and $\neg \lor$ is $\land$.

Any advice is greatly appreciated.

Answer 1

De Morgan's laws:

$\neg(P \lor Q)\iff(\neg P)\land(\neg Q)\tag1$ $\neg(P \land Q)\iff(\neg P)\lor(\neg Q)\tag2$

This question is the first case, with $P=\neg p$ and $Q=\neg Q$.

You can use truth table to prove your question.

$$\begin{array}{c|c|c|c} p&q&\neg(\neg p \lor \neg q)&p\land q\\\hline T&T&T&T\\\hline T&F&F&F\\\hline F&T&F&F\\\hline F&F&F&F \end{array}$$

Therefore $\neg(\neg p \lor \neg q)\iff(p\land q)$.

$\blacksquare$

Answer 2

The tautology is correct, but:

$$\neg(\neg p \lor \neg q) \not\implies \neg\neg p (\neg\lor) \neg\neg q$$

because the distribution law only applies to the operands, and NOT to the operator.