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

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