Obtaining one of DeMorgan's laws from the other

Question

How can:

$\lnot(x \lor y \lor z)=\lnot x \land \lnot y \land \lnot z$

be obtained from:

$\lnot (x \land y \land z) = \lnot x \lor \lnot y \lor \lnot z$   

such that $x, y, z $ are logical variables.

I have tried some manipulation of the equations such as:

\begin{align*} \lnot (x \land y \land z) &= \lnot x \lor \lnot z \lor \lnot y \\ \lnot x \lor \lnot (y \land z) &= \lnot(x \land y \land z) \\ \lnot x \lor \lnot y \lor \lnot z &= \lnot(x \land y \land z) \\ \end{align*}

However this doesn't seem to be the right direction to go.

Answer 1

You can prove the result by taking advantage of the fact that double negation is the identity function. \begin{align*} \lnot x \land \lnot y \land \lnot z &= \lnot\lnot(\lnot x \land \lnot y \land \lnot z) && \text{(by law of double negation)}\\ &= \lnot[\lnot(\lnot x \land \lnot y \land \lnot z)] \\ &= \lnot[\lnot(\lnot x) \lor \lnot(\lnot y) \lor \lnot(\lnot z)] && \text{(by the other of DeMorgan's Laws)} \\ &= \lnot[x \lor y \lor z] && \text{(by law of double negation)}\\ \end{align*}

Thus, $\lnot[x \lor y \lor z] = \lnot x \land \lnot y \land \lnot z$