Using general laws, prove that $\lnot(P\leftrightarrow Q)=P\leftrightarrow \lnot Q$

By definition $P \rightarrow Q = (\lnot P \lor Q)$ and $P \leftrightarrow Q = (P \rightarrow Q) \land (Q \rightarrow P)$. Hence \begin{align} \lnot(P \leftrightarrow Q) &= \lnot ((\lnot P \lor Q) \land (\lnot Q \lor P))\\ &= \lnot ( (\lnot P \land \lnot Q) \lor (P \land Q))\\ &= (P \lor Q) \land (\lnot P \lor \lnot Q) \qquad \mbox{de Morgan}\\ &= (\lnot P \lor \lnot Q) \land (Q \lor P) \qquad \mbox{Commutativity}\\ &= P \leftrightarrow \lnot Q \end{align}

Edit
For the step between first to second line. By distributivity \begin{align} (\lnot P \lor Q) \land (\lnot Q \lor P) &= (\lnot P \land \lnot Q) \lor (\lnot P \land P) \lor (Q \land \lnot Q) \lor (Q \land P) \end{align}

but $\lnot P \land P = 0, \lnot Q \land Q = 0$. Hence the second line.