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

I know it can be done by truth tables, but here the question is asked to be answered with general laws like (De Morgan, absorption, negation, double negation, distributive laws). Do you get it? 
 A: 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.
A: For a change, I will propose a different approach:
If we assume:
$$\lnot(P\iff Q)\equiv P\iff \lnot Q$$
$$\implies\lnot(\lnot(P\iff Q))\equiv\lnot (P\iff \lnot Q)$$
$$\implies P\iff Q\equiv \lnot((P\implies \lnot Q)\land(\lnot Q\implies P))$$
$$\implies P\iff Q\equiv ((P\land Q)\lor(\lnot Q\land\lnot P))\equiv (P\land Q)\lor\lnot(P\lor Q)$$
But in detail if you want to entertain yourself experimenting (which is a good way to conclude in advance because your brain will accumulate more and faster):
$$(P\land Q)\lor(\lnot P\land\lnot Q)\equiv((P\land Q)\lor\lnot P)\land((P\land Q)\lor \lnot Q))$$
$$\equiv\underbrace{(P\lor \lnot P)}_{1}\land(\lnot P\lor Q)\land(P\lor \lnot Q)\land\underbrace{(Q\lor \lnot Q)}_{1}$$
$$\equiv(P\lor \lnot Q)\land(Q\lor\lnot P)\equiv\neg((\lnot P\land Q)\lor(\lnot Q\land P))\equiv\neg((P\land \lnot Q)\lor(Q\land\lnot P))\equiv\lnot(\lnot(P\iff Q))\equiv \underbrace{P\iff Q}_{again}$$
Hope it is fun!
