Proving that $\neg((P\vee Q)\to R) \iff (P\wedge \neg R)\vee(Q\wedge\neg R)$ without a truth table 
Prove, using propositional calculus, that
  $\neg((P\vee Q)\to R) \iff (P\wedge \neg R)\vee(Q\wedge\neg R)$.

My work: 
$$(\neg (P\vee Q) \vee\neg R) 
\\
((\neg P\wedge \neg Q) \vee\neg R)   
\\
(P\wedge\neg R)\vee (Q\wedge\neg R)
$$
Is that correct ?
 A: As Graham Kemp pointed out, your argument is wrong at the outset, but the correct argument is not too difficult to construct. There are basically two rules being emphasized in this exercise: DeMorgan's and distributive. Also, you must become comfortable with the fact that $$\ell\to\eta\equiv\neg\ell\lor\eta.\tag{1}$$ This equivalence is used repeatedly in propositional logic. That being said, see if you can follow this argument:
\begin{align}
\neg[(P\lor Q)\to R]&\equiv \neg[\neg(P\lor Q)\lor R]\tag{by $(1)$}\\[0.5em]
&\equiv \neg[(\neg P\land\neg Q)\lor R]\tag{DeMorgan}\\[0.5em]
&\equiv \neg(\neg P\land\neg Q)\land\neg R\tag{DeMorgan}\\[0.5em]
&\equiv (P\lor Q)\land\neg R\tag{DeMorgan}\\[0.5em]
&\equiv (P\land\neg R)\lor (Q\land\neg R)\tag{distributivity} 
\end{align}
A: No.  The first step is wrong.  Errors compound from there.
Start with: $\neg ((P\vee Q)\to R) \iff (P\vee Q)\wedge \neg R$
A: The implication
$$
(P\vee Q)\to R
$$
is logically equivalent to
$$
\neg((P\vee Q)\wedge\neg R).
$$
This is intuitive since the implication says that if the hypothesis $P\vee Q$ is true then the conclusion $R$ should also be true, that is, it must NOT be the case that we have $P\vee Q$ and NOT $R$.
Now, simplifying is easy:
$$
\neg(\neg((P\vee Q)\wedge\neg R))
$$
is logically equivalent to
$$
(P\vee Q)\wedge\neg R
$$
and distributivity gives
$$
(P\wedge \neg R)\vee(Q\wedge\neg R),
$$
as was to be shown.
Alternatively, one could construct a truth table to prove the equivalence.
