How do I show a statement is a tautology in the Booleans algebra I want to show through Boolean algebra that the statement 
$$\neg Q\Rightarrow (R\Rightarrow \neg (P\land Q))$$
is equivalent to the tautology $Q\lor(\neg Q)$?
 A: Don't know exactly what you mean by "Boolean algebra", given that implications are not usually part of the language of Boolean algebras. If I translate $A\Rightarrow B$ to the logically equivalent $\neg A \lor B$, then it is straightforward:
\begin{align*}
\neg Q\Rightarrow (R\Rightarrow (\neg (P\land Q)) &\equiv \neg(\neg Q)\lor (R\Rightarrow (\neg (P\land Q)))\\
&\equiv Q \lor (\neg R \lor (\neg(P\land Q)))\\
&\equiv Q\lor (\neg R \lor (\neg P \lor \neg Q))\\
&\equiv Q \lor \neg R \lor \neg P \lor \neg Q\\
&\equiv (Q\lor \neg Q) \lor (R\lor \neg P)\\
&\equiv Q\lor \neg Q.
\end{align*}
Added. In the comments you ask how to prove it by contradiction: here we use the fact that $\neg(A\Rightarrow B) \equiv A\land \neg B$.
\begin{align*}
\neg\Bigl(\neg Q\Rightarrow (R\Rightarrow (\neg (P\land Q))\Bigr) &\equiv
\neg Q \land \neg\Bigl(R\Rightarrow (\neg (P\land Q))\Bigr)\\
&\equiv \neg Q \land \Bigl( R \land \neg(\neg(P\land Q))\Bigr)\\
&\equiv \neg Q \land (R \land (P\land Q))\\
&\equiv \neg Q \land R \land P \land Q\\
&\equiv (\neg Q\land Q) \land R \land P\\
&\equiv (\neg Q\land Q).
\end{align*}
Since the negation of the proposition is equivalent to a contradiction, the original proposition is a tautology, i.e., true.
