TITLE: How should I proceed in proving this tautology?
QUESTION [7 upvotes]: I know that following is a tautology because I've checked its truth table. I am now attempting to prove that it is a tautology by using the rules of logic, which is more difficult. How should I proceed?
$(p\land(p\implies q))\implies q$
$(p\land(\lnot p \lor q))\implies q$
$(p\land \lnot p) \lor (p\land q) \implies q$ This step is where I'm getting stuck at. I know that $(p\land \lnot p)$ is false. So it seems to me that the truth value of everything to the left of the $\implies$ operator depends on the truth value of $(p\land q)$ So what I want to do is this:
FALSE $\lor (p\land q) \implies q$ which reduces to 
$(p\land q) \implies q$
Is my thinking correct so far? If so, then I want to rewrite $(p\land q) \implies q$ as
$\lnot(p \land q) \lor (p \land q)$ by using the identity $p\implies q \equiv \lnot p \lor q$
Am I on the right track?

REPLY [4 votes]: $$
\begin{align}
(p\land(p\rightarrow q))\rightarrow q &\Longleftrightarrow (p\land(\neg p\lor q))\rightarrow q\\
 &\Longleftrightarrow ((p\land \neg p)\lor (p\land q))\rightarrow q\\
 &\Longleftrightarrow (F\lor (p\land q))\rightarrow q & \text{Negation law}\\
 &\Longleftrightarrow \neg(F\lor (p\land q))\lor q\\
 &\Longleftrightarrow (T\land \neg(p\land q))\lor q \\
 &\Longleftrightarrow (T\land(\neg p\lor \neg q))\lor q  &\text{DeMorgan's law}\\
 &\Longleftrightarrow (\neg p\lor \neg q)\lor q &\text{Domination law}\\
 &\Longleftrightarrow \neg p\lor (\neg q\lor q) \\
 &\Longleftrightarrow \neg p\lor T\\
 &\Longleftrightarrow T\\
\end{align}
$$