How should I proceed in proving this tautology? 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?
 A: Use De Morgan's law on $\neg(p\land q)$ near the end. This should give you a disjunction which should easily be seen as tautological by the law of excluded middle. Also, remember that $(p\land q)\implies q\equiv \neg(p\land q)\lor q$, not $\neg(p\land q)\lor(p\land q)$ as you wrote.
A: $$
\begin{align}
(p\land(p\rightarrow q))\rightarrow q &\Longleftrightarrow (p\land(\neg p\lor q))\rightarrow q\\
 &\Longleftrightarrow ((p\land \neg p)\lor (p\land q))\rightarrow q\\
 &\Longleftrightarrow (F\lor (p\land q))\rightarrow q & \text{Negation law}\\
 &\Longleftrightarrow \neg(F\lor (p\land q))\lor q\\
 &\Longleftrightarrow (T\land \neg(p\land q))\lor q \\
 &\Longleftrightarrow (T\land(\neg p\lor \neg q))\lor q  &\text{DeMorgan's law}\\
 &\Longleftrightarrow (\neg p\lor \neg q)\lor q &\text{Domination law}\\
 &\Longleftrightarrow \neg p\lor (\neg q\lor q) \\
 &\Longleftrightarrow \neg p\lor T\\
 &\Longleftrightarrow T\\
\end{align}
$$
A: Your reasoning is correct so far. It seems to me that since $a\to b$ is equivalent to $\lnot a\lor b$, then $(p\land q)\to q$ is equivalent to $\lnot(p\land q)\lor q$, which is different from what you wrote in the last line. To continue, I guess you want to use that $\lnot(p\land q)$ is equivalent to $\lnot p\lor \lnot q$. 
A: Using natural deduction notation 
Modus ponens:
$$\backslash\!\!\!\!\!{A}$$ $$\overline{B}$$ $$\overline{A\to B}$$
If from the assumption $A$ can deduce $B$, then $A\to B$ is deducible.
Law of Conjunction:
$$A\wedge B$$ $$\overline{\qquad A \qquad}$$ and 
$$A\wedge B$$ $$\overline{\qquad B \qquad}$$ and 
If $A\wedge B$ holds, then both $A$ and $B$ also holds.
To prove the statement using ND, start by breaking up the first implication - and you soon figure out what to do. (If there is any doubt just ask.)
