How to prove the following propositional logic equivalence? The question asked me to prove the following equivalence.
$$（r \rightarrow p)\rightarrow (p\land q)\equiv (\lnot r \rightarrow p)\land(\lnot r \rightarrow q)\land (p \rightarrow q)$$
And I did the followings:
For L.H.S., 
$（r\rightarrow p)\rightarrow (p\land q)\\\equiv \lnot (\lnot r\lor p)\lor (p\land q)\\ \equiv (r \land \lnot p) \lor (p \land q)$
For R.H.S., $(\lnot r \rightarrow p)\land(\lnot r \rightarrow q)\land (p \rightarrow q)\\ \equiv (r\lor p) \land (r \lor q) \land (\lnot p \lor q)\\ \equiv (r\lor (p\land q))\land (\lnot p\lor q)\\ $
Now, I'm a bit stuck as to how to show that LHS = RHS.
 A: I’ve done the proof and here is the answer.

L.H.S. =
$
（r\rightarrow p)\rightarrow (p\land q)\\
\equiv \lnot (\lnot r\lor p)\lor (p\land q)\\
\equiv (r \land \lnot p) \lor (p \land q)\\
\equiv (r\lor p)\land (\lnot p \lor p)\land (r\lor q)\land (\lnot p \lor q)\\
\equiv(r\lor p)\land T\land (r\lor q)\land(\lnot p\lor q)\\
\equiv(\lnot r\rightarrow p)\land (\lnot r\rightarrow q)\land (p\rightarrow q)\\
\equiv R.H.S.
$

Thx for all your helps !
A: After writing out the left hand side using $\;\phi \rightarrow \psi \;\equiv\; \lnot \phi \lor \psi\;$ and DeMorgan, it is an $\;\lor\;$ of $\;\land\;$'s.  Similarly, the right hand side becomes a $\;\land\;$ of $\;\lor\;$'s.
How do you get from one to the other?  Through distribution.
So you choose one of the two sides, and distribute $\;\land\;$ over $\;\lor\;$, or $\;\lor\;$ over $\;\land\;$, and work towards the other side.

After you completed the proof, look up conjunctive normal form and disjunctive normal form and see how those terms relate to your proof.
