Proving equivalence of $(P \Rightarrow Q) \land (Q \Rightarrow P) \equiv (P \lor Q) \Rightarrow (P \land Q)$ I am a little stymied by the following:$\def\imp{\Rightarrow}$
$(P \imp Q) \land (Q \imp P) \equiv (P \lor Q) \imp (P \land Q)$ 
Working with the RHS I have:
$\neg(P \lor Q) \lor (P \land Q) $
$( \neg (P \lor Q) \lor P) \land (\neg (P \lor Q) \lor Q))$ by distributive
$((\neg P \land \neg Q) \lor P) \land ((\neg P \land \neg Q) \lor Q)$ by de Morgan's Law
$((P \lor \neg P)\land(P \lor \neg Q)) \land ((Q \lor \neg Q)\land(Q \lor \neg P))$ by distributive
$(P \lor \neg P) \land (Q \imp P) \land (Q \lor \neg Q) \land (P \imp Q)$
Assuming this is correct so far, I am very close to $(P \imp Q) \land (Q \imp P)$, however I am unsure of how to deal with the $X \lor \neg X$ 's in the above, have I made an eror in the above expansions?
 A: I chose this way :
$(p \rightarrow q) \land (q \rightarrow p ) \equiv (p \lor q) \rightarrow (p \land q) \equiv (\lnot p \lor q) \land (\lnot q \lor p) \equiv ((\lnot p \lor q) \land \lnot q) \lor (\lnot p \lor q) \land p)$
$\equiv ((\lnot p \land \lnot q) \lor (q \land \lnot q)) \lor ((p \land \lnot p) \lor (p \land q)) \equiv (\lnot p \land \lnot q) \lor(p \land q)$  
$(p \lor q) \rightarrow (p \land q) \equiv \lnot(p \lor q) \lor (p \land q) \equiv (\lnot p \land \lnot q) \lor (p \land q)$  
So the two things that u mentioned are equivalent.
EDIT : About your mistake... nothing was wrong ... just notice that $p \lor \lnot p $ is equivalent to $1$ and $ q \lor \lnot q$ is equivalent to $1$ . 
A: $(P \Rightarrow Q)$ $\land$ $(Q \Rightarrow P) $ $\equiv$ $(\lnot P \lor Q)$ $\land$ $(\lnot Q \lor P)$ 
The distributive law gives you (note that two of those four expressions that you get simply vanish because they represent contradictions)
($\lnot P$ $\land$ $\lnot Q)$ $\lor$ ($P \land Q)$ $\equiv$ $\lnot (P \lor Q)$ $\lor$ $(P \land Q)$ $\equiv$ $(P \lor Q)$ $\Rightarrow$ $(P \land Q)$,
while the last step is just the application of 
$\lnot P \lor Q $ $\equiv$ $P \Rightarrow Q$.
