$(P \wedge Q \rightarrow R) \leftrightarrow (P \wedge \lnot R \rightarrow \lnot Q)$ I guess the biggest issue for me is not knowing how to work with $\wedge$.
I'm not new to the concept of proof... just formal proofs.  
Here is my attempt:
Show: $(P \wedge \lnot R \rightarrow \lnot Q)$ 
$P \wedge \lnot R$
Show $\lnot Q$
$Q$.  (assumption for contradiction)
$(P \wedge Q) \rightarrow R$
now I'm stuck....  Really I just don't know how to work with $\wedge$, any help appreciated.  Or maybe if I could see half the proof, then I could work the other direction.  Thank you
 A: Almost there!
Isolate the $P$ from $P \land \neg R$ by $\land$ Elim.
Combine this with the $Q$ from the Assumption to get $P \land Q$ using $\land $ Intro
Now you can use $\rightarrow$ Elim with $(P \land Q) \rightarrow R$ and $P \land Q$ to get $R$, and that will contradict the $\neg R$ that you can isolate from the $P \land \neg R$ by $\land$ Elim.
Here it is more formally (and inside the larger $\leftrightarrow$ proof):

A: Proof for $\rightarrow$ part
To show that $(P \wedge Q \rightarrow R) \leftrightarrow (P \wedge \sim R \rightarrow Q)$.
We can assume $(P \wedge Q \rightarrow R)$ and $P \wedge \sim R$ and then show $Q$.
Or, we can assume $(P \wedge Q \rightarrow R)$, $P \wedge \sim R$ and $Q$ and arrive at a contradiction.
$P \wedge \sim R \rightarrow P$ and $P \wedge \sim R \rightarrow \sim R$.
Also, $P \wedge Q \rightarrow R$.
So, we have both $R$ and $\sim R$ and this is a contradiction.
I hope you can do the other half yourself.
A: I only give you the proof of the forward direction i.e. "$\Rightarrow$" so you could have the chance to work on the proof for the backward direction.
Proof. ($\Rightarrow$) Suppose $P∧Q \rightarrow R$. Now we prove the contrapositive. Suppose $Q$. Now we have two cases:
Case 1. $\lnot P$, ergo we obviously have $\lnot P \lor R$.
Case 2. $P$. Since we had $Q$, then from $P$ and $Q$ and "$P∧Q \rightarrow R$", by modus ponens we get R. Hence $\lnot P \lor R$.
Since from both cases we had $\lnot P \lor R$, then $Q \rightarrow \lnot P \lor R$ which is equivalent to $P∧\lnot R \rightarrow \lnot Q$. Therefore, if $P∧Q \rightarrow R$, then $P∧\lnot R \rightarrow \lnot Q$.
Hope this helps.
