Put $(a \leftrightarrow b) \wedge c$ in DNF $$(a \leftrightarrow b) \wedge c$$
I'm having problems with this. If I do:
$$(a \rightarrow b) \wedge (b \rightarrow a) \wedge c$$
then
$$(\neg a \vee b) \wedge (\neg b \vee a) \wedge c$$
But now I'm stuck.
I can do this but it gets too complicated... I need the "ORs of AND terms where every AND term consists of all variables."
I can distribute the $c$, but then I get a conjunctive form.
$$[(\neg a \wedge c) \vee (b \wedge c)] \wedge [(\neg b \wedge c) \vee (a \wedge c)]$$
Am I making this too complicated?
 A: Hint.  Try starting with
$$a\leftrightarrow b\quad\equiv\quad (a\wedge b)\vee(\neg a\wedge \neg b)\ .$$
A: David's hint is the best way to go, but I'll also mention that you are making the distribution harder than it needs to be.  You could just go from 
$(\neg a \lor b) \land (\neg b \lor c) \land c$ 
to
$(\neg a \lor b) \land ( [\neg b \land c] \lor [a\land c])$
And from there you can distribute the remaining conjunction as a big long ugly thing, but it's a margin better than the distribution you would have to do in the next step of what you've done.
A: Lets take a look at a truth table for your expression
\begin{array}{c|c|c|c}
A&B&C&\left(A\Longleftrightarrow B\right)\wedge C\\
\hline 0&0&0 &0\\
0&0&1 &1\\
0&1&0 &0\\
0&1&1 &0\\
\hline 1&0&0 &0\\
1&0&1 &0\\
1&1&0&0\\
1&1&1&1\\
\end{array}
Your expression is true if and only if 


*

*$A$ is false and $B$ is false and $C$ ist true or 

*$A$ is true and $B$ is true and $C$ ist true


Thus your DNF is given by
$$\left(\lnot A\wedge \lnot B \wedge C\right) \vee\left(A\wedge B\wedge C\right) $$
