Simplify a proposition I can not come up with anything concrete,
$$ [\overline{(p  \wedge q)} \wedge r] \vee   [p \wedge \overline{( q  \wedge r)}]  \Leftrightarrow \,  ?    $$
Thanks!
 A: First, I'll use negation $\lnot$: $\lnot (p \land q)$ to denote $\overline{(p\land q)}$, etc. 
So First let me write the equivalent expression  to your expression, using $\lnot$ (the lhs of $\equiv$).
$$[\lnot(p\land q)] \land r] \lor [p \land \lnot (q \land r)]\equiv [\overline{(p  \wedge q)} \wedge r] \vee   [p \wedge \overline{( q  \wedge r)}]   $$
We need DeMorgan's twice: e.g., $\lnot (p \land q) \iff \lnot p \lor \lnot q, \quad $ or $$\overline{(p \land q)} \iff \overline p \lor \overline q$$
We need the Distributive rule twice: $(\lnot p\lor \lnot q) \land r \iff (\lnot p \land r) \lor (\lnot q \land r)\quad $ Or  
$$ (\overline p \lor \overline q)\land r \iff (\overline p \land r) \lor (\overline q \land r)$$
That get's us to, using your notation:
$$(\overline p\land r)\lor(\overline q\land r)\lor(p\land\overline q)\lor(p\land\overline r)$$
This is in disjunctive normal form.
There are many expressions that can be equivalent to this; what exactly counts as simplified, needs more specification. 
If you need to determine the truth-value of the proposition, then you need to use something like a truth-table to determine and show the values of all possible 8 truth-value assignments of $p, q, r$. That is, your proposition is not "inherently true" for all $p, q, r$ (it is not a tautoloty), nor is it inherently false (it is not a contradiction).
A: I'm not sure the following equivalent form counts as a simplification, but I like it: $(p\lor r)\land\neg(p\land q\land r)$.
