Prove this logical equivalence. I'm determining whether this logical proposition is a tautology or a contradiction. I'm stuck in the middle of this equation, and cannot move further using only logical equivalences. The proposition is:
 -(-p -> -(-x v -y)) -> -(x ^ y))

Can someone brief me on the laws/steps it takes to solve this proposition?
What I have so far:
 (-p -> -(-x v -y)) v -(x ^ y)   implication and double negation
 (-p -> (x ^ y)) v (-x v -y)     demorgans twice
 (-p -> x) ^ (-p -> y) v (-x v -y) implication
 p v(x ^ y) v (-x v -y)          distributive

Stuck.
 A: I'll start off where you left off. I'll simply carry the disjunct $p$ along for the ride: we'll see in the end that the proposition is a tautology regardless of whether $p$ is true or false. 
You've gotten to $(1)$
$$\begin{align} p \lor (x \land y) \lor (\lnot x \lor \lnot y) & \equiv p\lor [(x \land y) \lor \lnot x \lor \lnot y] \tag{1}\\ \\
& \equiv p \lor [(x\lor \lnot x \lor \lnot y) \land (y \lor \lnot x \lor  \lnot y)]\tag{2} \\ \\
& \equiv  p \lor \{[(x \lor \lnot x) \lor \lnot y] \land [(y \lor \lnot y) \lor \lnot x]\}\tag{3}\\ \\
& \equiv p \lor [(T \lor \lnot y) \land (T \lor \lnot x)] \tag{4}\\ \\
& \equiv p \lor (T \land T) \tag{5}\\ \\ 
& \equiv p \lor T \tag{6}\\ \\
& \equiv T\tag{7}\end{align}$$
Thus we have a tautology: It is true regardless of  truth-values of $p, \, x\, \text{ or}\; y$.
$(1)$ just follows by associativity of $\lor$. Note we distribute to get $(2)$, use associativity and commutativity of $\lor$ in $(3)$.
We also make use of the law of the excluded middle: $q \lor \lnot q\equiv T$ in $(4)$. In $(5), (7)$, $T \lor q\equiv T$ (domination) and in $(6)$, $T \land T$ always evaluates to true, by the definition of $\land$. 
A: Your formula (there were parentheses that weren't matching, I'm guessing by your derivations)
 -(-p -> -(-x v -y)) -> -(x ^ y)

or in usual notation
$$\neg(\neg p \to \neg(\neg x \lor \neg y)) \to \neg(x \land y)$$
could be transformed into contrapositive
$$(x \land y) \to (\neg p \to \neg(\neg x \lor \neg y))$$
further using De Morgan's laws to
$$(x \land y) \to (\neg p \to (x \land y))$$
and finally substituting $\alpha = x \land y$ we arrive at
$$\alpha \to (\neg p \to \alpha).$$
This could be seen as tautology by itself, but if you require further reduction, we could uncurry it
$$(\alpha \land \neg p) \to \alpha$$
and weaken the premise to get the identity axiom
$$\alpha \to \alpha.$$
I hope this helps $\ddot\smile$
