# Use logical equivalencies to classify as tautology, contradiction, or contingency.

Classify the following as tautologies, contradictions or contingencies using logical equivalences.

Can anyone let me know what I'm missing or doing wrong? I got stuck, here is what I have so far:

$\lnot(\lnot x \rightarrow (y \rightarrow z)) \rightarrow (y \rightarrow(x \lor z))$

$≡ \lnot(\lnot x\rightarrow (y \rightarrow z)) \rightarrow (\lnot y \lor(x \lor z))$ Implication Equivalence

$\equiv \lnot(\lnot x \rightarrow (\lnot y \lor z)) \rightarrow (\lnot y \lor (x \lor z))$ // Implication Equivalence

$≡ ¬(\lnot\lnot x\lor (\lnot y \lor z)) \rightarrow (\lnot y \lor x \lor z)$ Implication Equivalence, Associativity

$≡ \lnot\lnot (\lnot \lnot x \lor \lnot y \lor z) \lor (\lnot y \lor x \lor z)$ Implication Equivalence, Associativity

$≡ (x \lor \lnot y \lor z) \lor (\lnot y x \lor z)$ Double Negation, twice used.

$≡ (x \lor x) \lor (\lnot y \lor \lnot y) \lor (z \lor z)$ Associative and Commutative Law

$≡ x \lor \lnot y \lor z$ // Idempotent law

I just don't know what rules to use or what to do from here.

Thanks

Now we need to interpret the result, with the last line logically equivalent to the original proposition, we can look at the last line, $x \lor \lnot y \lor z$, to evaluate if and when it is true/false. This can be done easily by using a truth-table (you'll find one row evaluates to false, the rest true).
The proposition is false when $x, z$ are both false and $y$ is true. (Can you see that $x \lor \lnot y \lor z$ would be false under that truth value assignment?)
Otherwise, it is true. That is, it is true if $x$ is true (independently of the truth value assignment of $y$ or $z$.) Likewise, it is true if $y$ is false, and/or it is true when $z$ is true.