Determine if tautology, contingency or contradiction I have to determine if the statement is a tautology, contradiction or contingency. Been at it for days but didn't get too far.
The original question is
$$\left((\lnot p\vee z)\wedge(p\vee q)\right)\rightarrow(z\vee q)$$
I got to here
$$\left((p\wedge\lnot z)\vee(\lnot p\wedge\lnot q)\right)\vee(z\vee q)$$
Any suggestions? Not allowed to use double distribution.
Can't use truth tables, just rules of logic.
 A: Starting from where you ended:
$\left((p\wedge\lnot z)\vee(\lnot p\wedge\lnot q)\right)\vee(z\vee q) \\
\lnot(\lnot p \vee z) \vee \lnot(p \vee q) \vee z \vee q \\
(\lnot(\lnot p \vee z) \vee z) \vee (\lnot(p \vee q) \vee q) \\
((\lnot p \vee z) \rightarrow z) \vee ((p\vee q) \rightarrow q) \\
((p \rightarrow z) \rightarrow z) \vee ((\lnot p \rightarrow q) \rightarrow q)$
We know that either $p$ or $\lnot p$ is true. 
Assume $p$ is true. Then $(p\rightarrow z) \rightarrow z$ simplifies to $z \rightarrow z$, which is tautologically true.
Assume $\lnot p$ is true. Then $(\lnot p \rightarrow q) \rightarrow q$ simplifies to $q \rightarrow q$, which is tautologically true.
Thus the entire statement is equivalent to a tautology since it is always true.
A: It's a tautology.  We can prove it algebraically (using the Boolean logic rules). 
First, let's apply distributivity to the antecedent:
$$(\neg p\vee z) \wedge (p\vee q) \equiv ((\neg p \vee z) \wedge p ) \vee ((\neg p \vee z) \wedge q).$$
Using distributivity again:
$$(\neg p \vee z) \wedge p ) \equiv (\neg p \wedge p) \vee (z \wedge p)$$
and then the identity rule:
$$(\neg p \wedge p) \vee (z \wedge p) \equiv (z\wedge p).$$
Now, the antecedent can be rewritten like this:
$$ (z\wedge p)\vee ((\neg p \vee z) \wedge q).$$
Again, we apply distributivity, to get:
$$ (z\wedge p)\vee ((\neg p \vee z) \wedge q)\equiv ( (z\wedge p)\vee(\neg p \vee z)) \wedge ( (z\wedge p)\vee q)$$
and again: 
$$( (z\wedge p)\vee(\neg p \vee z)) \wedge ( (z\wedge p)\vee q) \equiv ( ((z\wedge p)\vee \neg p) \vee ((z\wedge p)\vee z)) \wedge ( (z\wedge p)\vee q).$$
I'm going to skip the next few applications of distributivity, but you should be able to conclude:
\begin{eqnarray}
((z\wedge p) \vee \neg p) & \equiv & (z\vee \neg p) \\
((z\wedge p)\vee z)) & \equiv & z \\
((z\wedge p)\vee \neg p) \vee ((z\wedge p)\vee z) & = & (z\vee \neg p) \\
\end{eqnarray}
Then you just keep on in this vein.
A: It's not in terms of the rules that you linked, but it might be nice to see this in a natural deduction form, too.  Since the formula is provable, it's a theorem (and thus a tautology).
\begin{equation}
\begin{array}{l}
\begin{array}{|l}
(\lnot p \lor z) \land (p \lor q) \hspace{1cm}\mbox{Assume} \\
\hline 
\lnot p \lor z \hspace{1cm}\mbox{$\land$ elimination} \\
p \lor q \hspace{1cm}\mbox{$\land$ elimination}\\
\begin{array}{|l}
p \hspace{1cm}\mbox{Assume}\\
\hline
\begin{array}{|l}
\lnot p \hspace{1cm}\mbox{Assume}\\
\hline
\bot \hspace{1cm}\mbox{$\bot$ introduction}\\
z \hspace{1cm}\mbox{$\bot$ elimination}
\end{array} \\
\begin{array}{|l}
z \hspace{1cm}\mbox{Assume} \\ \hline z \hspace{1cm}\mbox{Reiteration}
\end{array} \\
z \hspace{1cm}\mbox{$\lor$ elimination} \\
q \lor z \hspace{1cm}\mbox{$\lor$ introduction}
\end{array} \\
\begin{array}{|l}
q \hspace{1cm}\mbox{Assume} \\ \hline q \lor z \hspace{1cm}\mbox{$\lor$ introduction} 
\end{array} \\
q \lor z \hspace{1cm}\mbox{$\lor$ elimination} 
\end{array} \\
((\lnot p \lor z) \land (p \lor q)) \to q \lor z \hspace{1cm}\mbox{$\to$ introduction}
\end{array}
\end{equation}
A: The formula 

$((\lnot p \lor z) \land (p \lor q)) \rightarrow (z \lor q)$

is a "convoluted" version of :

$((\lnot q \rightarrow p) \land (p \rightarrow z )) \rightarrow (\lnot q \rightarrow z)$ 

that is a tautology, being a version of Hypothetical syllogism.
A: Note: though the other answers looked good, I tried in this one to use only the rules given in the notes referred to in a comment by the OP.
From your correct steps you have
$$\left((p\wedge\lnot z)\vee(\lnot p\wedge\lnot q)\right)\vee(z\vee q).$$
By using the commutative and associative laws (from the page you refer to) this can be rearranged to get
$$((p\wedge\lnot z)\vee z )\vee ((\lnot p\wedge\lnot q)\vee q).$$
Now the distributive laws may be applied in each half of the main $\vee.$ Technically they are written as distribution of a left term over two right ones, but with commutative laws there is no problem about that. We get to
$$((p \vee z) \wedge (\lnot z \vee z))\vee ((\lnot p \vee q) \wedge (\lnot q \vee q)).$$
Now in your notes it says one can use $A \vee \lnot A=T.$ Then by the identity law the two $T$ values may be dropped in the last line, and get to
$$(p \vee z) \vee (\lnot p \vee q).$$
From here we can bring the $p$ and $\lnot p$ together and get to $T \vee (z \vee q)$, and then by domination we arrive finally at $T$.
A: Here is yet another way to do this, a bit similar to this earlier answer.
Let's use the rules of (classical propositional) logic to simplify the statement, as follows:
\begin{align}
& (\lnot p \lor z) \land (p \lor q) \rightarrow (z \lor q) \\
= & \qquad \text{"rewrite $\;P \rightarrow Q\;$ to $\;\lnot P \lor Q\;$ -- this gives more freedom to manipulate"} \\
& \lnot ((\lnot p \lor z) \land (p \lor q)) \lor z \lor q \\
= & \qquad \text{"DeMorgan (three times) and double negation -- to simplify"} \\
& (p \land \lnot z) \lor (\lnot p \land \lnot q) \lor z \lor q \\
= & \qquad \text{"use negation of $\;z\;$ on other side of $\;\lor\;$, and the same for $\;q\;$"} \\
& (p \land \lnot \text{false}) \lor (\lnot p \land \lnot \text{false}) \lor z \lor q \\
= & \qquad \text{"simplify"} \\
& p \lor \lnot p \lor z \lor q \\
= & \qquad \text{"excluded middle"} \\
& \text{true} \lor z \lor q \\
= & \qquad \text{"simplify"} \\
& \text{true} \\
\end{align}
In other words, the original statement is a tautology.
