Propositional calculus logical proof help I am asked to prove:
(Y =&gt; Z) V X    ≡    (Y V X) =&gt;(Z V X)
This is what I have so far,
(Y V X) =&gt;(Z V X)
(Definition of =&gt; X,Y := Y V X, Z V X)
(Y V X) V (Z V X)≡ (Z V X)
(V SYMMETRY twice )
(X V Y) V (X V Z)≡ (Z V X)
(V/V X,Y,Z := X,Y,Z)
X V (Y V Z) = (Z V X)
How do I get rid of the (Z V X) so that I have the full proof Picture of the proofs I am using
$$\begin{align} (Y \to Z) \lor X &amp;\equiv (Y\lor X) \to (Z\lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &amp;\equiv \lnot (Y \lor X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &amp;\equiv (\lnot Y \land \lnot X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &amp;\equiv (\lnot Y \lor Z \lor X) \land \underbrace{(\lnot X \lor Z \lor X)}_{true}\\ \\\iff (\lnot Y \lor Z \lor X) &amp;\equiv (\lnot Y \lor Z \lor X)\end{align}$$
I leave it to you to justify the steps.  If you are unsure of any of them, feel free to comment below the answer.  I've made use of the equivalence between $(P\to Q)\equiv (\lnot P \lor Q)$ (and more than once).  I've also used DeMorgan's law, and the fact that $\lnot X \lor X$ is always true ($\top$).  I used the fact that $\top \lor Z$ is always true, and the fact that $\top \land P \equiv P$.
$$\begin{align} (Y \to Z) \lor X &amp;\equiv (Y\lor X) \to (Z\lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &amp;\equiv \lnot (Y \lor X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &amp;\equiv (\lnot Y \land \lnot X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &amp;\equiv (\lnot Y \lor Z \lor X) \land \underbrace{(\lnot X \lor Z \lor X)}_{true}\\ \\\iff (\lnot Y \lor Z \lor X) &amp;\equiv (\lnot Y \lor Z \lor X)\end{align}$$
I leave it to you to justify the steps.  If you are unsure of any of them, feel free to comment below the answer.  I've made use of the equivalence between $(P\to Q)\equiv (\lnot P \lor Q)$ (and more than once).  I've also used DeMorgan's law, and the fact that $\lnot X \lor X$ is always true ($\top$).  I used the fact that $\top \lor Z$ is always true, and the fact that $\top \land P \equiv P$.