Show that -p → (q + r) and q→ (p V r) are logically equivalent.

Let us find the truth table for  $q\to (p \lor r)$:

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & p\lor r & q\to(p\lor r)\\ \hline 0 & 0 & 0 & 0 & 1 \\ \hline 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 0 & 0 \\ \hline 0 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 1 & 1 \\ \hline 1 & 0 & 1 & 1 & 1 \\ \hline 1 & 1 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

It follows that the truth value of $|q\to (p \lor r)|=0$ if and only if $|p|=|r|=0$, $|q|=1$. On the other hand, for $|p|=|r|=0$, $|q|=1$ we have that $|q+r|=1+0=1$, and therefore by definition of implication, $|-p → (q + r)|=1$. Consequently, the formulas $-p → (q + r)$ and $q→ (p\lor r)$ are not logically equivalent.