Question

p ↔ ¬ (r˅q)

Answer 1

Let us construct the trush table for the formula $p ↔ ¬ (r˅q):$

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

We conclude that this formula is neither tautology, nor contradiction.