**p ^ ~ (q ^ r)**

The **Answer to the Question**

is below this banner.

**Here's the Solution to this Question**

Let us construct the trush table for the formula $p \land \sim (q \land r):$

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

We conclude that this formula is neither tautology nor contradiction.