**(π β π) β (π β¨ ~π)**

The **Answer to the Question**

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

Let us construct the trush table for the formulaΒ $(π β π) β (π β¨ \sim π):$

$\begin{array}{||c|c||c|c|c|c||} \hline\hline p & q & π β π & \sim π & π β¨ \sim π & (π β π) β (π β¨ \sim π)\\ \hline\hline 0 & 0 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 0 & 1 &1\\ \hline\hline \end{array}$

We conclude that this formula is tautology.