**((π β π) β§ (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 truth table of the following proposition.

$f=((π β π) β§ (π β π)) β (π β π)$

$\ \begin{array}{||c|c|c||c|c|c|c|c||} \hline\hline p & q & r &π β π & π β π & (π β π) β§ (π β π) & π β π & f\\ \hline\hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1\\ \hline 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1\\ \hline 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline\hline \end{array}$

It follows that the formulaΒ $((π β π) β§ (π β π)) β (π β π)$Β is a tautology.