**Determine whether ( πβ¨π)β§(πβπ)β§( πβπ )βπβ¨π is a Tautology or a contradiction**

The **Answer to the Question**

is below this banner.

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

Let us prove thatΒ $( πβ¨π)β§(πβπ)β§( πβπ )βπβ¨π$Β is a tautology using proof by contraposition. Suppose that the formula is not a tautology. Then there existsΒ $(p_0,q_0,r_0,s_0)\in\{T,F\}^4$Β such thatΒ $|( π_0β¨π_0)β§(π_0βπ_0)β§( π_0βπ _0)βπ_0β¨π _0|=F.$Β The definition of implication implies thatΒ $|( π_0β¨π_0)β§(π_0βπ_0)β§( π_0βπ _0)|=T$Β andΒ $|π_0β¨π _0|=F$.

The definitions of conjunction and disjunction imply thatΒ $| π_0β¨π_0|=|π_0βπ_0|=|π_0βπ _0|=T$Β andΒ $|π_0|=|π _0|=F.$Β It follows fromΒ $|π_0|βF=|π_0|βF=T$Β thatΒ $|p_0|=|q_0|=F.$Β Consequently,Β $|p_0\lor q_0|=F\lor F=F$Β and we have a contradiction withΒ $|p_0\lor q_0|=T.$Β Therefore, our assumption is not true, and we conclude that the formulaΒ $( πβ¨π)β§(πβπ)β§( πβπ )βπβ¨π$Β is a tautology.