**Given the following 2 premises, 1. πβ(πβ¨π) 2. πβπ Prove πβ(πβ¨π ) is valid using the Proof by Contradiction method.**

The **Answer to the Question**

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

truth table has a row where the conclusion column is FALSE while every premise column is TRUE

so,Β $\neg (πβ(πβ¨π ))$Β is invalid

then, by contradiction,Β $πβ(πβ¨π )$Β is valid