Given the following 2 premises, 1. 𝑝→(𝑞∨𝑟) 2. 𝑞→𝑠 Prove 𝑝→(𝑟∨𝑠) is valid using the Proof by Contradiction method.

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

so, $\neg (𝑝→(𝑟∨𝑠))$ is invalid

then, by contradiction, $𝑝→(𝑟∨𝑠)$ is valid

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