Answer to Question #285073 in Discrete Mathematics for triple

Question

Given the following 2 premises, 1. 𝑝→(𝑞∨𝑟)

2. 𝑞→𝑠

Prove 𝑝→(𝑟∨𝑠) is valid using the Proof by Contradiction method.

Answer 1

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