**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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Solution:

Proof by Contradiction Method:

- $p\rightarrow (q\lor r)$ Premise
- $q\rightarrow s$ Premise
- $\neg (p\rightarrow (r\lor s))$ Premise, proof by contradiction
- $\neg (\neg p\lor (r\lor s))$ 3, Definition of $\rightarrow$
- $p\land \neg (r\lor s)$ 4, DeMorgan’s law
- $p$ 5, Specialization
- $\neg (r\lor s)$ 5, Specialization
- $\neg r\land \neg s$ 7, DeMorgan’s law
- $\neg r$ 8, Specialization
- $\neg s$ 8, Specialization
- $\neg q$ 2, 10, Modus Tollens
- $\neg q\land \neg r$ 9, 11
- $\neg (q\lor r)$ 12, DeMorgan’s law
- $\neg p$ 1, 13, Modus Tollens
- False 6, 14, proof by contradiction

Premise $\neg (p\rightarrow (r\lor s))$ was false, so $p\rightarrow (r\lor s)$ must be true.