0
$\begingroup$

Prove P ∨ Q, Q ∨ R, P → ¬R |- Q with natural deduction.

  1. P ∨ Q, premiss
  2. Q ∨ R, premiss
  3. P → ¬R, premiss ... ... Conclusion, Q

I dont know how to properly solve this question?

I know somewhat how to solve similar questions like below, but on this specific question above I dont know where to start.

$\endgroup$
2
  • 1
    $\begingroup$ This should be asked on Mathematics, not here. $\endgroup$
    – user16034
    Sep 25, 2023 at 11:38
  • $\begingroup$ The idea is to do OR-elimination. The intuition behind OR-elimination is that you're going to do a case analysis. If you choose to eliminate $p \lor q$, you will first consider the possibility that $p$ is true, and then consider the possibility that $q$ is true. The meaning of the rule is that regardless of whether $p$ or $q$ is true, you must reach the same conclusion either way. That's the idea of OR-elimination. So that's your first step. And that's what you see in the proof in lines 3 and 5. Reply if you have other questions. $\endgroup$
    – ShyPerson
    Sep 29, 2023 at 1:55

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.