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(p → q) ∧ ¬p ⇒ ¬q

I have tried to use the rules of inference in many ways to prove this argument to be valid. However, I cannot seem to find a way to validate the argument, but I am afraid I have missed something.

Using a truth table I find that when p and q is 0, then p → q is 1, ¬p is 1 and ¬q is 1. Would that not make the argument valid?

I apologize in advance if I have misunderstood the whole concept. Any answer would be much appreciated. Thank you

  • 1
    What are your rules of inference? – 2017-01-14
  • 2
    In fact, your proposition is not valid. – 2017-01-14

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Note: $$(p\rightarrow q) \Leftrightarrow \neg p \lor q$$ Thus: \begin{align*} (p\rightarrow q)\land q &\Leftrightarrow (\neg p \lor q) \land q \\&\Leftrightarrow (q\land \neg p) \lor (q\land q) \\ &\Leftrightarrow (q\land \neg p) \lor q \\ \end{align*} which is not the same statement as the one you proposed. For example, take $q = 1, p = 0$. Then $(q\land \neg p)$ evaluates to 1, while $\neg q$ evaluates to 0.