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If I prove the implication ¬P ⇒ ¬R∧R, and then I prove the implications ¬R ⇒ Q, R ⇒ ¬Q, is it valid to say ¬P ⇒ Q∧¬Q? I am unsure because while it is the case that Q and ¬Q both follow from P, it seems that we may have assumed that R is both true and not true in order to conclude this last implication, which violates the law of the excluded middle; and constructing a contradiction from a contradiction seems like we have a very un-solid base from which to conclude this last contradiction. We have ¬P ⇒ (¬R ⇒ Q) and ¬P ⇒ (R ⇒ ¬Q), and to hold them simultaneously being true feels counterintuitive for me, but it seems on a logical level this may be the same process we do when we construct ¬P ⇒ ¬R∧R from ¬P ⇒ ¬R and ¬P ⇒ R. However, I would appreciate some justification/clarification on whether we can say ¬P ⇒ Q∧¬Q in this situation or not.

thanks

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  • $\begingroup$ See Ex Falso: with this rule, from a contradiction everything follows. $\endgroup$
    – Mauro ALLEGRANZA
    May 28 at 6:54

1 Answer 1

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If $A\implies B$ and $C\implies D$ then $(A\land C)\implies(B\land D).$

This general rule has no exception. It holds even when $C=\neg A$ and $B=\neg D.$

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