Consider we have the following statement to prove: $P \implies Q \wedge R$. For a proof by contradiction, we assume $P \wedge (\neg Q \lor \neg R)$.

How would one go about this? Typically to prove a statement of the form $(A \lor B) \implies C$, we show $A \implies C$ and then $B \implies C$. Do we break down such a proof by contradiction into such cases?


  1. Does it suffice to reach separate contradictions by considering the cases $P \wedge \neg Q$ and $P \wedge \neg R$ separately?

  2. Must one also consider the case when $(P \wedge \neg Q)$ and $(P \wedge \neg R)$ are both assumed to hold "simultaneously"?

Many thanks in advance for your help.

  • $\begingroup$ If you want to derive a contradiction (if any) from $P \wedge (\neg Q \lor \neg R)$, you have to consider the two cases forming the disjunction. $\endgroup$ – Mauro ALLEGRANZA Jan 26 '20 at 9:54
  • $\begingroup$ @MauroALLEGRANZA, yes that I know - I meant it 'loosely' i.e. in the context of the proof technique. Thank you for pointing it out - removed the sentence for clarity. $\endgroup$ – Xandru Mifsud Jan 26 '20 at 9:55
  • $\begingroup$ @MauroALLEGRANZA, so if I understood correctly, you first show contradiction of $P \wedge \neg Q$ independent of $P \wedge \neg R$, and then similarly for $P \wedge \neg R$ independent of $P \wedge \neg Q$. $\endgroup$ – Xandru Mifsud Jan 26 '20 at 9:57
  • $\begingroup$ See Proof by cases: if you show that (P∧¬Q) implies a contra and that (P∧¬R) implies a contra, it is fine. $\endgroup$ – Mauro ALLEGRANZA Jan 26 '20 at 9:58
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    $\begingroup$ Thanks @MauroALLEGRANZA, your last comment clarifies the matter for me. Please, how can I mark it as solved by you? $\endgroup$ – Xandru Mifsud Jan 26 '20 at 10:00

If we want to derive a contradiction (if any) from $P ∧ (¬Q∨¬R)$, we have to consider the two cases forming the equivalent disjunction:

$(P ∧ ¬Q) ∨ ( P ∧ ¬R)$.

See Proof by cases: if we show that $(P∧¬Q)$ implies a contradiction and that $(P∧¬R)$ implies a contradiction, it is done.

  • $\begingroup$ You should have said or, not and. $\endgroup$ – Ben W Jan 28 '20 at 20:13
  • $\begingroup$ @BenW --- NO. Se the Wiki entry regarding Didjunction Elim linked above: if we have $P \to R$ and $ \to R$, we can conclude that $R$ follows from $P \lor Q$. $\endgroup$ – Mauro ALLEGRANZA Jan 28 '20 at 20:18

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