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Using proof cases with cases $p$ and $~\neg p$, carry out a deductive sequence proof:

$$[(p\to q)\land(r\to q)\land(p\to r)]\to\neg(p\land r\land\neg q)$$

I have been trying for about 30 minutes now, but I do not know how to go about this. I start by assuming $p$ is true, but I'm not sure exactly what this provides for me.

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    Proof by cases ? Assume $(p \& r \& \lnot q)$ and derive a contradiciton, – 2017-02-09

1 Answers 1

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Case p: If $p$, then you can get $q$ ... so it won't be true that $p \land r \land \neg q$ (do a proof by contradiction for that)

Case $\neg p$: Then it won't be true that $p \land r \land \neg q$ (do a proof by contradiction for that)