How to prove this using propositional logic operations? looks to be equvalent to !p | q
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$\lambda f:P\to Q.\lambda k:\neg Q.\lambda a:P.k(f(a))$ – Derek Elkins left SE Apr 05 '18 at 11:04
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Assume $p$; assume $\lnot q$; derive a contradiction and then derive $\lnot p$ by negation-introduction, discharging temporary assumption $p$. – Mauro ALLEGRANZA Apr 05 '18 at 11:05
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Thanks, it was helpful – Tomasz Kosiński Apr 05 '18 at 12:31