I am having trouble finding the correct laws to use to prove the following.

$\lnot (p \land q) \land (p \lor \lnot q) \equiv \lnot q$

I understand DeMorgans Law of:

$ \lnot (p \land q) \equiv \lnot p \lor \lnot q$

Am I on the right track? Is DeMorgans Law relevant in this case? I am only just getting into Logical Equivalence etc.

All help is appreciated.


Yes, De Morgan law is relevant in this case, but alone it is not enough, you should use also other rules. More precisely:

\begin{align} \lnot (p \land q) \land (p ∨ \lnot q) &\equiv (\lnot p \lor \lnot q) \land (p ∨ \lnot q) & &\text{by De Morgan law}\\ &\equiv (\lnot p \land p) \lor \lnot q & &\text{by distributivity of $\lor$ over $\land$} \\ &\equiv \lnot q & & \text{by the identity law for $\lor$} \end{align}

We used the identity law for $\lor$ because $\lnot p \land p$ is a contradiction (i.e. a formula that is false for any truth assignment).

A list of the main logical equivalences in propositional calculus is here.

  • $\begingroup$ Thank you very much. Your explanation has helped me greatly! $\endgroup$ – KeptAwakeByCoffee Jul 26 '18 at 10:25

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