True or false: Suppose $p$ and $q$ are propositions. Then $\lnot(p\implies q) \equiv p \land q.$

I am not very familiar with truth tables but I think that the $$\lnot$$ should get distributed among both $$p$$ and $$q$$ making the problem $$\lnot p \implies \lnot q$$ which does is not the same as $$p\land q$$ making the statement false.

I know that $$\lnot q \implies \lnot p$$ is the contrapositive of $$p \implies q$$ which is also equivalent to $$\lnot p$$ or $$q$$, and if we switch the $$p$$ and $$q$$ it will still make it false.

If anyone can confirm my answer or give more of an explanation that would be great as I am very lost!

Thank you to all of the help in advance, it is very appreciated.

• Note that $p \implies q \equiv \neg p \vee q$. So it's negation is $\neg (\neg p \vee q) \equiv p \wedge \neg q$ by De-Morgan's law. Commented Jul 15, 2020 at 20:16
• @AnuragA Please post your comment as an answer, as it is correct and exactly what I would have answered. (Though I would also have said explicitly that the proposed equivalence by the OP is false.) Commented Jul 15, 2020 at 20:24
• To emphasize a part of @amWhy's comment: $\neg$ does not distribute over $\implies$. Commented Jul 16, 2020 at 17:55

It is false.

Consider when both $$p$$ and $$q$$ are true. Then the RHS is true, whereas, since $$p\implies q$$ is true, the LHS is false.