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

discrete mathematicslogicpropositional-calculussolution-verification

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.

Best Answer

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.