# Why need both converse and inverse in proofs.

Given the implication, If $p$, then $q$, translates to $p \implies q$. It has the truth table that makes the meaning/interpretation as: $\lnot p \lor q.$

If need find the converse, then it is given by: If $q$, then $p$, that translates to $q \implies p$, and interpretation in logic as $\lnot q\lor p.$

If need find the inverse, then it is given by: If not $p$, then not $q$, that translates to $\lnot p \implies \lnot q$, and interpretation in logic as $\lnot q \lor p.$

So, the converse and inverse are logically equivalent. Then, is it an ease in constructing proofs that necessitates the need for both converse & inverse to be used?

For example, if given that: if $p_1 \wedge p_2$, then $q$, then for finding the converse need prove:

if $q$, then $p_1 \wedge p_2$;

while the equivalent logic is provided by proving inverse as:

if $\lnot(p_1 \wedge p_2)$, then $\lnot q.$

"if $q$, then $p_1 \wedge p_2$" $\equiv$ "if $\lnot(p_1 \wedge p_2)$, then $\lnot q$".

So, is it the ease which allows us to choose which form to take for taking converse, to check if inverse is possible or not. If yes, then in which situations (like, possibly for compound propositions), this ease is helpful.

• I assume you know about the contrapositive of a logic statement. That is what gets commonly exploited in proof writing. – Jonathan Feb 15 '18 at 13:51
• @JonathanBrown Thanks for pointing that. I would request however, some examples to support my post. – jitender Feb 15 '18 at 13:53
• Just to be clear, you are asking for examples of proofs which assume $\lnot p$ and use that to prove $\lnot q$, and for examples of other proofs which assume $q$ and use that to prove $p$. Correct? – Lee Mosher Feb 15 '18 at 14:14
• Following up on the comment of @JonathanBrown, what have you found by looking for proofs in, say, your calculus book, which have the form $p \implies q$ and which are proved by assuming $p$ and proving $q$, and examples of other theorems of the same form but which are proved by assuming $\lnot q$ and proving $\lnot p$? It should be easy for you to find both. – Lee Mosher Feb 15 '18 at 14:51
• It is worth noting that in constructive/intuitionistic logic we do not have $(P\implies Q)\equiv(\neg Q\implies\neg P)$ (though we do have the left-to-right entailment). In such logics, it could be the case that $\neg Q\implies\neg P$ is provable while $P\implies Q$ is not. This in some sense indicates that the "inverse" of an implication is "easier" to "directly" prove than its converse. However, a "direct" proof of the converse may be more informative then a "direct" proof of the "inverse". – Derek Elkins Feb 15 '18 at 16:48