I'm seeing math logic and I have a question.
Let $p$ be a proposition. Let's suppose I have $\lnot p$.
By disjunction rule, this implies $\lnot p \vee q$, where $q$ is any proposition.
This is equivalent (looking at the truth tables) to $p \implies q$
Does this mean that we only have to prove $\lnot p$ in order to prove $p \implies q\;$?