Logical equivalence of ¬p→q Just wondering what other ways $\neg p \to q$ can be expressed.
I know that $p\to q$ is logically equivalent to $\neg p\lor q$, hence I think that $\neg p\to q$ has the same logical equivalence as $p\lor q$.
 A: As others have pointed out, truth tables will confirm the equivalence of $\neg p \to q$ and $p \lor q$.
But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from  $p \lor q$ you can infer $\neg p \to q$.
Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$.  So, in symbols, from $\neg p \to q$ you can infer $p \lor q$.
A: You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.
In this case, $(\lnot p) \rightarrow q \equiv \lnot(\lnot p) \lor q \equiv p \lor q$, as you stated correctly.
A: As you said, $p\Rightarrow q$ is logically equivalent to $\neg p\vee q$.
Then by double negation, $\neg p \Rightarrow q$ is logically equivalent to $\neg(\neg p)\vee q$, which amounts to $p\vee q$.
