Indirect proof of universal statement

Question

Can I prove $\neg R(x) \implies \neg R(y)$ by proving $\neg \neg R(y) \implies \neg \neg R(x)$?

Can I also prove that by proving $R(y) \implies R(x)$?

Answer 1

The statement $$\neg q \implies \neg p$$ is called the contrapositive of $$p \implies q.$$ Yes they are equivalent (at least in "classical logic" which it's 100% certain you're studying). What @avid19 says in his answer is true too (proving one of these implications and inferring the other is called proof by contraposition).

Also ("classically"), both $$\neg \neg p \implies p$$ and $$p \implies \neg \neg p$$ are theorems, whatever $p$ might be.

We don't know what system you're working in, what the rules are, etc., so we can't say if you can go from $\neg\neg p$ to $p$ in one step or whether it will take more. You can prove the last two implications mentioned above, and then use modus ponens.

But... do you really need the double negations? If your goal is to prove $\neg R(x) \implies \neg R(y)$, it's surely easier to start with $R(y) \implies R(x)$.

Answer 2

Yes you can, this is called proof by contraposition.