Well, there's an informal notion that if two theorems T1 and T2 are both true, yet T1 <=> T2 is much easier to prove than either T1 and T2, then the two are equivalent. (There's also the formal notion that two axioms are equivalent if assuming either one lets you prove the other, but I don't think that's especially relevant here.) That's not close to being the most confused part of the comment you're replying to.