Choose Your Paradox – the downside of the Axiom of Choice
&gt;But it also implies that ZF is consistent. This sounds nice, too, but is actually a disaster. It means that we can’t prove the consistency of AD with ZF (assuming the consistency of ZF).<p>I don&#x27;t think this is strictly speaking correct. For instance, Con(ZF) <i>itself</i> trivially implies that ZF is consistent (which is literally what it says). But we can prove that ZF+Con(ZF) is consistent, assuming that ZF itself is consistent. (For that matter, we can prove that ZF+not(Con(ZF)) is consistent, again assuming that ZF itself is consistent. Don&#x27;t ask.) So just because something implies Con(ZF) doesn&#x27;t mean we can&#x27;t prove that it plus ZF is consistent, assuming ZF is consistent.<p>If ZF is allowed access to Con(ZF) it can prove a great many things that ZF alone cannot, without generating any paradoxes, as far as I know.<p>I may be missing something, this always blows my brain.