vi21maobk9vp comments on Second-Order Logic: The Controversy - Less Wrong
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So after reading that, I don't see how it could be true even in the sense described in the article without violating Well Foundation somehow, but what it literally says at the link is that every model of ZFC has an element which encodes a model of ZFC, not is a model of ZFC, which I suppose must make a difference somehow - in particular it must mean that we don't get A has an element B has an element C has an element D ... although I don't see yet why you couldn't construct that set using the model's model's model and so on. I am confused about this although the poster of the link certainly seems like a legitimate authority.
But yes, it's possible that the original paragraph is just false, and every model of ZFC contains a quoted model of ZFC. Maybe the pair-encoding of quoted models enables there to be an infinite descending sequence of submodels without there being an infinite descending sequence of ranks, the way that the even numbers can encode the numbers which contain the even numbers and so on indefinitely, and the reason why ZFC doesn't prove ZFC has a model is that some models have nonstandard axioms which the set modeling standard-ZFC doesn't entail. Anyone else want to weigh in on this before I edit? (PS upvote parent and great-grandparent.)
Well, technically not every model of ZFC has a ZFC-modelling element. There is a model of "ZFC+¬Con(ZFC)", and no element of this monster can be a model of ZFC. Not even with nonstandard element-relation.
Linked impressive authority says the model has a ZFC-model-encoding element, plus enough nonstandard quoted ZFC axioms in-model that in-model ZFC doesn't think it's a ZFC-model-encoding element. I.e., the formula for "semantically entails ZFC" is false within the model, but from outside, using our own standard list of axioms, we think the element is a model of ZFC.
Ah, sorry, I didn't notice that the question is about model of ZFC inside a "universe" modelling ZFC+Con^∞(ZFC)