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Working in ZFC.

I've defined a function-like binary predicate $R$ on a proper class. It has to be recursive; i.e. $R(a,b)$ must usually depend on one or more $R(c,d)$ for some $c$s and $d$s calculated from $a$ and $b$. There is, however, no well-founded recursive, unary function that agrees with $R$ (i.e. there is no $F$ such that $y = F(x)$ iff $R(x,y)$).

So I've defined it like ZFC defines $\in$: I've added a new binary predicate symbol $R$ and defined axioms for it. Every axiom conforms to the schema $\forall x_1 . \dots \forall x_n . P_1(x_1) \wedge \dots \wedge P_n(x_n) \rightarrow Q(x_1,\dots,x_n) \rightarrow R(x_1,x_2)$. I think the salient facts are 1) that no axiom concludes that a set exists, only that $R$ relates two sets; and 2) $Q(x_1,\dots,x_n)$ is a formula that usually refers to $R$.

I'd like to prove that ZFC+R proves that no more sets exist than ZFC does. How can I do this?

(Also, is "conservative extension" the right term for this?)

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    Is it the case that in every model of ZFC you have an interpretation of $R$?2011-04-08
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    I'm sorry, but I don't quite understand what you mean by an interpretation of $R$. I'm teaching myself model theory as I need it, so I probably missed something fundamental. (If I'm reading you right, though, I don't think it's possible to translate sentences in ZFC+R directly into ZFC, since those $Q(x_1,\dots,x_n)$ are formulas that refer to $R$. I'll edit the question to make that clear, even if that's not what you're getting at.)2011-04-08
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    Here is a concrete example, to clarify what I mean: Say that $R$ is a well-ordering of the whole universe. Then it is not provable that there is such a class $R$ (in ZFC, say, were `classes' ought to be definable from parameters). However, using class forcing, you can add to any model of ZFC a predicate $R$ that is such a global well-ordering, and in a fashion that adds no new sets.2011-04-08
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    The point of the above is this: Suppose the theory $T=$ZFC+"$R$ is a well-ordering" proves a sentence $\phi$ in the language of set theory. Pick any model of ZFC. We can extend it to a model of $T$ in the manner described in the previous comment. In the extended model, $\phi$ holds. But that means that $\phi$ holds in the original model, because adding $R$ didn't add any sets. By the completeness theorem, $\phi$ is provable, because it is true in all models.2011-04-08
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    Oh, and yes: If $T$ has the property that any sentence in the language of ZFC it proves is actually ZFC-provable, then $T$ is said to be conservative over ZFC.2011-04-08
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    Wow, I didn't think the answers would go this far past my current knowledge... :) I've been googling "class forcing" and I think I get it now. So, to use class forcing, I would have to define $R$ without parameters. (Done.) $R$ would also have to be a partial order. Its reflexive closure is a partial order; call it $R'$. I think I could use $R'$ w.l.o.g. And then I would prove $R'$ is tame, therefore it preserves ZFC; QED. Is that right?2011-04-08
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    @Neil: Yes, that would be the approach. A different example in the same spirit is that Goedel-Bernays is conservative over ZF. The point is that given a model of ZF, we can extend it to a model of GB by taking as proper classes of the extension only the definable classes of the original model. But then, exactly the same argument as for the other example shows that we have conservation.2011-04-08
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    @Andres: Perhaps promote some of the comments here to an answer so the question will get off the unanswered list?2013-07-11
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    @AsafKaragila Sure. I'll post in a few minutes.2013-07-11
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    @YuvalFilmus Since you are the creator of the (transfinite-recursion) tag, I though it might be polite that I have started [a discussion on meta](https://math.meta.stackexchange.com/questions/28242/on-the-tags-transfinite-induction-and-transfinite-recursion) related to this tag.2018-04-15

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