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For an absolute Function F (meaning that the formula y=F(x) is absolute), is there a proof that $F(x) \in z$ is an absolute Formula? This Lemma from Kunen applied for $\Phi(w,z)$ being $w \in z$ and $G_1=F$ says there should be: [![Lemma from Kunen][1]][1]

However the proof of the lemma is not convincing so i tried to write one out for this specific example but i failed. Please help me with this example, so i can understand the lemma. [1]: https://i.stack.imgur.com/gLch5.png

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    $\begingroup$ $F(x)\in z$ is equivalent to $\forall y (y=F(x)\to y\in z)$ and $\exists y (y=F(x)\land y\in z)$, so $F(x)\in z$ is absolute between two models in which $\forall x\exists! y F(x)=y$ hold. $\endgroup$
    – Hanul Jeon
    Feb 21 at 23:41
  • $\begingroup$ @Hanul Jeon I agree to your unabbreviations of $F(x) \in z$. But how does absoluteness follow from that. $\endgroup$
    – Rubids
    Feb 21 at 23:48
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    $\begingroup$ The first one gives you the upward absoluteness, and the second one gives you the downward absoluteness. $\endgroup$
    – Hanul Jeon
    Feb 22 at 0:03
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    $\begingroup$ The proof that these are absolute uses exactly the same steps as the proof that ($\Pi$) $\Sigma$-formulas are (downwards) upwards absolute. You might want to look into that if you have never seen it. $\endgroup$ Feb 22 at 0:15
  • $\begingroup$ Thank you both, I was now able to solve it. My Mistake was to try to do it only by equivalence from the existential version. Small sidenote: The first one (forall version) in Hanuls first comment gives the downward absoluteness (not the upward) and the second the upward. $\endgroup$
    – Rubids
    Feb 22 at 0:39

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