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What is the least theory in which the following sentence is proved?

$ \exists M: M\text { is CTM(ZFC+ GCH)} \land \forall \kappa \in Card^M (\kappa > 1 \implies \\\exists N: N \text { is CTM(ZFC) } \land Card^N=Card^M \land \operatorname\neg \operatorname {CH}^N(\kappa))$

Where $Card^S=\{\kappa : S \models \exists x \, (\kappa=|x|)\}$

$\operatorname {CH}(\kappa) \iff \not \exists \lambda: \kappa < \lambda < 2^\kappa$

"$\operatorname {CH}^N..$" is the relativization of all quantifiers in the open expansion of "$\operatorname {CH}..$" by "$\in N$".

In English: There exists a countable transitive model $M$ of ZFC +GCH such that for every $M$-cardinal $\kappa$ (bigger than 1) there is a countable transitive model $N$ of ZFC such that the set of all $N$-cardinals is the set of all $M$-cardinals (so existence of injections between cardinals in both models is preserved), and such that $\kappa$ violate the continuum hypothesis in $N$.

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    $\begingroup$ Doesn't the standard forcing argument show that this holds as soon as there is some CTM of ZFC? $\endgroup$
    – Wojowu
    Commented Apr 27, 2022 at 20:51
  • $\begingroup$ @Wojowu, I'm not sure of that. This is my guess, that's why I asked this question. So, if this is correct then it is ZFC+ there is some CTM of ZFC? Or is it something more than that? $\endgroup$ Commented Apr 27, 2022 at 21:19
  • $\begingroup$ It should be literally just ZFC+there is some CTM of ZFC. Then you should be able to make GCH fail at $\kappa$ by adding more than $\kappa$ reals to the model, and this should moreover not change any cardinals. The condition on injections is equivalent to that, I don't know why you added it separately. I am not that well-versed in forcing so I will let someone more knowledgeable confirm all this and post an answer. $\endgroup$
    – Wojowu
    Commented Apr 27, 2022 at 21:41
  • $\begingroup$ @Wojowu, you mean the condition on injections is redundant? $\endgroup$ Commented Apr 27, 2022 at 22:13
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    $\begingroup$ You are literally requiring a CTM, so, yes. But it's just Cohen/Cohen-style forcing. Nothing more than that. $\endgroup$
    – Asaf Karagila
    Commented Apr 28, 2022 at 12:36

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