If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the Goldbach conjecture proved? If ZFC+CH implies Goldbach, and if the Goldbach turn out to be false, then it would mean that ZFC+CH is not consistent, but we know that ZFC+CH is consistent assuming that ZFC is consistent... What do you think?
1 Answer
Because the Goldbach conjecture is an arithmetic statement, it is absolute between any two models which agree on the natural numbers.
Now, given any model of $\sf ZFC$, $M$, there is a forcing extension $M[G]$ with the same ordinals (and in particular, the same natural numbers, which are the just the finite ordinals), in which $\sf CH$ holds. Or, better yet, simply consider $L^M$, which is an inner model with the same ordinals (and, again, the same natural numbers), in which $\sf CH$ holds.
Therefore, if you can prove Goldbach, Riemann, or the ABC Conjecture, assuming $\sf CH$, you may as well have proved it. Using $L$ will also tell you that using the Axiom of Choice was redundant, so in fact the proof is in $\sf ZF$ and not $\sf ZFC$.
So, to sum this up, if you prove that $\sf ZFC+CH$ implies Goldbach's conjecture, and then you prove that Goldbach's conjecture is false, you've proved that $\sf ZF$ is inconsistent. Which, to my taste, is a far bigger result than Goldbach's conjecture (although others may disagree).

1$\begingroup$ As I write in the third paragraph, if ZFC+CH proves Goldbach, then ZF proves it. $\endgroup$– Asaf Karagila ♦Commented Jan 5, 2022 at 15:49

2$\begingroup$ What do you mean by "true", though, if you're not specifying a universe (or at least a metatheory, in which we can interpret "true" as "provable", i.e. independent of the universe)? I feel like your question in this comment is an XY problem. You're asking about X, but you are really interested in an answer about Y. $\endgroup$– Asaf Karagila ♦Commented Jan 5, 2022 at 15:54

2$\begingroup$ I have questions that I don't know if they are the same question. So I was not sure how to formulate my orignial question. Let's say that I know that "if CH+ZFC implies Goldbach and ZFC is consistent, then Goldbach is true". I don't know if this is the same thing as "if CH+ZFC implies Goldbach and ZFC is consistent then there is a formal derivation of Goldbach in ZFC." $\endgroup$– JiuCommented Jan 5, 2022 at 15:59

2$\begingroup$ The standard model of PA depends on the universe you're working in. If you're working in a nonstandard model of ZFC, then its standard model of PA is not necessarily the same as the one of its metatheory. $\endgroup$– Asaf Karagila ♦Commented Jan 5, 2022 at 17:04

3$\begingroup$ @Jiu If your question is whether it is possible that Goldbach is true in the standard model of PA but false in some nonstandard model of PA, then yes, as far as we know, this is possible; it is equivalent to saying that Goldbach is independent of PA, which is unknown. The reason Asaf Karagila asked for clarification is that sometimes people use "real world" to refer to the metatheory. Also you didn't say whether, by "model," you meant model of PA or model of ZFC. $\endgroup$ Commented Jan 6, 2022 at 14:10