TITLE: If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the conjecture proved?
QUESTION [14 upvotes]: 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?

REPLY [32 votes]: 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).