Indeed, if we presume ZFC + Continuity to be consistent, then your situation cannot happen. However, it could be that ZFC + Continuity has Goldbach true, and ZF

Indeed, if we presume ZFC + Continuity to be consistent, then your situation cannot happen. However, it could be that ZFC + Continuity has Goldbach true, and ZFC + Forcing has goldbach false. I should say I recall reading that if the Riemand hypothesis is false, then it is provably so in ZFC, hence my inclusion of the Goldbach conjecture. [1, 2] [1] https://mathoverflow.net/questions/79685/can-the-riemann-hypothesis-be-undecidable https://mathoverflow.net/questions/79685/can-the-riemann-hyp... [2] https://www.youtube.com/watch?v=O4ndIDcDSGc&ab_channel=Numberphile https://www.youtube.com/watch?v=O4ndIDcDSGc&ab_channel=Numbe...