TITLE: If $\phi$ holds for all standard models of ZF and ZF proves this, then does ZF prove $\phi$?
QUESTION [5 upvotes]: I apologize if this is a nonsensical question. 
Suppose $\phi$ holds in all standard models of ZF. Suppose further that ZF proves this. Then does ZF prove $\phi$?

REPLY [8 votes]: No (assuming there are standard models at all). $\operatorname{Con}\sf(ZF)$ is always true in standard models, and $\sf ZF$ proves that $\operatorname{Con}\sf(ZF)$ is always true in standard models. But of course that $\operatorname{Con}\sf(ZF)$ is not provable from $\sf ZF$ itself. 
The reason is that standard models agree with the universe about the natural numbers, and therefore about basic arithmetic statements like the example above. 
If there are any models of $\sf ZF$, then $\operatorname{Con}\sf (ZF)$ is true. And so, if there are no standard models at all, then vacuously, $\operatorname{Con}\sf (ZF)$ holds in all of them; and if there is a standard model, then $\operatorname{Con}\sf (ZF)$ is true in the universe, and therefore in the standard model. So indeed $\sf ZF$ proves that in any standard model of $\sf ZF$, $\operatorname{Con}\sf (ZF)$ must be true.