In "The Modal Logic of Forcing", Joel David Hamkins and Benedikt Löwe show that the ZFC-provable forcing principles are exactly those of the modal logic S4.2 (interpreting $\Diamond \phi$ as asserting that $\phi$ is forceable).

Do the $\mathsf{ZF}$-provable principles of forcing obey a different modal logic?

  • $\begingroup$ You're asking two different questions here. $\endgroup$
    – Asaf Karagila
    Commented Nov 27, 2015 at 20:43
  • $\begingroup$ @AsafKaragila Removed the second and reworded the first. $\endgroup$
    – Dennis
    Commented Nov 27, 2015 at 20:46
  • 2
    $\begingroup$ (Let me remark on the second that forcing in ZF is the same as forcing in ZFC, only you don't have the mixing lemma (or maximality principle or fullness (it has many different names)); and chain conditions can and will fail you; and other usual things may fail. But the essence of forcing does not hinge on the axiom of choice.) $\endgroup$
    – Asaf Karagila
    Commented Nov 27, 2015 at 20:47

1 Answer 1


The definition of forcing is the same with and without the axiom of choice. And the truth lemma holds with and without the axiom of choice. Namely,

$$p\Vdash_\Bbb P\varphi\iff\text{For every }V\text{-generic } G\subseteq\Bbb P\text{ with }p\in G: V[G]\models\varphi$$

You can also consider iterations without choice, at least with a two-step iteration this holds absolutely no difficulties (compared to all sort of non-finite supports and so on). So the proof that $\sf S4.2$ is a subset of the modal logic of forcing in $\sf ZF$ is immediate.

In the other direction, clearly $\sf ZF$ cannot prove more than $\sf ZFC$. So you get $\sf S4.2$ immediately as a result.

  • 1
    $\begingroup$ Simple and to the point. Great answer. $\endgroup$
    – Dennis
    Commented Nov 27, 2015 at 21:08
  • $\begingroup$ You're welcome. $\endgroup$
    – Asaf Karagila
    Commented Nov 27, 2015 at 21:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .