Why God doesn't stop satan?

[Angrboda]

The question at issue is whether P1 AND not-P1 are BOTH consistent with P2 and P3.  They cannot be consistent with both, therefore P1 is necessarily true.  If P1 is necessarily true, then she necessarily chose flavor F and her choice was not free.  Any choice that is necessarily the case is not a free one.   Since both P1 and not-P1 cannot both be consistent with P2/P3, her choice was not free, P1 is not sufficient for free will, and free will is inconsistent with them.

Okay this may get messy. Let's take it one at a time. As I suspected, you're not in disagreement with me that P2 and P3 are logically equivalent. You instead have a problem with (P1 compatible with P2/P3). Is this your point ?

I am not sure you're being precise here, obviously a (proposition AND its negation) can't be consistent with ANYTHING.

You're right, I haven't been sufficiently clear. Let me try to clarify.

They can't both be true in the same world, but they can be consistent with many things and they are both consistent with many possible worlds (God exists in all possible worlds). P1 and not-P1 are not consistent with each other, but both are consistent with her having a free choice. Free will necessarily entails that a proposition and its negation both "could be true", not that they are both true. It simply says that both P1 and not-P1 are possible. But if not-P1 is possible, then P2 and P3 are possibly false. But that's not possible, because P2/P3 are necessarily true. There's your inconsistency.

Let me illustrate:

P1 == Daughter D could choose flavor F
not-P1 == Daughter D could choose a flavor other than F

P1 and not-P1 are consistent with each other if and only if free will exists;

But, Daughter D cannot choose a flavor other than F if God knows that daughter D chooses F;

Therefore, if God knows that she chooses F, then Daughter D cannot choose a flavor other than F [not-P1] (as doing so violates his omniscience);

But, not-P1 and P1 must both be true and consistent for free will.