RE: Why God doesn't stop satan?
June 25, 2021 at 7:03 pm
(This post was last modified: June 25, 2021 at 7:11 pm by Angrboda.)
(June 25, 2021 at 6:58 pm)Klorophyll Wrote:(June 25, 2021 at 6:46 pm)Angrboda Wrote: And as I pointed out, her choice wasn't free unless she could have made a diffferent choice. But if she made a different choice, then P2 and P3 are false. That's an inconsistency. If P1 assumes free will, it's not necessarily consistent with P3 and the argument fails. All you're doing is dressing up the assertion that free will and omniscience are compatible in logic -- it remains a bald assertion unless you demonstrate that free will can exist after assuming omniscience. P1 doesn't demonstrate free will.
I think you should re-read the argument, you're just confused now. I am not trying to demonstrate free will, at all. I assert there is an agent D with free will, I assert there is a foreknower then prove by equivalence that these two assertions are logically compatible.
I know that the consistency between P1 and P3 is not clear, that's what I and Nudger are arguing about. But proving that P2 and P3 are logically equivalent is really easy.
Once this is done, the argument is as follows : If (P2 is logically equivalent to P3) AND (P1 and P2 are logically compatible) THEN (P1 and P3 are logically compatible). QED.
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.
As noted originally, P1 says nothing about free will. You're just dressing up an assertion in logic.
Answer me this, can she choose any other flavor than F and P2/P3 still be true?
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