(June 25, 2021 at 5:20 pm)Klorophyll Wrote:

(June 25, 2021 at 5:16 pm)Angrboda Wrote: None of this says anything about free will.  You've created an irrelevant argument about nothing.  P1 being true doesn't mean daughter D has free will.  Nor does P1 and P3 being consistent.

It does. P1 is a sufficient condition for free will. An agent has free will == he's able to choose (at least once in his life) from a set of actions. 

P1 means D freely chose her flavor of ice cream today, therefore, by definition, D has free will.

P1 is not a sufficient condition for free will unless you include the proviso that choosing in P1 refers to freely choosing. However if it does, it begs the question and the argument is invalid. So P1 cannot be a sufficient condition for free will without begging the question. Free will implies that both P1 and not-P1 could be true. However P3 rules out the possibility of not-P1, so P3 rules out free will.

(June 25, 2021 at 5:46 pm)Angrboda Wrote: P1 is not a sufficient condition for free will unless you include the proviso that choosing in P1 refers to freely choosing.  However if it does, it begs the question and the argument is invalid.  So P1 cannot be a sufficient condition for free will without begging the question.  Free will implies that both P1 and not-P1 could be true.  However P3 rules out the possibility of not-P1, so P3 rules out free will.

P1 obviously entails free will. There is no begging involved, I am not trying to prove free will based on P1, I already assumed D has free will by asserting P1.

And of course P1 is not a necessary condition of free will, let's not play this silly game, I am giving one example of free will (choosing some flavor of ice cream), obviously one can have free will and never eat ice cream in their life.

You say "P3 rules out free will", which is a claim you didn't prove. P3 is exactly equivalent to P2 as I proved before. omniscience mean knowledge of all true propostions. If P3 is true, then P1 is true at all times, and a fortiori, ten years ago, therefore P2. Inversely if P2 is true, then because God is omniscient, God knows about P1, therefore P3.

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(June 25, 2021 at 5:34 pm)The Grand Nudger Wrote: I don't feel the need to repeat a simple explanation for how free will and foreknowledge are, by definition, mutually exclusive.  

If you believe in the author of creation, and you've said that you do - you believe in fate.  You may have negative ideations about "fatalists" - as your shaman tells you they are, but by expressing such a position you have identified yourself as a fatalist no matter how much you think fatalist heretics deserve the sword.  

Know thyself.

In truth, whether we have free will or there is some author or knower of creation is unimportant to me.  Neither state of affairs has anything to say about why I don't bend the knee to your trashgod.

If we don't have free will...I just can't worship trashgod, because I'm not enough of a garbage person to do so.
If we do have free will, and even if I were a garbage person, I still have the constitutional volition to say no to at least one bad thing.

Worshipping a deity is an entirely separate issue. But if a deity exists and instructs us to worship it, we can safely assume it's also a moral authority, which means worship is a good moral imperative by definition.

(June 25, 2021 at 6:13 pm)Klorophyll Wrote:

(June 25, 2021 at 5:46 pm)Angrboda Wrote: P1 is not a sufficient condition for free will unless you include the proviso that choosing in P1 refers to freely choosing.  However if it does, it begs the question and the argument is invalid.  So P1 cannot be a sufficient condition for free will without begging the question.  Free will implies that both P1 and not-P1 could be true.  However P3 rules out the possibility of not-P1, so P3 rules out free will.

P1 obviously entails free will. There is no begging involved, I am not trying to prove free will based on P1, I already assumed D has free will by asserting P1.

And of course P1 is not a necessary condition of free will, let's not play this silly game, I am giving one example of free will (choosing some flavor of ice cream), obviously one can have free will and never eat ice cream in their life.

You say "P3 rules out free will", which is a claim you didn't prove. P3 is exactly equivalent to P2 as I proved before. omniscience mean knowledge of all true propostions. If P3 is true, then P1 is true at all times, and a fortiori, ten years ago, therefore P2. Inversely if P2 is true, then because God is omniscient, God knows about P1, therefore P3.

If P1 assumes free will, the argument is invalid. Let me put it another way. Let P1(b) be the proposition that the choice in P1 is fully determined and not free. Now one of two things is true: a) P1(b) is consistent with P1 and therefore P1 is not about free will and the argument fails, or b) P1(b) is inconsistent with P1 and therefore P1 assumes free will, thus begging the question, and making the argument invalid. You can't simply "assert" free will.

You don't seem to have the first clue about free will. Color me not surprised.

(June 25, 2021 at 6:32 pm)Angrboda Wrote:

(June 25, 2021 at 6:13 pm)Klorophyll Wrote: P1 obviously entails free will. There is no begging involved, I am not trying to prove free will based on P1, I already assumed D has free will by asserting P1.

And of course P1 is not a necessary condition of free will, let's not play this silly game, I am giving one example of free will (choosing some flavor of ice cream), obviously one can have free will and never eat ice cream in their life.

You say "P3 rules out free will", which is a claim you didn't prove. P3 is exactly equivalent to P2 as I proved before. omniscience mean knowledge of all true propostions. If P3 is true, then P1 is true at all times, and a fortiori, ten years ago, therefore P2. Inversely if P2 is true, then because God is omniscient, God knows about P1, therefore P3.

If P1 assumes free will, the argument is invalid.  Let me put it another way.  Let P1(b) be the proposition that the choice in P1 is fully determined and not free.  Now one of two things is true: a) P1(b) is consistent with P1 and therefore P1 is not about free will and the argument fails, or b) P1(b) is inconsistent with P1 and therefore P1 assumes free will, thus begging the question, and making the argument invalid.  You can't simply "assert" free will.

You don't seem to have the first clue about free will.  Color me not surprised.

Yes I can assert free will. Because this is not an argument for free will, but for compatibility between foreknowledge and free will. Meanwhile, keep playing mind games with P1.

(June 25, 2021 at 6:37 pm)Klorophyll Wrote:

(June 25, 2021 at 6:32 pm)Angrboda Wrote: If P1 assumes free will, the argument is invalid.  Let me put it another way.  Let P1(b) be the proposition that the choice in P1 is fully determined and not free.  Now one of two things is true: a) P1(b) is consistent with P1 and therefore P1 is not about free will and the argument fails, or b) P1(b) is inconsistent with P1 and therefore P1 assumes free will, thus begging the question, and making the argument invalid.  You can't simply "assert" free will.

You don't seem to have the first clue about free will.  Color me not surprised.

Yes I can assert free will. Because this is not an argument for free will, but for compatibility between foreknowledge and free will. Meanwhile, keep playing mind games with P1.

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.

Is P2 and P3 true if she chooses something else? No they are not. So she couldn't have chosen from any "set" of actions and still be consistent with P2 and P3. Her choice is thus determined.

(June 25, 2021 at 6:46 pm)Angrboda Wrote:

(June 25, 2021 at 6:37 pm)Klorophyll Wrote:

Yes I can assert free will. Because this is not an argument for free will, but for compatibility between foreknowledge and free will. Meanwhile, keep playing mind games with P1.

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.

(June 25, 2021 at 6:46 pm)Angrboda Wrote: Is P2 and P3 true if she chooses something else?  No they are not.  So she couldn't have chosen from any "set" of actions and still be consistent with P2 and P3.  Her choice is thus determined.

I think you have an issue with something else - the compatibility between P1 and P2. This is known as logical fatalism, it can be found in the literature that it has been resolved, I will try to find a reference to a solution. 

My arguments rests on the equivalence between P2 and P3. The compatibility between P1 and P2 is proven, but I'll need to think futher before giving a detailed answer.

(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?

I have a PS4 but want a PS5.

Am I doing this right?

(June 25, 2021 at 7:03 pm)Angrboda Wrote: 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.

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(June 25, 2021 at 7:03 pm)Angrboda Wrote: Answer me this, can she choose any other flavor than F and P2/P3 still be true?

Your question is the object of the entire argument. So yes, she can. To see why, you have to follow all the steps above, I admit that it's not clear that she can.

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(June 25, 2021 at 7:05 pm)Foxaire Wrote: I have a PS4 but want a PS5.

Am I doing this right?

I have a not-PS

(June 25, 2021 at 7:11 pm)Klorophyll Wrote:

(June 25, 2021 at 7:03 pm)Angrboda Wrote: 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.