EPTCool stuff. I think we have free will, logical determinism be damned.
Formal Proof About Classical Free Will - Page 5
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hp.Shell
United States2527 Posts
Cool stuff. I think we have free will, logical determinism be damned. | ||
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spinesheath
Germany8679 Posts
On May 29 2014 09:09 Lixler wrote: TL;DR of the following: for your reductio, you assume that the antecedent of the conditional is true. but the conditional can be true without the antecedent being true. if we assume, as we logically ought to, that our conditional as a whole is true, no contradiction follows. you have simply proven that the antecedent must be false (viz. that B could not be impossible) let's fix your formalization somewhat: instead of using B, we'll just use ~A. we'll also introduce quantificational and modal symbols so classical free will is: a -> ◇~a or more properly, since classical free will only says that some of our actions are freely willed: Ǝa (a -> ◇~a) the contrapositive (for our assumed a) is now ~◇~a -> ~a or □a -> ~a we can conclude a few things from this, but first let's run through your logical proof with our new symbols. our assumption, in attempt of the reductio, is that ~a was not possible, or rather that □a. from □a -> ~a we conclude both (this is trivially true) a ~a and obviously this is a contradiction. well, what went wrong? this: in attempt to prove that your conditional led to a contradiction, you assumed that the antecedent was true. but we don't assume the truth of our antecedent in order to prove that our conditional was false. we assume that our conditional was true! and there are three ways for our conditional to be true. they are, as follows □a ^ ~a (which leads to a contradiction) ~□a ^ ~a ~□a ^ a now our second set of truth values would lead to a contradiction, since we already assumed that a (we assumed there was some action such that we could have not done it). we are forced to conclude that ~□a ^ a. unfortunately for your proof, these two values do not lead to a contradiction. Look, I really don't know this notation and didn't bother trying to figure it out, but I think it's just saying this: On May 29 2014 06:17 spinesheath wrote: So you have a contradiction if you assume that there is some B that was not possible. Then obviously everything is possible. | ||
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zf
231 Posts
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bookwyrm
United States722 Posts
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Boblion
France8043 Posts
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Deleted User 137586
7859 Posts
On May 29 2014 09:29 MichaelDonovan wrote: a) It is at this point in the proof where I am able to get rid of the need for any modal operators. I only need ~<>B* to show ~B*. And yes, I should have included that somewhere in my proof, but I am assuming that B being impossible entails that B* cannot be true. b) Sure, I don't see any problem with saying it like that. Modal logic gets kind of messy though sometimes, so I wanted to try to do away with it. c) This objection would be problematic if it were not for the fact that all of my statements in this arguments are based only on the logical structure of the proposition. That is, we don't really care if it makes sense empirically. What we're showing is that the classical free will thesis is logically incoherent. This doesn't say much about free will. It just says that the way the classical thesis is written leads to contradictions. So, the fact that being blown up this morning makes it impossible for me to be sitting in my office isn't really important. If it were, we would include that statement in our argument, and the contradiction would be clear. I'm not sure if this a sufficient response to your objection though, so hammer it a bit more if you're dissatisfied. d) That's a fun example. I'll have to think about that for a bit. But I don't think it's a problem for my argument because we either have free will or we don't. We either do action A or we do action not A. It's not like If you do action B, then you probably didn't do A. Or something like that. When you have mutually exclusive sets, the probability of their intersection is zero since their intersection is the empty set. I dunno if that addresses the problem completely, so again, hammer it some more if you're not satisfied. Let's address (b) as that's the most important one. Diamond A wedge Diamond B isn't a contradiction, you won't be able to derive Diamond B and ~Diamond B from it unless you introduce a formula which is a contradiction itself. But then the problem is with that assumption and not free will. Basically, as others have said, your formalism fails because you're allowing moves which modal logic would not allow. *** Regarding (c), I can say that free will is linked to causation. We can either influence the chain of events or we cannot. If you represent causation with material implication, you can derive things which we know do not correspond to causation itself. This is basically the same as (d), as the logical formalism makes counter-intuitive predictions about causation, we should identify the problem with the logic, and not with free will. And besides not using modal logic, the biggest problem is the use of material implication instead of more refined accounts of implication such as strict implication, relevant implication, or Kratzer's context-sensitive conditionals. | ||
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radscorpion9
Canada2252 Posts
I think the biconditional statement needs to be modified, because as it stands it assumes that A* and B* are both possible. But the contrapositive of your proposition indicates that if B* is impossible, then A* is impossible. So there should be the possibility that A* and B* are both impossible. So to be clear, B is any distinct action that is not A. The biconditional should be modified to this: If B and A both exist as possible actions such that B* and A* are both possible, then: (3) B* <-> ~A* This way it makes sense, because B needs to be possible before an agent performs the action (B*), and the contrapositive requires that A be possible before A* is. So then when you start with the assumption that B is impossible, you do indeed get ~B* as a consequence, but before you continue on to using the biconditional I think you would need to establish that A is actually possible (and consequently A*), which you can't really get from anything else unfortunately. Thus it would be invalid to use the biconditional in your fifth step, and the argument just ends there uncompleted. So that's the main thing: The biconditional assumes both actions are always possible, while the contrapositive to your proposition considers the possibility that both may be impossible. The biconditional thus needs to be modified to allow for these possibilities edits: Grammar. By the way I hope you enjoyed your counterstrike session!! I used to love that game | ||
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Lixler
United States265 Posts
On May 29 2014 14:46 spinesheath wrote: Look, I really don't know this notation and didn't bother trying to figure it out, but I think it's just saying this: On May 29 2014 20:47 zf wrote: Lixler pointed out the major problem with your proof - namely, you're using first order logic, not modal logic. But if you do try to rewrite it in modal logic, I'd be wary about using contraposition. Ernest Sosa has a famous argument against contraposing subjunctive conditionals, which I think applies to your case as well. I don't think I'm saying either of these things. Zf's point is fairly obvious - we ought only deal with possibility using modal logic - but the exact problem I'm pointing out in OP's proof is not rooted specifically in his choice of logic. Specifically, he sets up his reductio falsely. Quoting it now for ease of reading: Contrapositive of the proposition: If it was not possible for the agent to have done B, then the agent did not do A. We will attempt to disprove the proposition by showing that its logically equivalent contrapositive leads to a contradiction. So we assume (for reductio) that it was not possible for the agent to have done B. If the proposition is true, then it should follow that the agent did not do A. We will show that this is not the case. When we attempt a reductio of some proposition, we assume that proposition is true and work to derive a contradiction. OP thinks he is doing this when he assumes "that it was not possible for the agent to have done B," but in fact he needs to assume "If it was not possible for the agent to have done B, then the agent did not do A." There are three possible ways for this latter conditional to be true. The first, which OP demonstrates is contradictory, is for it to have been possible for the agent to have done B and for the agent to not have done A. The second, which also will not work (as I demonstrated in the symbolization) is for it to have been possible for the agent to have done B and for the agent not to have done A. The third, which leads to no contradiction, is for it to have been possible for the agent to have done B and for the agent to have done A. When these two statements hold (i.e. when the antecedent of the conditional is false while its consequent is true), the conditional is true (as we assumed), yet no contradiction follows. So OP has failed to demonstrate that the classical free will thesis is contradictory. The specific move where OP goes wrong involves a fairly common mistake made by people learning formal logic. Namely, he assumes that the only way for a conditional to be true is for both the antecedent and the consequent to be true. But this is not so. As it turns out, we can derive a contradiction from any conditional and its contrapositive if we don't assume this (specifically because the equivalence of a conditional with its contrapositive relies on the fact that the conditional is true when both its parts are false, but whatever). Let's take the following example. If I love my wife, I will buy her a gift for our anniversary. Contrapositive: If I will not buy my wife a gift for our anniversary, I do not love her. Now let us assume 1) that a conditional can only be true when both its parts are true and 2) that our contrapositive is true. This means that we have to assume that it is true 1) that I will not buy my wife a gift for our anniversary and 2) that I do not love her. But if both these are the case, and it is further the case that a conditional can only be true when both parts are true, then it follows that our initial conditional is false. But this leads us to a contradiction, since the contrapositive (which we assumed to be true) is supposed to be logically equivalent to the initial conditional, i.e. true when it is true, false when it is false. We cannot simultaneously maintain that a conditional is only true when both its parts are true and that a conditional is logically equivalent to its contrapositive. Since OP does this in his proof, it is invalid. Indeed, we could prove anything we wanted if we tried to maintain both these things. Hopefully it is clear now where and why OP's argument goes wrong, although obviously there are problems with his symbolization that don't have anything to do with my qualm. | ||
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MichaelDonovan
United States1453 Posts
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EatThePath
United States3943 Posts
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Nemesis
Canada2568 Posts
On May 29 2014 10:22 2Pacalypse- wrote: Uhm, no, it's the other way around. When we demonstrated with empirical evidence that the electron can be at the two places at once, something that seems logically impossible, we didn't dismiss it due to our sense perception misleading us. Sure, you can try to "update" logic to fit with the reality, but you can't update reality to fit with your logic. While I think that what he said is incorrect about dismissing empirical evidence over logic, quantum mechanics is not really illogical. It has its own rules which it follows, which just seems counter intuitive to everyday human experiences. Logic is simply a way of connecting one statement to another. It's not that logic dictates that it's impossible for the electron to be at two places at once, it's the assumption of how a particle should act that is wrong. Initial assumptions from which the logic is derived can be wrong, but logic itself is not wrong. But at the same time, logic is just a play on words and doesn't really apply to reality unless you actually have something like empirical evidence to start from which you know is correct in the real world. What I'm talking about is really all kinds of logic such as mathematical logic, logical probabilism, and whatever other kinds of valid logic you can think off. If he's only talking about classical logic, then sure feel free to bash him  | ||
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ninazerg
United States7291 Posts
On May 29 2014 09:45 MichaelDonovan wrote: Well... I'm not really satisfied with the way I worded it in plain English. It's kind of hard to say it properly because it's not all intuitive. Trying to explain why 1 + 1 = 2 in English without using formal logic to derive this truth from definitions leads to an explanation that won't really be satisfying and is probably full of holes. And I'm not sure I put the logic into words correctly there either... Not having stepped forward is a necessary condition for having stood still. Was it possible for you to have stood still given that you stepped forward? No, because not having stepped forward is a necessary condition for having stood still. So given that you have stepped forward, it is impossible that you could have stood still. I dunno. All that tells me is that I can't change an action I made in the past, but has very little implications for what I'm going to do in the future, unless you mean to say that you're going to do what you do because you can't not do what you're going to do. But if that's the case, all that means is that I'm going to make decisions, and it is impossible for me to not make decisions, because if I make a decision not to decide, I'm still making a decision and cancelling out my decision to not decide anything. | ||
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