I think this argument doesn’t deserve anywhere near as much thought as you’ve given it. Caplan is committing a logical error, nothing else.
He probably reasoned as follows:
If determinism is true, I am computable.
Therefore, a large enough computer can compute what I will say.
Since my reaction is just more physics, those should be computable as well, hence it should also be possible to tell me what I will do after hearing the result.
This is wrong because “what Caplan outputs after seeing the prediction of our physics simualtor” is a system larger than the physics simulator and hence not computable by the physics simulator. Caplan’s thought experiment works as soon as you make it so the physics simulator is not causally entangled with Caplan.
I don’t think fixed points have any place in this analysis. Obviously, Caplan can choose to implement a function without a fixed point, like f:x↦−x (edit: rather x↦¬x), in fact he’s saying this in the comment you quoted. The question is why he can do this, since (as by the above) he supposedly can’t.
See also my phrasing of this problem and Richard_Kennaway’s answer. I think the real problem with his quote is that it’s so badly phrased that the argument isn’t even explicit, which paradoxically makes it harder to refute. You first have to reconstruct the argument, and then it gets easier to see why it’s wrong. But I don’t think there’s anything interesting there.
I agree that Caplan is actually committing a logical error, but I don’t think the way I present the argument is uninteresting. I don’t think Caplan actually considered any of what I bring up in the post.
If you don’t make the physics simulator causally entangled with Caplan then his argument doesn’t work at all. It’s true that you don’t run into these problems of a simulator having to simulate a larger system than itself, but on the other hand the compelling first-hand intuition that “I could just do the opposite of what I’m told” melts away once you’re not actually being told what you will do. That’s why the causal entanglement is crucial to the argument.
Also, x→−x is a function with a fixed point, namely x=0. I assume your minus sign is meant to be a logical complement?
Right, I was just thinking negation, but yeah the x↦−x formalization doesn’t make a lot of sense. (Also, there was some frustration with Caplan bleeding into my comment, apologies.)
If you don’t make the physics simulator causally entangled with Caplan then his argument doesn’t work at all.
Yes, because it doesn’t work at all!
Bryan is saying he can avoid fixed points. This is trivially true in the deterministic setting, and your post is asking whether it’s also true in the probabilistic setting. But it doesn’t matter whether it’s true in the probabilistic setting! The entanglement argument shows that even if you have a function without fixed points, this still doesn’t give you LFW.
(I think you got confused in the post and thought that having fixed points shows LFW, but if anything it’s the opposite; Bryan’s point is that he can avoid fixed points. So the deterministic setting is the more favorable one for his argument, and since it doesn’t work there, it just doesn’t work period.)
(I think you got confused in the post and thought that having fixed points shows LFW, but if anything it’s the opposite; Bryan’s point is that he can avoid fixed points. So the deterministic setting is the more favorable one for his argument, and since it doesn’t work there, it just doesn’t work period.)
Not at all. Having fixed points proves LFW wrong, not right.
The whole point of my post is that LFW advocates would say that they can avoid fixed points, while if some theory such as hard determinism or compatibilism is correct then this argument shows that there’s a situation in which you can’t avoid fixed points.
Bryan is saying he can avoid fixed points. This is trivially true in the deterministic setting, and your post is asking whether it’s also true in the probabilistic setting. But it doesn’t matter whether it’s true in the probabilistic setting! The entanglement argument shows that even if you have a function without fixed points, this still doesn’t give you LFW.
The point of my post is not that not having fixed points implies LFW, it’s that LFW implies not having fixed points. Obviously there are other arguments for why you would not have fixed points, e.g.g might fail to be continuous.
Not at all. Having fixed points proves LFW wrong, not right.
Okay; I got the impression that you had it backward from the post, one reason was this quote:
Fortunately for the argument, the assumption of continuity for g is plausible in some real-world settings
Where I thought “the argument” refers to Caplan, but continuity is bad news for Caplan, not good news.
Anyway, if your point is that showing the existence of fixed points would disprove LFW (the contrapositive of LFW → no fixed points], then I take back what I said about your post not being relevant for LFW. However, I maintain that Caplan’s argument has no merit either way.
I think we agree that Caplan’s argument has no merit. As I’ve said in another comment, the reason I quote Caplan is that his remark in the podcast is what prompted me to think about the subject. I don’t attribute any of my arguments to Caplan and I don’t claim that his argument makes sense the way he conceived of it.
Zero could be considered a degenerate case, like how one could bash the target person over the head with a shovel and then say “you’re going to still be lying there unconscious in five seconds.”
I think this argument doesn’t deserve anywhere near as much thought as you’ve given it. Caplan is committing a logical error, nothing else.
He probably reasoned as follows:
If determinism is true, I am computable.
Therefore, a large enough computer can compute what I will say.
Since my reaction is just more physics, those should be computable as well, hence it should also be possible to tell me what I will do after hearing the result.
This is wrong because “what Caplan outputs after seeing the prediction of our physics simualtor” is a system larger than the physics simulator and hence not computable by the physics simulator. Caplan’s thought experiment works as soon as you make it so the physics simulator is not causally entangled with Caplan.
I don’t think fixed points have any place in this analysis. Obviously, Caplan can choose to implement a function without a fixed point, like f:x↦−x (edit: rather x↦¬x), in fact he’s saying this in the comment you quoted. The question is why he can do this, since (as by the above) he supposedly can’t.
See also my phrasing of this problem and Richard_Kennaway’s answer. I think the real problem with his quote is that it’s so badly phrased that the argument isn’t even explicit, which paradoxically makes it harder to refute. You first have to reconstruct the argument, and then it gets easier to see why it’s wrong. But I don’t think there’s anything interesting there.
I agree that Caplan is actually committing a logical error, but I don’t think the way I present the argument is uninteresting. I don’t think Caplan actually considered any of what I bring up in the post.
If you don’t make the physics simulator causally entangled with Caplan then his argument doesn’t work at all. It’s true that you don’t run into these problems of a simulator having to simulate a larger system than itself, but on the other hand the compelling first-hand intuition that “I could just do the opposite of what I’m told” melts away once you’re not actually being told what you will do. That’s why the causal entanglement is crucial to the argument.
Also, x→−x is a function with a fixed point, namely x=0. I assume your minus sign is meant to be a logical complement?
Right, I was just thinking negation, but yeah the x↦−x formalization doesn’t make a lot of sense. (Also, there was some frustration with Caplan bleeding into my comment, apologies.)
Yes, because it doesn’t work at all!
Bryan is saying he can avoid fixed points. This is trivially true in the deterministic setting, and your post is asking whether it’s also true in the probabilistic setting. But it doesn’t matter whether it’s true in the probabilistic setting! The entanglement argument shows that even if you have a function without fixed points, this still doesn’t give you LFW.
(I think you got confused in the post and thought that having fixed points shows LFW, but if anything it’s the opposite; Bryan’s point is that he can avoid fixed points. So the deterministic setting is the more favorable one for his argument, and since it doesn’t work there, it just doesn’t work period.)
Not at all. Having fixed points proves LFW wrong, not right.
The whole point of my post is that LFW advocates would say that they can avoid fixed points, while if some theory such as hard determinism or compatibilism is correct then this argument shows that there’s a situation in which you can’t avoid fixed points.
The point of my post is not that not having fixed points implies LFW, it’s that LFW implies not having fixed points. Obviously there are other arguments for why you would not have fixed points, e.g.g might fail to be continuous.
Okay; I got the impression that you had it backward from the post, one reason was this quote:
Where I thought “the argument” refers to Caplan, but continuity is bad news for Caplan, not good news.
Anyway, if your point is that showing the existence of fixed points would disprove LFW (the contrapositive of LFW → no fixed points], then I take back what I said about your post not being relevant for LFW. However, I maintain that Caplan’s argument has no merit either way.
I think we agree that Caplan’s argument has no merit. As I’ve said in another comment, the reason I quote Caplan is that his remark in the podcast is what prompted me to think about the subject. I don’t attribute any of my arguments to Caplan and I don’t claim that his argument makes sense the way he conceived of it.
Zero could be considered a degenerate case, like how one could bash the target person over the head with a shovel and then say “you’re going to still be lying there unconscious in five seconds.”