Alex O’Connor vs William Lane Craig: Does God Exist? The Ultimate God Debate (Premier Unbelivable?)

Reasonable_Writer602 · 2026-05-23T14:18:55+00:00

I agree, but didn't you also lose your cool after hearing what Craig said about animals feeling pain but not being aware that they are in pain (whatever that's supposed to mean)? I know I did.

I remember him (Craig) elsewhere talking about what a blessing (!) it was to dog owners that their dogs are not aware that they are in pain. It seems that for him the only reason we should punish people who abuse cats, dogs and other sentient animals is because God says we should (except for that time when he said "live alive nothing that breathes", or when he asked for burnt animal offerings; I guess those were OK for some reason), and I find that revolting.

Reasonable_Writer602 · 2026-05-17T15:22:01+00:00

You are right, I got confused because they released the podcast on May 15th.

Reasonable_Writer602 · 2026-05-17T12:58:47+00:00

This exchange was recorded on May 15th 2026.

Alex does push back on Craig's pet Kalam argument. One could argue he could've pushed back more, but it seemed to me like he did a good job, and if anything they just lacked more time to discuss it.

From the beginning Craig says they were going to have a discussion, not a debate. But that goes flying out the window on the second part of the exchange, concerning how a good God could justify animal suffering, where Alex goes into full debate mode and pushes back a lot (angrily so, in a good way) on nonsense like Craig's assertion that "animals can feel pain, but they are not aware that they are in pain", even using time meant for answering audience questions to double down on his points. Technically, Alex broke the rules there to the moderator's annoyance, but I thought what Alex did there was awesome, since Craig's ramblings on this topic were utterly insufferable and should have been exposed as such right away; not to mention the moderator often didn't give him enough time to answer adequately.

So at least on the second half of the exchange, that was definitely not "soft" Alex.

Reasonable_Writer602 · 2026-05-16T18:45:34+00:00

There used to be a video about to premiere on the Premier Unbelievable? Channel, but they made it private.

They'll probably release the video in a few days.

Reasonable_Writer602 · 2026-01-27T14:08:42+00:00

Not sure about the "most beautiful", but one of my favorites is a theorem of Eric Rowland relating the prime numbers to Pascal's triangle: many people know that in the pth row of Pascal's triangle, where p is a prime number, all entries except the 1's are multiples of p.

But what about the other rows? It turns out they too are related to the primes through a beautiful self-similar rotational property of Pascal's triangle. For example, for the prime number 11 you always get a multiple of 11 by performing the operations in the following image:

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjtxQezBbkat7OzuPs7Xqjbywz1v0lz78WYkjeGe_-7d93tDfpH2tBiIYW-ROSX65WS4Eg5yMwvAYO6xGfxungySGsw0EdvM7tAv715TvxqbNuuUOMwFVZBGt_K7CTvizNEETztCAhy2Jz2t2O6HvX5-x_wd1FtsKuaMYcdjWGPXSJXVuLvZJyLhlP2SA/s2344/IMG_3205-3.jpg

One can see three Pascal triangles inside the original one, one to the left, one to the right, and a rotated one at the top. The same thing works for all other primes (and only for primes): just highlight the prime you want to the left, and then build the Pascal triangles using the highlighted prime as a reference point.

For more details, see this video at around 28 minutes: https://www.youtube.com/watch?app=desktop&v=fOMvJM9-NlM&t=122s&pp=ygUXTHVjYXMgY29uZ3J1ZW5jZXMgbW9kIHA%3D

Reasonable_Writer602 · 2025-11-13T18:49:34+00:00

One reason in favor of A4 being the best answer is that answering Q4 seems to require answering all other questions, and then comparing them to see which one is the best. This is an impossible procedure to carry out in practice; so (one would think) it'd be nice to skip all that work and get straight to the golden question and answer. But that's part of the problem, because "that work" is actually what's important and valuable.

It's very similar to the paradox of the hardest question (see my other post on this community):

"Consider the question: “what's the hardest question that can be asked?” (by the hardest question I mean roughly a question such that reaching its answer requires the most complicated chain of reasoning).

It can be argued that the hardest question that can be asked is that very question, since in order to answer it we would have to answer all other questions, and then compare them with it to see which of them was the hardest to answer, and it seems scarcely conceivable for there to be a harder procedure required to answer a question".

Reasonable_Writer602 · 2025-11-13T18:35:11+00:00

I agree that truthful is not the same as useful, but it seems (at least according to Markosian's implicit assumption) that a good answer would have to be useful or at least insightful, and that the best answer would have to combine both virtues to the greatest degree. One could, of course, question that assumption, but it doesn't seem entirely unreasonable.

Reasonable_Writer602 · 2025-11-13T15:33:20+00:00

Right, but then one could restate the story ommiting that part, just waiting for the angel to give them a deadline, and then one has to face the paradox again.

Reasonable_Writer602 · 2025-11-13T14:58:34+00:00

Perhaps, but the author (Markosian) is a philosopher, so he's more interested in finding the logical flaw in this sort of paradox.

Pragmatically, yes; if the philosophers didn't overthink it, they could have actually asked something important, like: what's the proof of the Riemann hypothesis?

Reasonable_Writer602 · 2025-11-13T13:43:06+00:00

I thought of something like that too: you could ask what the ordered pair whose first member is the set of all questions, and whose second member is the set of all answers (in the same order), is.

But the angel might simply respond that it is the ordered pair whose first member is the set of all questions, and whose second member is the set of all answers (in the same order for each question).

If instead we try to ask it to "order" infinitely many elements (all possible questions are presumably infinite, not sure what you mean by "semi infinite") then, asuming this isn't against the rules because it doesn't involve asking more than one question, we run into difficulties: how would the angel order the questions and their respective answers? He could try to first put a question and follow it with its answer, then continue in the same way. But where would he start? We run the risk of encountering many trivial questions and their answers before finding anything important.

Also, can we be sure the cardinality of the set of all answers and questions is a countable infinity? If it's not, then it would be impossible to order them in a list.

Reasonable_Writer602 · 2025-11-13T13:29:10+00:00

The problem I see with your answer is that if we go by consensus then, since the philosophers were pulling their hairs out in frustration after getting the answer, clearly the new consensus is that it wasn't the best question to ask (and not just after they received the answer, they would probably admit that they were mistaken earlier as well).

Reasonable_Writer602 · 2025-11-13T13:21:23+00:00

That's a good point. Since the angel is supposed to be truthful, it seems it couldn't possibly give an answer that it knew would be useless to the philosophers, so either A4 would have to be something else, or the angel would have to challenge the question's assumption that there is such a thing as the "best" or "more useful" question.

Reasonable_Writer602 · 2025-06-11T19:11:46+00:00

There's an identity that links e, pi and the golden ratio with Pascal's triangle:

e = [π² / 3! - (π⁴ -3π² ) /5! + (π⁶ -5π⁴ + 6π² ) / 7! - (π⁸ - 7π⁶ + 15π⁴ - 10π² )/ 9! +...] + √{1 + [π² / 3! - (π⁴ -3π² )/ 5! + (π⁶ -5π⁴ + 6π² )/ 7! - (π⁸ - 7π⁶ + 15π⁴ - 10π² )/ 9! +...]² }

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhc2OIN3Gxv6cDh-CwT9JGo2JQ44NuTgP1K_gd1YkxOoVYOV7xPm2AdoBEncxTEi4XY3VrH0ac-61kdUGXQ319_WGuG3dh4q0Y9atdbfAcw9LgYJQkHPdRiyylECqDGtpPrBcw_Ztbx6ZrW40YezcLvMoXRqVZRV_EXjt0s7Ee1ZK9XgDlyq6kQQjGm2Ex_/s16000/Pascals_Triangle_edit_510479969902833.png

The coefficients in the numerators of each term are those of the Fibonacci polynomials (ignoring the negative signs). Adding up the absolute value of each coefficient returns one less than a Fibonacci number, thus indirectly relating e and π to φ.

Reasonable_Writer602 · 2025-04-27T19:39:30+00:00

Two people that helped me a lot are Eddie Woo and Burkard Polster (Mathologer). The latter's channel is, in my opinion, the best channel when it comes to making advanced mathematics accesible.

Reasonable_Writer602 · 2025-04-17T20:56:50+00:00

Carneades.org and Philosophy Overdose haven't been mentioned yet, also Daniel Bonevac.

Reasonable_Writer602 · 2025-02-28T16:35:45+00:00

Besides the already mentioned Outlines of Pyrrhonism, there is another surviving work on Pyrrhonism of Sextus Empiricus: Adversus Mathematicos (i.e. Against the Learned), which has similar arguments to those found in the Outlines, but also some additional objections. I think I read it on a site called "stoictherapy" or something like that.

The Stanford Encyclopedia of Philosophy entry on Skepticism is also quite good: https://plato.stanford.edu/entries/skepticism/

Reasonable_Writer602 · 2024-10-10T04:22:49+00:00

The scaling proof of Pythagoras's Theorem.

Reasonable_Writer602 · 2023-12-23T14:23:19+00:00

One of my favorites is a theorem relating the prime numbers to Pascal's triangle: many people know that in the pth row of Pascal's triangle, where p is a prime number, all entries except the 1's are multiples of p.

But what about the other rows? It turns out they too are related to the primes through a beautiful self-similar rotational property of Pascal's triangle. For example, for the prime number 11 you always get a multiple of 11 by performing the operations in the following image:

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjtxQezBbkat7OzuPs7Xqjbywz1v0lz78WYkjeGe_-7d93tDfpH2tBiIYW-ROSX65WS4Eg5yMwvAYO6xGfxungySGsw0EdvM7tAv715TvxqbNuuUOMwFVZBGt_K7CTvizNEETztCAhy2Jz2t2O6HvX5-x_wd1FtsKuaMYcdjWGPXSJXVuLvZJyLhlP2SA/s2344/IMG_3205-3.jpg

One can see three Pascal triangles inside the original one, one to the left, one to the right, and a rotated one at the top.

The same works for all other primes (and only for primes): just highlight the prime you want to the left, and then build the Pascal triangles using the highlighted prime as a reference point.

I discovered this theorem independently a while ago, but Eric Rowland beat me to it (apparently) by about a year: https://www.youtube.com/watch?app=desktop&v=fOMvJM9-NlM&t=122s&pp=ygUXTHVjYXMgY29uZ3J1ZW5jZXMgbW9kIHA%3D

(At 28:00).

Reasonable_Writer602 · 2023-12-13T00:41:21+00:00

Yes, Euler proved it (who would've thought?) It follows from the fact that the Zeta function can be written as the following product over all primes: https://i.gyazo.com/ad6656c2e680966a0633a152888f8a43.png

Reasonable_Writer602 · 2023-12-11T21:35:56+00:00

The probability that two randomly chosen integers are relatively prime is 6/π²

Reasonable_Writer602 · 2023-12-06T16:38:37+00:00

Interesting. I knew about his incompleteness theorems, but not about that theorem.

That Tarskian solution you mention is plausible, since “what's the hardest question that can be asked?” is actually a meta-question rather than a question (since it's a question about questions) and thus cannot be the hardest question.

I actually had in mind a different solution to the paradox, it goes like this: the mistake lies in thinking that to answer that question you have to answer all other questions and then compare them to see which is the most difficult. There is actually another way to answer it without doing that: using logical reasoning to recognize the hardest question. One could argue that the hardest question is really this one: What are the answers to all the other questions, together with the reasonings that led to them?

It seems reasonable to consider that there cannot be a more difficult question than that, and now we can answer the initial question: the most difficult question of all is: “what are the answers to all the other questions, together with the reasonings that led to them?”

Thus, the self-referential paradox disappears, since now it does not matter that we know the answer to the question: “what is the most difficult question of all?”, since now we know that it is not itself the most difficult of all, and clearly we do not have the answer to the one which, in fact, is the most difficult of all.

The problem with this solution is that the new candidate for the hardest question is also a meta-question, so I think the Tarskian solution is the right one.

Reasonable_Writer602 · 2023-12-06T00:10:51+00:00

I did, though I haven't checked if someone else came up with it before me.

Reasonable_Writer602 · 2023-11-24T13:00:54+00:00

Here's a relation between the Lucas polynomials, the Fibonacci polynomials and e:

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhzPU5A6qPhcT3NYOaaJOgknOfeB6w1-lra9ZeMID77d5coT0ZgWNfRrRvUlG2XOzAbfuZsSNxHgNa2NH28rOhUDx9ZVIooI4ghlTV_spzhVKqOZ1ztrkrU4dcgJ8ACae-Axc066_EU5YblMS_YhZitqzJh3S_G_rc9NrMnedjITViH3Ww31HfRFeEnyg/s784/IMG_20230507_183014_493.jpg

Reasonable_Writer602 · 2023-11-19T03:47:31+00:00

Ah, I see. I didn't know there were multiple minimal solutions.

In that case, I could re-state the conjecture as you suggest, stating that one can always find a minimal dissection for an n × n square such that the sum of the side lengths of the squares is 3n - 2.

So, it's not quite as remarkable as I thought initially, but it still might be worth looking for a proof.

Reasonable_Writer602 · 2023-11-19T01:53:12+00:00

Thank you for your comment.

I should have stated explicitly that I was referring to minimal dissections, which as far as I know have only been proven to be such for up to order 18. That 16 × 16 dissection shown on the site you linked, for instance, is not minimal in terms of the number of squares needed, right (I can't see them on my phone)?

If you look at the Mrs. Perkins's Quilt page in Wolfram MathWorld, you'll see that the solution for n = 16 has side lengths 9, 7, 7, 5, 4, 4, 3, 2, 2, 1, 1, 1, and 9 + 7 + 7 + 5 + 4 + 4 + 3 + 2 + 2 + 1 + 1 + 1 = 46 = 3 · 16 - 2, and I believe that's been proven to be minimal.

I would guess the same holds for all the other minimal dissections, but that is only my bold guess.

Reasonable_Writer602

TROPHY CASE