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I've recently been learning about non-classical logics and skepticism regarding the law of excluded middle.

This has been hindering by progress in classical mathematics because it's made me slightly depressed when doing proof by contradiction.

I am basically looking for reason to continue using this proof technique and continuing my study of classical mathematics.

Update: After viewing This math overflow thread, and discussion relating to top answers on this question, my concerns have essentially been solidified. My only motivation for doing classical mathematics now is that classical mathematics can give us new research areas / theorems to explore, but ultimately, this is only before they are transformed into a constructive version, and applied elsewhere, in 99% of cases.

Frazer
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Apr 05 '24 at 17:23
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    Have you considered that most proofs by contradiction can be converted to using the contrapositive? Have you considered that a theorem proved by contradiction is still true so its results are applicable everywhere they would be if the proof had been more to your liking? – John Douma Apr 06 '24 at 16:47
  • @JohnDouma have you considered the possible un-soundness of mathematics? Non-constructive proofs that cannot be converted to constructive ones pose questions – Frazer Apr 07 '24 at 00:15
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    @FraserPye That's ridiculous. – John Douma Apr 07 '24 at 02:08
  • @FraserPye, the issue of soundness/unsoundness is not an issue that separates classical from intuitionistic mathematics, because there are equiconsistency results between them. – Mikhail Katz Apr 07 '24 at 09:36

6 Answers6

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I assume that software development counts as sufficiently distinct from mathematics.

The correctness of algorithms are often proved by contradiction. For an algorithm to be correct, we need to show that it produces a correct output for all valid inputs. This can be done by supposing there is some input with a wrong output, and then reaching a contradiction. This is often easier than a direct proof, because

  • We get to make a stronger assumption (that the input is valid and the output is not), and
  • The goal is to derive any contradiction, rather than just the specific statement that the output is correct.

Furthermore, the lack of constructivity is no issue. We don't need a proof that shows the algorithm is correct and that some concrete object satisfying some property exists. We just want to know that the algorithm is correct.

Consider also a proof that an algorithm is incorrect, i.e. a proof that some valid input exists for which the output is incorrect. If this proof is constructive then we can use the concrete counterexample to write a test case, which will be useful for testing the software after attempting to fix the bug. The concrete counterexample might also be useful for stepping through the algorithm with a debugger, in order to find the exact point at which the algorithm "goes wrong" and hence understand the bug. So a constructive proof is definitely preferable here.

On the other hand, even a non-constructive proof (i.e. a proof that the algorithm is incorrect, but without exhibiting a concrete input which it fails on) can still be enlightening, since we can see which features or parts of the algorithm are used in the proof, and that could help to isolate the bug to one of those features or parts. And a proof at all is useful for telling us that a bug definitely exists (and work is needed to address it), which we wouldn't know otherwise.

I realise that the above argument in favour of proofs by contradiction is, ironically, non-constructive, since I didn't provide any specific algorithm which is easier to prove (or refute) by contradiction. However, the question as written asks for "reason[s] to continue using this proof technique", which I believe I have provided.

I've also referred only to algorithms which are "easier" to prove by contradiction, rather than those which "have only been" proved by contradiction. However, the nature of software development is that one often comes up with an algorithm for a particular purpose (which might be based on a standard algorithm, but nonetheless needs justification), and then the developer's proof of that algorithm is the only one written for that specific algorithm. Hence if they prove it by contradiction, it will have only been proved that way.

kaya3
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  • Do you happen to have a link to a worked example of an algorithm being proved correct by contradiction? – Eugene Apr 03 '24 at 22:32
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    @Eugene Here's an example of a proof by contradiction for Prim's algorithm (which finds a minimum spanning tree of an undirected graph). Here's one for breadth-first-search (which finds distances in an unweighted graph). These are famous algorithms so there will be many proofs by other methods, of course. – kaya3 Apr 03 '24 at 22:48
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    A lot of questions in computer science are decidable (in a constructive sense) to begin with. That Prim's Algorithm example doesn't need full LEM. The question "is S minimum weight" is decidable, since there are a finite number of trees and therefore a finite number of possible weights, so even a constructivist would accept that proof. – Silvio Mayolo Apr 04 '24 at 00:09
  • @SilvioMayolo Sure, but the question isn't only about whether LEM is really needed to do mathematics; it's about whether proofs by contradiction are applicable outside of mathematics. I already said there are proofs by other methods for these algorithms. – kaya3 Apr 04 '24 at 00:12
  • @kaya3 no I was looking for only non-constructive theorems, not theorems that are also proven constructively – Frazer Apr 04 '24 at 21:59
  • Which is why I've always had a problem with constructivism and computer science; it's all about excluding stuff like LEM which are provably true in finite mathematics. – prosfilaes Apr 04 '24 at 22:45
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    @FraserPye: For many algorithms, particularly of the divide-and-conquer type, a constructive proof may be somewhere between very hard and impossible to write. Most proofs for such algorithms use infinite descent, and it is not immediately obvious how you would rewrite such a proof to avoid the use of infinite descent or an equivalent technique that also tacitly relies on LEM. Computer scientists typically do not care about constructivism for these proofs, so you probably won't find such proofs in the literature, either. – Kevin Apr 05 '24 at 16:53
  • Note that in at least some kinds of constructive mathematics, "not not not" reduces to "not" fine; and a proof that "not X" is true via proving "not not X implies False" (a proof by contradiction) is a "not not not X" reduction to "not X" - ie, perfectly constructive. – Yakk Apr 05 '24 at 21:42
  • @Kevin there are settings where LEM is acceptable. For example, consider logic circuitry within computers. This is a perfect model of classical logic, and if we take the inputs to be propositions, I have no quarms about proof by contradiction. We just don't really know if propositions behave like this because they are not physical. "If it works it works", and it has been working, so may as well just keep going with it. – Frazer Apr 05 '24 at 22:10
  • @Kevin Proof by infinite descent appears to be the statement $(∀n. P(n) → ∃m<n. P(m)) → ∀n. ¬P(n)$. I know wikipedia says it is, "proof by contradiction," but it is being sloppy in doing so. This depends on induction, not on excluded middle. Here is a (constructive) proof in Agda. – Dan Doel Apr 07 '24 at 03:10
  • @DanDoel: Can you use that axiom without double negation elimination? The $P(n)$ you are thinking of is already a negated statement in its own right ("The algorithm does not work for $n$"), so $\neg P(n)$ is a double-negative. – Kevin Apr 07 '24 at 12:59
  • @Kevin What is an example of a divide-and-conquer algorithm that can only be proved correct by proving it isn't incorrect for every input by this method? The ones that come to my mind tend to have an obviously correct base case, and what you do after 'dividing' preserves the correctness of the subdivisions. Those are provable by induction on the same structure that the algorithm uses. – Dan Doel Apr 07 '24 at 15:50
  • In my experience, it's quite common for many people to automatically begin every/most proof with, "assume the contrary," because they've been (unintentionally) trained to do so. This happens even in cases where direct proofs would be clearer (in my opinion, at least). So the fact that many algorithmic proofs are written that way does not mean that they are necessary or even easier to write/understand that way. – Dan Doel Apr 07 '24 at 15:57
  • @DanDoel: If your algorithm starts "from the large end" and shrinks to smaller cases, how do you prove that you always arrive at the base case? I find it hard to believe you can do that with direct induction. I would like to see a proof for e.g. quicksort, or any other algorithm that can't be expressed in bottom-up terms. – Kevin Apr 09 '24 at 23:54
  • Divide and conquer algorithms are typically proved by induction. The same proofs can be rephrased as proofs by contradiction, usually starting with "suppose X is a minimal counterexample..." and then showing there must exist a smaller counterexample, contradicting the supposed minimality. Some people might find the latter easier to think about; induction is famously a difficult topic for students. – kaya3 Apr 10 '24 at 00:20
  • Quicksort recursively calls itself on a strictly smaller portion of the array for each sub-problem, until it ends up at slice sizes that are trivially sorted. This is justified by induction on the size of the portions. Here is quicksort in Agda. The Acc argument is enabling induction along arbitrary < steps. It doesn't prove the sorting properties, but the same argument is required to make Agda accept that the function terminates at all. Proofs of other properties would follow the same structure as the recursive algorithm. – Dan Doel Apr 10 '24 at 03:26
  • Even the answerer acknowledges that this answer does nothing to answer the question. But it has 27 upvotes. Ridiculous. – FCardelle Feb 27 '25 at 10:09
  • @FCardelle Please could you quote the part of my answer where you think I said that my answer does nothing to answer the question? – kaya3 Feb 27 '25 at 12:52
  • "I didn't provide any specific algorithm which is easier to prove (or refute) by contradiction" – FCardelle Feb 27 '25 at 13:33
  • @FCardelle The question is not asking for specific algorithms. How did you interpret it as asking for that? It asks if theorems which have been proven only by contradiction are useful outside of mathematics. It doesn't ask "please provide examples of such theorems". – kaya3 Feb 27 '25 at 22:34
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In 1964, Hohenberg and Kohn proved by contradiction that the ground-state energy of a molecule or material is (up to a constant) a unique functional of its density $\rho$. This is surprising: the ground-state energy is the minimal eigenvalue of the time-independent Schrödinger equation $H \lvert \Psi \rangle = E \lvert \Psi \rangle$, and the wavefunction $\lvert \Psi \rangle$ is 3$N$-dimensional, where $N$ is the number of electrons. It is also quite powerful: $\rho$ is a spatial density, three-dimensional for any $N$.

This theorem is not strictly "within mathematics", and it is also not especially deep or complicated. The universal functional it prescribes remains unknown sixty years later. But it almost instantly (bolstered by a workable approximation developed just one year later) created the field of density functional theory, which is one of the most popular methods for electronic structure in chemistry and materials science.

elutionary
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The Penrose-Hawking singularity theorem in astrophysics (and therefore "outside mathematics") was proved using nonconstructive techniques relying on classical logic (and the law of excluded middle). This was pointed out by historian Geoffrey Hellman. For a related study, see

Katz, K.; Katz, M. "Meaning in classical mathematics: Is it at odds with Intuitionism?" Intellectica 56 (2011), no. 2, 223-302. https://arxiv.org/abs/1110.5456 and https://www.persee.fr/doc/intel_0769-4113_2011_num_56_2_1154

Mikhail Katz
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  • I’m not sure if this is the same thing that you refer to (I don’t know enough to judge), but near the end of Bryce DeWitt’s review of The Large Scale Structure of Space-Time by Hawking/Ellis [Science (N.S.) 182 #4113, 16 November 1973, pp. 705−706] (JSTOR) DeWitt criticizes a proof in Hawking/Ellis (on pp. 249−251) for making use of Zorn’s lemma, with DeWitt ending with: (continued) – Dave L. Renfro Apr 03 '24 at 11:56
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    The physicist’s job is not done until he can show, by an explicit algorithm or construction, how one could in principle always select a member from every such set of developments. Failing this he has proved nothing. See also this mathoverflow answer. Three papers that deal with this issue are (there are other papers, but these three seemed to me the most relevant when I was briefly looking into this issue a few months ago): (continued) – Dave L. Renfro Apr 03 '24 at 11:56
  • Time machines, maximal extensions and Zorn’s lemma by Low (2012) AND On the existence of a maximal Cauchy development for the Einstein equations: A dezornification by Sbierski (2016; arXiv version) AND On the inextendibility of space-time by Manchak (2017; copy at Manchak’s homepage). – Dave L. Renfro Apr 03 '24 at 11:56
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    @DaveL.Renfro, You are certainly right: other nonconstructive aspects of classical mathematics are also routinely used in physics, and not merely the law of excluded middle. An example would be the notion of a Calabi-Yau manifold, which is fundamental in certain branches of theoretical physics. To prove the existence of such manifolds, one uses high-level PDE techniques that certainly rely on ZFC and classical logic (this example was also mentioned in the article in intellectica that I linked in my answer). – Mikhail Katz Apr 03 '24 at 11:59
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    It could be that the physicists' work is "not finished" until they produce a more constructive proof; nevertheless, they do seem to use nonconstructive techniques in a routine fashion. @DaveL.Renfro – Mikhail Katz Apr 03 '24 at 12:05
  • @MikhailKatz you say in a "routine fashion" but I think it couldn't be farther from routine. DeWitt essentially rejects this method entirely for example. Also most fields of physics are experimental and are not even remotely concerned with mathematical proofs. Maybe fields like fluid dynamics we might see it occasionally pop up, but it's rare. – Frazer Apr 03 '24 at 22:07
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    @FraserPye, As you point out, this depends on which the area of physics you are in exactly. If you work with Calabi-Yau manifolds, you are certainly using the ZFC background (+classical logic) routinely. – Mikhail Katz Apr 04 '24 at 11:16
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Does "$\forall x : \mathbb{R} . \ x \geq 0 \lor x \leq 0$" count as a theorem? This is not provable constructively (it's equivalent to LLPO), but is used freely by physicists and computer scientists the world over. In fact, if you ask the people who are really working with "applications", like the biologists and engineers for instance, I guarantee you that they'll ask you what's the use of your math that doesn't allow dichotomy of $\leq$!

Just because there are constructive versions of lots of interesting theorems (usually under some extra definability conditions on the inputs) doesn't mean that the "nonconstructive" results aren't worthwhile! And this is coming from someone who spends a lot of time thinking about constructive math. See this recent blog post, for instance, or any number of my answers on this website. I try quite hard to make my proofs constructive whenever possible, provided it doesn't overcomplicate exposition. All that to convince you it's not defensiveness or a dislike of constructive math that makes me say: Frankly, your question comes off as quite immature (not to mention dismissive of "regular" mathematicians). I'm sure that's not your intent, it's just something to be aware of.

Going back to the mathematics, these extra definability conditions can be really hard to pin down! It wasn't until quite recently that an algorithm for computing gröbner bases (which was already used in practice!) was actually proven to always terminate. Moreover, as you'll find by just searching for "contradiction" in the linked pdf, this proof crucially uses classical logic in multiple places (not to mention casework, which almost certainly violates LEM, but I haven't checked). Again, the actual practitioners (by which I mean the software engineers developing mathematica, sage, etc) don't care at all about constructive versus classical proofs! Now, of course, some version of the correctness of this algorithm is almost certainly provable constructively. But I highly doubt anybody knows what the right definability conditions on the ideals should be! Also, in case you doubt the utility of this algorithm, you should know that gröbner bases are roughly the analogue of gaussian elimination for arbitrary systems of polynomials. There's a reason that the mathematica team was so eager for a more efficient algorithm!

More concerningly still, constructively the notion of "finite" breaks down entirely. Indeed, quoting from Blass's An induction principle and pigeonhole principles for k-finite sets:

THEOREM $4$. Assume that, for all finite $X$ and all $f : X+1 \to X$, there exist $x$ and $y$ in $X+1$ with $f(x) = f(y)$ but $x \neq y$. Then the law of the excluded middle holds

The pigeonhole principle, I'm sure you'll agree, is used constantly in applications. For transparency, I should say that the pigeonhole principle is constructively provable if $X$ has decidable equality, but I guarantee you that people working in applications absolutely don't consider this subtlety.

Lastly, I actually have a question for you (and I'm sincerely interested in your answer!): What do you think of the various ways one can nonconstructively show that a constructive proof exists (though we may not be able to produce such a proof)?

For example, Robertson-Seymour prove that for each graph $H$ there exists an algorithm taking in a graph $G$ and deciding whether $H$ is a minor of $G$ (indeed, the algorithm runs in time $O(|G|^3)$!). However, their proof is nonconstructive, and for even quite simple $H$ we have no idea what this algorithm is!

In your mind, does this count as a constructive result? The algorithm is certainly constructive... And we know an algorithm exists... We just can't get our hands on it!

Similarly, how do you feel about theorems like the following: "If $\varphi$ is a geometric formula which is classically provable, then $\varphi$ is also constructively provable". This theorem is, itself, nonconstructive (since it relies on the completeness theorem for topos semantics), but it asserts the existence of a constructive proof. Would you consider $\varphi$ "constructively true" for your purposes?

I hope this helps ^_^

Chris Grossack
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    I really like the first half of this answer, but I find the stuff below the line very confusing. Proving that "$X$ is provable" is not sufficient to conclude $X$, even classically, much less constructively. – Z. A. K. Apr 05 '24 at 08:57
  • @Z.A.K. -- We're using the fact that topos semantics are complete for (let's say geometric) intuitionistic logic. That is, intuitionistic provability is the same thing as truth in all topoi. If we're able to show through some semantic argument that $\varphi$ is true in every topos, completeness tells us that it must be intuitionistically provable, even though it doesn't furnish us with a proof. Now given any topos, we can use Barr's Theorem to find a "nearby topos" (in the sense that it agrees on geometric truth) which satisfies AC and LEM. – Chris Grossack Apr 05 '24 at 14:37
  • So to show that $\varphi$ is intuitionistically provable one can argue as follows: 1. Give a classical proof. This tells us that $\varphi$ is true in all boolean topoi. 2. Take an arbitrary topos $\mathcal{E}$, which admits a surjection from a boolean topos $\mathcal{B}$ by Barr's theorem. Now since $\mathcal{E}$ and $\mathcal{B}$ agree on truth of geometric formulas and we know $\mathcal{B} \models \varphi$, we must have $\mathcal{E} \models \varphi$ too! So $\varphi$ is true in all topoi, thus intuitionistically provable! Of course, this doesn't tell us what the constructive proof is! – Chris Grossack Apr 05 '24 at 14:44
  • And the reason I was asking is because I think one's opinion on the validity of these "nonconstructive proofs that a constructive proof exists" can clarify their philosophy on constructive math. See, for instance, Simon Henry's answer to this MO question – Chris Grossack Apr 05 '24 at 14:49
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    Personally, and assuming I understand the example, this is exactly the sort of reason that I think classical computability and the effective topos are fundamentally flawed from the perspective of caring about having a mathematics where all the results allow computing actual answers/examples. I suspect the majority of computability arguments would still go through, but these, "solved by an unknowable algorithm," results are exactly the ones I don't want. – Dan Doel Apr 05 '24 at 17:40
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    @DanDoel -- I think that's a perfectly reasonable stance! I much prefer having an actual constructive proof-in-hand compared to merely knowing a constructive proof exists. Of course, I also think it's a cute party trick to be able to prove constructive theorems nonconstructively :P – Chris Grossack Apr 05 '24 at 20:30
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    @HallaSurvivor I am still skeptical about how you "know the algorithm exists". What if it doesn't? If you really question what you "know" you will realize that you don't really know anything for certain. It's a facet of the universe, we live in mystery, I will believe it when I see it. – Frazer Apr 05 '24 at 22:04
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    @HallaSurvivor: Thanks for engaging. Sorry for being unclear in my previous comment. I understand Barr's thm & the completeness results for geometric logic, that's not the part confusing me. I'm confused by you going from "a constructive proof of $\varphi$ exists" to asking whether OP would consider "$\varphi$ constructively true" . A yes answer to this question cannot be mathematically reasonable, and this has nothing to do with constructivism: even going from "I have shown that there is a proof of $\varphi$" to "I have shown that $\varphi$ holds", without any adjectives, is not reasonable.1/ – Z. A. K. Apr 06 '24 at 01:57
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    @HallaSurvivor: If you manage to prove in ZFC that "there is a ZFC proof of X", you can't conclude "therefore, X". Completeness theorems can only give you "therefore M\models X$ for every ZFC-structure $M$", which is not X, but something much weaker. You especially can't make inferences like these if what you've shown is not "ZFC shows that there's a ZFC proof of X" but "some other T shows that there's a ZFC-proof of X" (think of T= PA+~Con(PA)). You say knowing the answer to your question can "clarify their philosophy", and I'm confused because there seems to be only one possible answer. 2/ – Z. A. K. Apr 06 '24 at 02:09
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    @HallaSurvivor: OP gets this right in his comment, although naively: just because some theory T claims that a number with certain properties exists, does not mean that such a number really does exist. And "really" is not a philosophical question about natural reality, it's purely mathematical: it's essentially the observation that you can't conclude $\bot$ from PA+~Con(PA) if PA itself is consistent, even though PA+~Con(PA) certainly claims that PA+~Con(PA) proves $\bot$. 3/3 – Z. A. K. Apr 06 '24 at 02:15
  • @Z.A.K. I only half understand what you are saying in your last comment. Can you please dumb it down? We can’t conclude X from peano arithemetic and non contradiction (consistency) of Peano arithmetic? What is your take on the usefulness of proof by contradiction? – Hasan Zaeem Apr 06 '24 at 11:08
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    @HasanZaeem: Please see this question from earlier this year, it has multiple good answers, so should be enough to get you started on the things I mention above, or to pin down which points you find confusing. As for proof by contradiction: I don't think I said anything about it in this comment chain. – Z. A. K. Apr 06 '24 at 11:25
  • @Z.A.K. so we rely on the assumption that ZFC and PA is sound? I.e just because we can “prove” the graph exists we really are relying an assumption that the graph actually exists? That’s crazy – Hasan Zaeem Apr 06 '24 at 12:09
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    @Z.A.K. -- Aaaah, thanks for clarifying! And sorry for underestimating your knowledge of logic and misinterpreting your comment. I was using the word "true" in a much less precise way than the technical (and subtle!) version you're using. You're, of course, completely right that if a formal system proves $T$ proves that "$\varphi$ is provable", that tells us nothing about $\varphi$'s truth "in the real world. – Chris Grossack Apr 06 '24 at 17:02
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    But I'm making the (common, if abusive) pair of implicit assumptions that 1. the informal logic mathematicians use to reason is sound for "the real world", whatever that means and 2. this informal logic is well approximated by something formal that we can show (again in the informal "daily logic" of mathematicians) is sound and complete for topos semantics. As you've said, this is literally not true, but for what I do it's not a distinction that I've never needed to come to grips with (which is another reason it wasn't on my radar as something to think about when I first read your comment) – Chris Grossack Apr 06 '24 at 17:03
  • @HallaSurvivor so you’re saying everyday mathematics relies on assumptions about the consistency and soundness of mathematics, which may turn out to be false? I.e we may never find the graph you reference – Hasan Zaeem Apr 06 '24 at 21:40
  • @HallaSurvivor: No worries at all, it's my fault for not being clear about my point of confusion. Plus, your explanation will benefit readers seeking more depth on why a geometric formula that is provable classically is also provable constructively. Thus, it surely aid some curious minds! – Z. A. K. Apr 07 '24 at 15:12
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True proof by contradiction is captured by the deductive rule ( ( ¬A ⇒ ⊥ ) ⊢ A ), which allows you to deduce A if you can deduce that ¬A implies a contradiction. Besides your chosen axioms, this rule is essentially the only non-constructive assumption you need in your foundational system for mathematics. It is equivalent over the other (constructive) deductive rules to LEM (excluded middle). I mention this because viewing the issue using LEM instead of proof by contradiction (despite being equivalent) is helpful for furnishing many natural examples. Some of my examples below are indeed literal instances of LEM.

Hex has a winning strategy for the first player

The (currently known) proof that the first player has a winning strategy has two parts:
  (1) The second player has no winning strategy.
  (2) At least one player will win.
Although the proof of (2) is constructive, the proof of (1) is not, and hence the proof of the full theorem is not constructive either. Till today, nobody knows a winning strategy for the first player, and it is infeasible to compute it even though it is in theory possible, since the number of possible game plays is astronomically gigantic.

Brouwer's fixed-point theorem

This theorem is non-constructive in an essential way; there is a computable continuous operator on the unit square that has no computable fixed-point! Hence any proof of this theorem must be non-constructive!

This theorem is also an easy consequence of the Borsuk-Ulam theorem, which implies that the latter theorem also cannot be constructively proven. And that latter theorem implies that an ideal teddy bear covered completely with fur cannot have its fur combed completely flat. Is that application enough?

From here on we consider programs in a fixed Turing-complete programming language.

The halting problem is well-defined

Given any program P and input X, either P halts on X or P does not halt on X. Well? Do you agree? If you do, then you must have used a non-constructive proof! The reason is simple; there is no program to compute the answer from P,X.

I think this fact is concrete enough that it should be accepted just for its own intrinsic meaning.

Kolmogorov complexity is well-defined

For every string X, there is a minimum m∈ℕ such that some inputless program of length m outputs X. Isn't this obviously true? But if you want to prove it, you would necessarily need a non-constructive proof. The reason is that if you have a constructive proof, then you can construct a program K that computes this m from X, and construct an inputless program S that outputs the (length-lexicographically) first string Y such that K(Y) > length(S), which must exist since there are only finitely many possible outputs by programs of each length. Note that S can obtain its own length, by the standard quine technique. But then length(S) ≥ K(Y) by definition of K, which is impossible.

Why is Kolmogorov complexity relevant outside Mathematics? We like to have good compression algorithms, and in a very precise technical sense Kolmogorov complexity is a lower bound on compression (up to a constant additive factor). And this theorem tells us that there is absolutely no systematic way to obtain the best compression possible, even in principle.

Minimal plane-tiling extensions of Wang tile sets exist

Given any set T of Wang tiles, there is a minimum number c of Wang's special 11 tiles (for an aperiodic tiling) that need to be added to T to get a set that can tile the plane. But there is no non-constructive way to prove this, otherwise there would be a program that solves the halting problem.

This fact is extremely concrete and is not directly expressing any computation-related statement.

user21820
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  • Now that I think of it, the halting problem is proven by contradiction yes, which now makes me skeptical. – Frazer Apr 04 '24 at 21:57
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    @FraserPye: Anyone can be skeptical of anything. It doesn't change the facts. The real world continues existing and obeying classical FOL whether you like it or not. And so other people who use classical FOL will continue to deduce truths about the real world starting from true assumptions, again whether you like it or not. As Pilcrow said, your ability to use this website is due to an incredible amount of physics, engineering and computer science, all of which has been built by people based on classical FOL. – user21820 Apr 05 '24 at 02:02
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    The halting problem theorem is not proved 'by contradiction.' It is a constructive proof of a (partially) negative statement. "For every Turing machine $H$, there (constructively) exists a Turing machine $M$ and input $I$ such that $H$ does not report the halting behavior of $M(I)$." – Dan Doel Apr 05 '24 at 05:36
  • @DanDoel: Indeed. Note also that FP conflated "halting problem answer" with "halting problem unsolvability". The former is a non-constructive object, whereas the latter (as per your comment) is a constructive theorem. – user21820 Apr 05 '24 at 05:58
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Every area of math has a set of underlying assumptions. Which assumptions you choose will determine which conclusions can be derived.

As an example consider, Euclidean geometry assumes parallel lines never cross. Many students never learn about non-Euclidean geometry. When they do learn about non-Euclidean geometries, it doesn't make Euclidean geometry invalid. Because the initial assumptions differ, this creates a divergence into two or more entirely different spheres of math. Indeed, many would say Euclidean geometry seems to be the most broadly applicable geometry. This despite the fact that other geometries which break Euclidean underlying assumption(s) have existed for hundreds of years. Euclidean geometry describes classical physics perfectly well. However, to grasp relativity non-Euclidean geometries become essential.

  • Each sphere has its use and applications.*

The set of assumptions chosen should be appropriate for the practical application or context to which one hopes to apply the results.

A rural reader
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nickalh
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  • Feel free to correct or clarify my definitions. Instead of parallel lines I should use the parallel postulate. – nickalh Apr 04 '24 at 10:15
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    You wrote "The set of assumptions chosen should be appropriate for the practical application or context to which one hopes to apply the results" but this is precisely the OP's question: is classical logic appropriate for applications? Your answer does not address the issue. – Mikhail Katz Apr 04 '24 at 15:30