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The answers to the question Can every true theorem that has a proof be proven by contradiction? show that if a theorem can be proven directly, it can be proven by contradiction.

The main arguments presented there are of the form:

Let P be a proof of the theorem. Assume the theorem is false. However we can exhibit P, which contradicts the assumption

However this argument assumes that we can rely on the validity of the direct proof in our proof by contradiction.

My question is then, if you have a theorem that can be proven directly, is it possible to prove the theorem by contradiction, but without using the fact that there exists a direct proof in your proof by contradiction?

Bernard
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Kenshin
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    Yes, assume not $P$, then prove $P$, and you have a contradiction to not $P$, thus proving not not $P$. – Angina Seng Aug 12 '18 at 10:38
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    That's a question that does not seem to be easy to formalize. What's the motivation for it ? (If there is any, besides curiosity, which is a valid reason for asking of course) – Maxime Ramzi Aug 12 '18 at 10:38
  • @LordSharktheUnknown, yes but the bold of my question states you cannot use the direct proof in your argument. – Kenshin Aug 12 '18 at 10:39
  • @LordSharktheUnknown : look at the specific question, which is asking if we can do it "without using a proof of $P$" – Maxime Ramzi Aug 12 '18 at 10:39
  • I think the answer is no. Remember that Godell already said that some theorems are not even probable by the maths we have. So in order to prove that those theorems are true the only possible thing we can do is prove that they are unprovable. Otherwise, they would have a counterexample which we could find. – Erik T. Aug 12 '18 at 10:57
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    @ErikT., to be clear, I'm only referring to the theorems that can already be proven directly. – Kenshin Aug 12 '18 at 11:00
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    If you prove a theorem then do you really use the fact that a direct proof exists? It is more that you show or confirm that a direct proof exists. – drhab Aug 12 '18 at 11:13

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Not necessarily -- but for even less interesting reasons than the argument you don't like!

The following is fact:

First-order Peano Arithmetic proves $\forall x\forall y( x\cdot Sy = x\cdot y + x )$.

The proof is very simple -- since the claim to be proved is one of the axioms of PA, simply stating it with "(axiom)" next to it constitutes a proof.

On the other hand, the other axioms of PA are not sufficient to prove the claim -- to see this, observe that we get a model of those other axioms by taking the universe to be $\mathbb N$, $S$ to be the successor function, $+$ to be addition, and $\cdot$ to be the constant function that always returns $0$. This model does not satisfy the claim we want to prove, so since first-order logic is sound, the claim does not have a proof.

In other words: Every proof of $\forall x\forall y( x\cdot Sy = x\cdot y + x )$ in PA must at some point prove $\forall x\forall y( x\cdot Sy = x\cdot y + x )$ directly -- because that is the only way the proof can depend on that axiom, as it must.

hmakholm left over Monica
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  • (+1) I've heard that there is a theory of topology of proofs - this must be an illustration of what zero-dimensional looks like! – Arnaud Mortier Aug 12 '18 at 11:17
  • But this can be proven by contradiction. First assume that your axiom is wrong, but since it contradicts the axiom, it must be true by contradiction. I note that this has a similar structure to the argument I wish to avoid, but in this case I'm using the axiom rather than the "direct proof" in my contradiction which is permitted. – Kenshin Aug 12 '18 at 11:18
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    @Kenshin: That's exactly the reasoning you say in your question you want to exclude. – hmakholm left over Monica Aug 12 '18 at 11:19
  • @HenningMakholm, not quite, I wish to avoid relying on the proof. If it so happens an axiom is the same as it's proof, I can use the axiom. I can edit the question for clarity in this regard. – Kenshin Aug 12 '18 at 11:20
  • @HenningMakholm, otherwise I couldn't use any axiom, as one could argue that all axioms are proofs (trivially). I wish to allow axioms in the proof by contradiction, just not direct proofs. – Kenshin Aug 12 '18 at 11:20
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    @Kenshin: If you want to ask a different question than the one you have asked here, ask it as a new question rather than moving the goalposts on the one you have already asked and gotten an answer to. – hmakholm left over Monica Aug 12 '18 at 11:20
  • @HenningMakholm, no problem, happy to do that also :) – Kenshin Aug 12 '18 at 11:21
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    @Kenshin: And when you do so, be sure to provide an explicit definition of your (apparently extremely nonstandard) concept of "direct proof". – hmakholm left over Monica Aug 12 '18 at 11:22
  • @HenningMakholm, I think you have misinterpreted my comments. My definition of direct proof is standard, it is my restrictions on what I wish to allow in the proof by contradiction that I want to adjust. – Kenshin Aug 12 '18 at 11:24
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    @Kenshin: It sure sounds like you are using a concept of "direct proof" according to which pointing out that such-and-such is an axiom does NOT constitute a "direct proof". Such a concept is necessarily non-standard, because in any remotely sane and standard way of using the words, pointing to an axiom is the most direct way one can conceivably prove anything. – hmakholm left over Monica Aug 12 '18 at 11:26
  • @HenningMakholm, you have misinterpreted because I agree with you that an axiom is a trivial direct proof. It is the restriction I want to adjust in my proof by contradiction, because I will allow axioms (i.e. trivial direct proofs) to be used in the proof by contradiction. – Kenshin Aug 12 '18 at 11:28
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    @Kenshin: Let me quote you: "I wish to allow axioms in the proof by contradiction, just not direct proofs." If you disallow direct proofs everywhere in the proof tree and you agree with the standard meaning of "direct proof", then you're necessarily also disallowing axioms, because mentioning an axiom constitutes a direct proof. – hmakholm left over Monica Aug 12 '18 at 11:29
  • @HenningMakholm, we can both read the comments, my comment there could have been clearer, but nevertheless, I think we both realise at this point we're in agreement. – Kenshin Aug 12 '18 at 11:32
  • @Kenshin: We are not in agreement if you think you can disallow subproofs that are (according to the standard meaning) direct proofs without also disallowing every use of an axiom. – hmakholm left over Monica Aug 12 '18 at 11:33
  • @HenningMakholm, I accepted your answer didn't I? I therefore do allow all direct proofs here. The new question will be subtly different, with new conditions as requested. – Kenshin Aug 12 '18 at 11:34
  • @Kenshin: I'm talking about the revised question you say you're going to ask. The one where you want to disallow direct proofs but allow axioms. – hmakholm left over Monica Aug 12 '18 at 11:34
  • @HenningMakholm, but cannot I constrain the question to only allow direct proofs to be used in the proof by contradiction where those direct proofs are axioms, or derivations using proof by contradiction from those axioms. – Kenshin Aug 12 '18 at 11:38
  • @HenningMakholm, perhaps this will clarify: Suppose all direct proofs are deemed invalid, yet axioms are valid. In this case, while axioms constitute direct proofs, they also exist independently as axioms. So it is wrong to prove the axiom from itself in this scenario, but it isn't wrong to prove the axiom using contradiction trivially. This will be the set up of the new question. – Kenshin Aug 12 '18 at 11:40
  • @Kenshin: Sure. (But it would be easiler to describe that as "the only allowed inference rules are proof-by-contradiction and axioms", and not mention direct proofs at all). – hmakholm left over Monica Aug 12 '18 at 11:40
  • @HenningMakholm yes I agree with you – Kenshin Aug 12 '18 at 11:40