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In a standard proof by contradiction of the statement p => q “p implies q”, we can suppose the statement is false. If we then derive an absurdity then the statement is shown to not be false, hence it must be true.

But it seems when we begin our proof by contradiction we begin with the claim:

the statement is either true or false. Suppose it is false...

Hence when we show the statement cannot be false, the only other possible option/case is true. But it seems there is another option in that the statement could be unprovable, so the original claim in yellow above should read:

the statement is either true or false or unprovable.

I am hoping someone could show me why a proof by contradiction still works and where I have gone wrong.

Also: I believe I remember my professor saying that if a statement is unprovable then it can be shown to both be true AND false. Is this correct?

Note: I am not acquainted with formal logic notation

H_1317
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  • What makes a statement unprovable? If a statement cannot be proven, it is due to the fault in the mathematician's efforts, not the statement itself. – Mr Pie Dec 30 '19 at 05:32
  • Ahh oops it cannot be proven true or false – H_1317 Dec 30 '19 at 05:33
  • Correct? @MrPie – H_1317 Dec 30 '19 at 05:34
  • Though i should add this to the post, i thought i remember my professor saying if a statement is unprovable it can be shown to be both true AND false – H_1317 Dec 30 '19 at 05:36
  • A statement in mathematics is either true or false. (That is the law of excluded middle) Whether it is provable is a different question altogether and depends on the assumptions you are willing to make, but it does not constitute a third option – aman Dec 30 '19 at 05:37
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    For example, consider the Goldbach conjecture: Every even number is the sum of two primes. Clearly, it is either true or false: If it is false, there is a counter-example, and if it is true, there is no counter-example. However, we don't know if the statement is PROVABLE if it is true. If we can assume it is false, and then arrive at an absurdity, that would constitute a proof – aman Dec 30 '19 at 05:40
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    No statement can be both true and false. However, we can show that it is impossible to prove it to be true or false within a particular axiomatic system – aman Dec 30 '19 at 05:42
  • If the statement was unprovable, would we be able to both prove it false and true? – H_1317 Dec 30 '19 at 05:42
  • @H_1317 No, that would be proving a contradiction, which is very bad. What you can show is that you can't derive a contradiction by assuming it true or assuming it false. – eyeballfrog Dec 30 '19 at 05:43
  • If a statement is un-provable, and it is ASSUMED to be true (added as an axiom), it might be possible to use that assumption to prove that it is also not true. However, this is not a proof by contradiction, because assuming that it is not true would result in proving its truth – aman Dec 30 '19 at 05:45
  • So to recap: if Statement is unprovable, and i attempt a proof by contradiction, i will never be able to prove it is false as unprovable means it can not be proven true or false — unprovable does not mean the statement can both be proven true and false. Correct? – H_1317 Dec 30 '19 at 05:51
  • Yes that is correct – aman Dec 30 '19 at 08:40

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In standard logic we accept all tautologies, which include the law of the excluded middle. For any statement $\phi$ we have $\phi \vee \lnot \phi$. If you start with $\phi$ and derive a contradiction, you can conclude $\lnot \phi$

Depending on your axioms, there may be many statements $\phi$ where we cannot derive either $\phi$ or $\lnot \phi$. This means the axioms are incomplete. In the metatheory this tells us that our axioms are consistent, as from inconsistent axioms you can derive anything. In this case you can assume $\phi$ and still cannot prove a contradiction. You can also assume $\lnot \phi$ and cannot prove a contradiction. You can then add either $\phi$ or $\lnot \phi$ to your axioms and still have a consistent set.

No, if you cannot prove $\phi$ or $\lnot \phi$ that does not mean that $\phi$ is both true and false. It means there are models of the axioms where $\phi$ is true and also models where $\phi$ is false.

Ross Millikan
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  • What do you mean by “models of the axioms”? – H_1317 Dec 30 '19 at 06:03
  • Given a set of axioms, a model is a set of elements, functions, and whatever is mentioned in the axioms that satisfies them. If the axioms are those of a group, every group is a model of the axioms. We cannot derive commutivity or its negation from the group axioms, so we can add that or its negation and not have a contradiction. Thus there are abelian and nonabelian groups. – Ross Millikan Dec 30 '19 at 06:13
  • Ok so a nonabelian group is a “model of the axioms” as it is an object that satisfies the group axioms, as is an abelian group. So we have 2 models, the abelian group and the non abelian group. Hence there is 1 model that satisfies the statement $\phi$ = “all groups are abelian” and 1 model that satisfied the negation. Is this an example of when you say “it means there are models of the axioms where $\phi$ is true also models where $\phi$ is false”? – H_1317 Dec 30 '19 at 06:24
  • Yes, exactly. Many theories we use are incomplete in that there are statements in the language that we can neither prove or disprove. Some, like the reals, are complete. Kurt Gödel famously showed that basic arithmetic is incomplete. – Ross Millikan Dec 30 '19 at 14:44
  • Though the $\phi$ i describe isnt an unprovable statement. So are u saying with $\phi$ unprovable in general that if we restrict the possible models in our space then $\phi$ will be true in 1 set of restrictions and false depending on other restrictions (ie eliminate all non abelian groups is one such restriction)? Is this restriction what u mean by “so we can add that or its negation” ie add an axiom saying all groups must be abelian or nonabelian and still have a consistent system. .... im trying to think of the continuum hypothesis — how is there a model in both cases for that statement? – H_1317 Dec 30 '19 at 15:24
  • The situation with CH is just like abelian groups. ZFC (assuming it is consistent) has many models. Some of them satisfy CH, some do not. The proof that CH is consistent with ZFC is usually done with Gödel's constructible universe, where you restrict the power set axiom as much as you can. The proof that not CH is consistent goes through forcing, which is not easy to learn. – Ross Millikan Dec 30 '19 at 15:42
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Proofs don't exist in vacuum of course. Proofs start from somewhere. A theory.

Truth also does not exist in vacuum. Truth is relative to a structure, usually one that satisfies a certain theory.

If you underlying logic, in this case I suppose first-order logic, is sound, then every provable statement is true in all the models of the theory. And if the logic is complete, then a statement true in all the models of a theory is also provable.

In some cases we say that something is true to mean that there is a concrete structure in which we measure that truth value. For example, in the case of arithmetic this would be $\Bbb N$, but this is far from the unique model of Peano's axioms. In other cases when we say true we really just mean provable (e.g. $\forall x,y:\ x\cdot y=y\cdot x$ is not true in the case of groups because there are groups where it is false; whereas $\forall x\exists y:x\cdot y=y\cdot x$ is true since we can take $y$ to be the identity, but it's not "true", instead it is provable from the axioms of a group).

And therein lies the rub. You are conflating truth and provability since you haven't set the context in sufficient details to distinguish them.

In the case of a proof by contradiction relying on an unprovable statement, you will either fail to complete the contradiction, or you end up proving that if the unprovable statement is true in your model, then the consequent is also true in that model.

Mind you, you still prove (by contradiction) the implication is true, but nobody promised you that the antecedent is true.

Asaf Karagila
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the statement is either true or false or unprovable.

'unprovable' is not mutually exclusive from 'true' or 'false'. I can tell you that the shirt I am wearing right now is either red or blue, but I am not telling you which it is. Then you could say:

your shirt is either red or blue or of unknown color ... and I pick 'unknown'!

but obviously, just because you don't know what color my shirt is does not mean that it can no longer be either red or blue

Same for the truth of statements. Just because the truth-value of some statement is unprovable and unknown does not mean that it is suddenly no longer true or false.

Also: I believe I remember my professor saying that if a statement is unprovable then it can be shown to both be true AND false. Is this correct?

Absolutely not. See my shirt

Bram28
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