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The standard proof by contradiction goes like

  1. It is known that $P$ is true.
  2. Assume that $Q$ is true.
  3. Using the laws of logic, deduce that $P$ is false.
  4. Rejecting this contradiction, we are forced to accept the falsity of $Q$.

In rejecting the contradiction we implicitly assume that mathematics is consistent. However, doesn't Godel's (Second) Incompleteness Theorem tell us that the consistency of mathematics cannot be proven? Does this pose a problem?

Antonio Vargas
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Herng Yi
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    If mathetmatics is inconsistent, we don't care what we can prove in it, since it is useless. So worrying about the case where an axiom system is not consistent is hardly of much value. We hope our axiom systems are consistent, and we work with that assumption every day, because inconsistency is a catastrophe. – Thomas Andrews Sep 05 '13 at 13:33
  • @ThomasAndrews Not really. – Red Banana Jan 20 '17 at 04:09

4 Answers4

19

Godel's Incompleteness Theorem does not apply to every mathematical system. However, let us suppose we are working in a system to which it applies. If our system is consistent, your proof of $\neg Q$ is meaningful. If our system is inconsistent, then everything can be proven from it. So your proof tell us that a theorem is a consequence of the axioms of our system.

Elchanan Solomon
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    Well... you also have to assume that the law of excluded middle holds (which is probably more to the point when considering proof by contradiction). – user642796 Sep 05 '13 at 12:58
  • @ArthurFischer do you mean that we must consider the possibility that our system is neither consistent nor inconsistent? – Herng Yi Sep 05 '13 at 13:03
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    @ArthurFischer Strictly speaking LEM is not needed for the outlined proof: if you want to prove $\lnot Q$, then it is constructively legitimate to show that assuming $Q$ leads to a contradiction. The trouble is if you want to prove $Q$ but only show that assuming $\lnot Q$ leads to a contradiction... – Zhen Lin Sep 05 '13 at 13:06
  • @ZhenLin if we show that assuming $\neg Q$ leads to a contradiction, then you say that it is "constructively legitimate" to conclude $\neg(\neg Q)$. Can't we then reduce that to $Q$ without LEM? – Herng Yi Sep 05 '13 at 13:09
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    No, to deduce $Q$ from $\lnot \lnot Q$ requires LEM. The axiom scheme $\lnot \lnot Q \to Q$ is equivalent to LEM. – Zhen Lin Sep 05 '13 at 13:10
  • @HerngYi: The Law of Excluded Middle simply says that every statement is either true or false. If you know that it is impossible for $Q$ to be false you (might) need this law to conclude that $Q$ is actually true. – user642796 Sep 05 '13 at 13:18
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    @ZhenLin: I should really bone up on my intuitionistic logic. You are absolutely correct. – user642796 Sep 05 '13 at 13:18
  • @ZhenLin: If a theorem can be deduced from a system of axioms via a proof by contradiction, can it also be proven formally from those axioms? – Elchanan Solomon Sep 05 '13 at 13:22
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    A proof by contradiction is a perfectly legitimate proof, if you accept LEM. But I reiterate: the proof strategy outlined in the OP is not proof by contradiction; rather, it is a constructively valid "direct" proof of a negative statement. In particular, if you prove that $Q$ leads to a contradiction, then $Q$ is false. You may have confused this with the constructivist stance that $Q$ being true is stronger than $\lnot Q$ being false. – Zhen Lin Sep 05 '13 at 13:25
  • @ZhenLin No, (¬¬Q→Q) is not equivalent to the law of the excluded middle. It comes as very well-known that in Lukasiewicz three-valued logic and Lukasiewicz infinite-valued logic also, that (¬¬Q→Q) hold, but the law of the excluded middle does not (which also qualifies an axiom scheme). – Doug Spoonwood Sep 06 '13 at 01:43
  • @ArthurFischer The law of the excluded middle doesn't say anything about statements as being either true or false and not having a third truth value, that's the principle of bivalence. Also, Zhen Lin isn't absolutely correct or even partially correct. You don't need LEM to deduce Q from ¬¬Q (or equivalently q from NNq). For instance, you could have {CpCqp, CCpqCCqrCpr, CCCpNppp, CCNpNqCqp} as your axiom set under substitution and detachment, and Apq := CCpqq. You can't deduce ApNp... that is CCpNpNp. But, you can deduce CNNqq, and thus infer q from NNq via detachment. – Doug Spoonwood Sep 06 '13 at 01:49
  • @DougSpoonwood I was not talking about those logics. In the presence of intuitionistic logic, it is well-known that the axiom scheme $\lnot \lnot Q \to Q$ is equivalent to LEM. – Zhen Lin Sep 06 '13 at 07:17
  • @ZhenLin In the presence of intuitionistic logic, (¬¬Q→Q) is not an axiom scheme, nor a theorem scheme. I'm not sure that I understand what you mean to assert here. Saying something like "in the presence of intuitionistic logic" means you have the whole apparatus for intuitionistic logic, which I'll denote here by IL axioms and rules. Do you mean to assert that "IL axioms and rules and (¬¬Q→Q)" implies LEM, and "IL axioms and rules and LEM" implies (¬¬Q→Q)? If so, that's not really the equivalence of well-formed formulas, but something more complicated. – Doug Spoonwood Sep 06 '13 at 16:27
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If logic is consistent, we have proven Q false.

If logic is inconsistent, then all statements are false (and true, simultaneously).

Either way, Q is false.

7

The utility of formal proofs implicitly assume that the formal system is consistent (except those weird paraconsistent logics). Proof by contradiction, and more generally resolution proofs, are explicitly propositional logic, which is consistent and complete. Incompleteness is introduced by rules of inference for manipulating quantifiers powerful enough to represent arithmetic, which is beyond mere proof by contradiction.

The incompleteness theorems tells us that formal systems that are powerful enough to represent Robinson arithmetic (or its stronger, more orthodox cousin Peano arithmetic) have undecidable statements, but propositional calculus isn't that powerful and is entirely decidable. Even Presburger arithmetic and extensions of it that involve multiplication by constants are SMT decidable, and are provably complete. So in short, resolution proofs themselves aren't handicapped by incompleteness.

Incompleteness means that its possible there's some new Russel's paradox out there that forces everyone to rebuild the foundations of systems strong enough to represent Robinson arithmetic. The consequences of such a hidden inconsistency is catastrophic, because then all statements can be proven (by contradiction, as you demonstrated) and the system becomes useless; But not necessarily irreparable, as we did recover from Russel's paradox. However, we can take comfort that for many useful weaker systems, they are on perfectly stable, decidable, complete ground.

dezakin
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    What do you mean by "proofs assume the formal system is consistent"? If, in a context where the base theory is supposed to be $\mathsf{ZFC}$, one assumes $\text{Con}(\mathsf{ZFC})$ in the course of proving a theorem, then one had better state it in the hypothesis because the notion of "theorem of $\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})$" is different than the notion of "theorem of $\mathsf{ZFC}$." – Trevor Wilson Sep 06 '13 at 16:30
  • If in the context where the base theory is supposed to be ZFC, one assumes Con(ZFC) if we assume a theorem of ZFC has any useful meaning at all, because if a system isn't consistent, you can prove anything from it and its useless. You have to note Con(ZFC) in certain meta-logical proof systems to formally declare what you're proving, like independence proofs by forcing.

    While its correct to say that the logical consequences of a proof don't assume consistency, a proof being useful does depend on it in most of the logics people are familiar with.

     – dezakin Sep 06 '13 at 19:28
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    However the statement is a bit overreaching, so I've edited it to be more clear: The utility of proofs implicitly assume consistency. – dezakin Sep 06 '13 at 19:36
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I think item $4$ in the question introduces an irrelevancy, namely "Rejecting this contradiction". Before you get to item 4, the available information is that $P$ is true and that $Q$ implies not $P$. So $Q$ implies the contradiction "$P$ and not $P$". In both classical and constructive logic, the negation of a statement is equivalent to "that statement implies a contradiction". So we have the negation of $Q$.

The role of the contradiction here is not to frighten us so that we reject it because of our belief in the consistency of mathematics. Rather it is to serve as the (standard) criterion for negation.

Andreas Blass
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