Does proof by contradiction assume that math is consistent? The standard proof by contradiction goes like


*

*It is known that $P$ is true.

*Assume that $Q$ is true.

*Using the laws of logic, deduce that $P$ is false.

*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?
 A: 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.
A: 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.
A: 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.
A: 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.
