Proofs of the form $(P\lor\neg P)\implies Q$ Suppose I have a statement $S$ along with two contidtional proofs:


*

*A proof that the Riemann hypothesis implies $S$, and

*Another proof that the negation of the Riemann hypothesis also implies $S$.


Can we say I proved $S$? Or did I leave out the possibility that the Riemann hypothesis is undecidable? Also it would bw interesting to know about instances of this (one might call it "undecidability dilemma").
 A: Even if $P$ is independent of your base theory $T$, this is still a valid proof (assuming classical logic, as mentioned by joro).
Given any model of $T\cup\{P\}$, we know that $Q$ holds by the first proof. Given any model of $T\cup\{\lnot P\}$, we know that $Q$ holds by the second proof. In any model of $T$ either $P$ holds or $\lnot P$ holds. So in every model of $T$ we know that $Q$ holds. Therefore $T$ proves $Q$.
And of course this is a long way for a short drink of water. We can take a shorter, syntactic route. $T$ proves that: $P\rightarrow Q$ and $\lnot P\rightarrow Q$, therefore it proves that $P\lor\lnot P\rightarrow Q$, and therefore $T$ proves $Q$.
A: Maybe what you are confused about is the following:
If you showed that a proof of the Riemann hypothesis would give you $S$ and that a proof of the negation of the Riemann hypothesis would also give $S$, then you cannot conclude that $S$ is true.
A: Decidability has nothing to do with it.  Issues of decidability in the context of classical logic involve propositions that cannot be proved from the axiomatic system you are working in.  For example if you work in ZFC you can't prove either the continuum hypothesis or its negation (by the work of Goedel and Cohen).
In classical logic "P or not P" is always true and therefore your $S$ is indeed proved to be true.
The intuitionists will not accept your argument but then again there are many arguments they won't accept, such as the proof of the extreme value theorem.
