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I was recently reminded of this proof technique in Wikipedia using the Riemann hypothesis along with the law of excluded middle, whether or not this hypothesis is true or not and may even be independent.

https://en.wikipedia.org/wiki/Riemann_hypothesis#Excluded_middle

Is this a logically sound step in proving things? Does the independence of a certain claim of any relevance for the application of the law of excluded middle along with it? It feels kinda fishy. On the other hand, I used it so many times along with claims that may or may not be independent, that maybe it's not so different after all.

Feelix
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Here's a silly, but hopefully helpful, example: Suppose I have some integer $p$. You don't know the value of $p$, but you want to figure out whether $p+p$ is even or odd.

You argue as follows:

  • Either $p$ is even, or $p$ is odd.

  • If $p$ is even, then $p=2k$, so $p+p=2k+2k=4k=2(2k)$ is even.

  • If $p$ is odd, then $p=2k+1$, so $p+p=2k+1+2k+1=4k+2=2(2k+1)$ is even.

  • So $p+p$ is even.

At no point did you learn anything about $p$, or the truth/falsity of the statement "$p$ is odd". In fact, the parity of $p$ may be independent from your favorite set of axioms (say, ZFC): for instance, what if $p$ is defined to be $1$, if the Riemann hypothesis is true, and $2$, if the Riemann hypothesis is false? Nonetheless, you can say with absolute certainty that $p+p$ is even.

That's what's going on here, just on a more complicated level.

Noah Schweber
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This is a valid proof technique. The key is that such statements are not only independant of GRH, but that they are true independent of GRH. Likewise, if both GRH and "not GRH" imply that a statement is false, then that statement is false.

Let me elaborate. When we say that $P$ is independent of $A$, we mean that the truth or falsehood of $P$ does not depend on the truth or falsehood of $A$.

When we say that $P$ is true independent of $A$, we mean that $(A\Rightarrow P)\wedge(\neg A\Rightarrow P)$. It is a formal principle of logic that $(A\Rightarrow P)\wedge(\neg A\Rightarrow P)\vdash P$

Stella Biderman
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