How could the Riemann Hypothesis be independent of ZFC? I have read: Can the Riemann hypothesis be undecidable? from MO, and If the Riemann Hypothesis were independent of ZFC, would this mean that $\sigma(n)\leq H_n+e^{H_n} \log(H_n)$  is also independent of ZFC? But how could such an inequality be independent of ZFC? Wouldn’t that just imply that it is true (because no counterproof can exist)? from Quora. The Quora answer (to the best of my understanding) tells me that supposing we have a stronger system than ZFC (abbr. "SS") in which we prove that the Riemann Hypothesis ("RH") is independent of ZFC, then this is a proof of RH in SS because if RH was false in SS, then there would be a counterexample to Robin's inequality, which would be detectable in ZFC (because Robin's inequality is just simple arithmetic), i.e. RH would be provably false in ZFC, contradicting our assumption that "there is a proof in SS that RH is independent of ZFC".
Is this understanding correct? It's just that I see the statement "Riemann hypothesis may be true but unprovable in ZFC", but have always been confused about it (true with respect to what system? unprovable w.r.t. what system?). Would a better formulation be: "RH may be unprovable in (i.e. independent of) ZFC (with respect to a stronger system SS), in which case the RH is true (i.e. provable) in SS; however in this case there will also be another stronger system SS2 (stronger than ZFC) such that RH is provably false in SS2 (because that is what being independent of ZFC means)"?
EDIT: but in this latter case, wouldn't a proof of the falsity of RH in SS2 translate to a counterexample in ZFC (which would again violate that we proved RH was independent of ZFC in SS, right?)? Unless SS2 is "non-constructive" in such a way that we could proof that RH is false in SS2 without ever being able to get a counterexample. I'm not sure how that would work.
Thanks for any further clarifications.
 A: I believe (but cannot find a source for this - someone help me out?) that the Riemann Hypothesis can be translated into a first-order statement about natural numbers.
That is, one can form a statement $P$ using only first-order logic and the fumction symbols $0$, $s$, $+$, and $\cdot$ which is logically equivalent to the Riemann Hypothesis in the sense that ZFC proves $\mathbb{N} \models P$ iff the Riemann Hypothesis is true.
Most mathematicians tend to agree that when you make a first-order statement about the natural numbers using only the symbols $0$, $s$, $+$, and $\cdot$, the statement has a "real meaning". That is, each such statement is "really" true or false when talking about the "real" natural numbers.
The truth value of the Riemann Hypothesis is, in a certain sense, meaningful.
But we can go even further. If I recall correctly, the statement $P$ is logically equivalent to a statement of the form $\forall n (f(n) = 0)$, where $f$ is a primitive recursive function. This means that if the Riemann Hypothesis is true in any model of $ZFC$, then $P$ is true in that model of $ZFC$. And this means that in this model of $ZFC$, the statement $\forall n (f(n) = 0)$ is true.
But consider some "actual natural number" $k$. Then in this model of $ZFC$, $f(k) = 0$. But we can calculate $f(k)$ ourselves, so our calculation must give us that $f(k) = 0$. That means that for all "actual numbers" $k$, $f(k) = 0$.
So if $P$ is true in any model of ZFC, it must be a true statement about "the real natural numbers". And therefore the Riemann Hypothesis must "really be true".
