@Student Also, it is important to note that by Gödel's completeness theorem, $\phi$ being provable from a theory $T$ is equivalent to $\phi$ being true in all models of $T$. So in this sense, when we quantify over all models, truth and provability become the same.

I'm saying that the model-theoretic notion of truth does not depend very much on the background universe.

Do you mean rather $|R/Q|\nleq |R|$?

@DavidWhite If double-blind leads to less open-access (or slower speed of access), I doubt it being overall a positive trend. That's a big trade-off.

@darijgrinberg I find that due to specialization, anonymity is practically impossible to maintain even for first impressions. I even worry that people are able to reliably guess the identity of the referee based on the way the comments are worded and what kinds of things they say.

The negation can be forced with Mitchell forcing up to an inaccessible, for example for $\kappa=\lambda=\omega_1$.

Just take a model where $2^{<\kappa}\leq\lambda$ and $2^\kappa>\lambda$.

Analytic determinacy is a good example. The original proof was assuming a measurable, but the argument can be carried out using only sharps, and we actually get an equivalence.

Keep reading. The answers have to do with the applications. For example, find the definition of zero sharp and find some theorem showing some equiconsistency with the existence of zero sharp. In the course of understanding this, you should see why there is a top measure and why we care about iterating it.

@AndrésE.Caicedo As there is no proof written in the referenced paper, do you have any idea of why Moschovakis cardinals do not exist?

Why is $\sup j[\lambda] \in \ran(k)$ if we select $\eta \not= \sup j[\lambda]$?