I’m not sure if it’s helpful, but the above-linked paper gives you the following. If you can write down some set of criteria you want your measure to satisfy, and prove that for every epsilon, there is a finite set on which all of your criteria are satisfied within epsilon, then there is a canonical “filter measure” that gives you a finitely additive partial probability measure satisfying the criteria. If you extend the filter to an ultrafilter, it will yield a total measure.

logicandanalysis.org/index.php/jla/article/view/403

So you take a club $C$ on which for $\alpha<\beta$ from $C$, $\beta$ is above the domain of $p_\alpha$. This is how you reduce compatibility to the lower parts. For the actual range, you need to enumerate your tree forcing and relevant subsets like I said. Sorry I don’t feel like giving more details. Please look up a proof of the stationary chain condition for Cohen forcing to get an idea.

I mean the part below $\gamma$ is not cofinal in $\gamma$.

Basically, the tree part of the forcing has only $\aleph_{\alpha+1}$ many possibilities. Assume $\lambda = \aleph_{\alpha+2}$. Let your regressive function be $f(p_\alpha) = p \restriction \alpha$. Since the domains of the second part have size $\leq\aleph_{\alpha}$, this domain is bounded below $\alpha$ when $cf(\alpha) \geq \aleph_{\alpha+1}$. So if two functions $p_\alpha,p_\beta$ end up having the same lower part, and their upper parts are disjoint, then they are compatible. For the general case, enumerate the ordinals in question in an $\aleph_{\alpha+2}$-sequence.

Pretend for a second that $\lambda = \aleph_{\alpha+2}$. Is the claim clear?

You should define SMP and GMA.

plato.stanford.edu/entries/axiom-choice

Would assuming a countably closed model of ZFC+I3 do the trick?

Why is it more complicated? Just assume there is a structure with the properties of axiom I3, from which you derive much information about finite Laver algebras. You never use the uncountability of $V_\lambda$ in the arguments, or the fact that it contains the real $\mathbb R$, do you?

Why doesn’t the existence of a countable transitive model of these axioms serve the same explanatory role?

@მამუკაჯიბლაძე Not among the “standard” ones, meaning those that can be characterized by definable elementary embeddings.

@JesseElliott It seems you’re saying something more. Empirically, there doesn’t seem to be an alternative system to large cardinals with similar properties, and this calls out for explanation. I think that’s a good point.

237. That’s the limit.

@JesseElliott Thanks, but honestly I think Joel’s answer is better. As to your challenge, what I am claiming in my answer is that there is a naturalistic explanation for the consistency of large cardinals, but I’m not claiming to explain why set theorist tend to favor them over competing hypotheses when it comes to adopting axioms. There may be other frameworks whose consistency can be explained in similar ways, but I’d say large cardinals have the strongest case, both in terms of empirical evidence and theoretical coherence. (Although $V=L$ is a different case.)

@JosephVanName Because we focused a lot of energy on finding the boundary between consistency and inconsistency. When we play around near the edge, we encounter “near inconsistency”. As for why the boundary lies exactly where it does (which we don’t know with certainty), I think that’s just a bare fact.