Amazing result! I'm surprised about how hard it is to see. Intuitively I would imagine that there is a clever encoding that will immediately give a way to recover the additive reduction of the original model. And I'm very surprised about the multiplicative result

@bof yeah, I think it is fine

@bof the original reference is in French I believe (I think it is this), but you can find a mention of this fact in "Antichain Decompositions of a Partially Ordered Set"

I noticed I made a mistake in my previous comment, the correct function is $f:P→T_2(κ)$ with $f\in 2^{ι(x){\boldsymbol{+1}}}$

The last Proposition is a theorem by Kurepa, who proved the more general form statement: For a cardinal $κ$ let $T_2(κ)=(2^{<κ}, <_{lex})$, if $P$ is a union of $κ$-many anti-chains, then it is embeddable into $T_2(κ)$. Let $(A_α\mid α<κ)$ be some well-ordering of an antichain decomposition (as we may assume they are disjoint), and let $ι(x)$ to be the unique $α$ with $x∈A_α$. Define the witness $f:P→T_2(κ)$ by $f(x)∈2^{ι(x)}$ with $f(x)_α=1$ iff there exists $y∈A_α$ with $y≤x$

Great! I will be waiting for any updates

@JoelDavidHamkins Thanks for the comments, which theorem by Friedman are you referring to?

Thanks for the answer, my intent in this question was the more general manner, indeed if we can show that given an additive cluster structure we can identify M up to isomorphism I would regard this as a positive answer

If the block is unbounded, then certainly the block, otherwise the limits

A good place to start looking at is the limit cardinals bellow $κ$, as most reflecting principle will reflect to some stationary subset of those cardinals

Although it appears that the operators above don't include $<$ in them, so it is not truely the same as the Skolem hull of the above Skolem functions

The ordinals are well ordered, so the definable elements are closed under the Skolem functions defined as "take the minimal witness of ∃yφ(x,y)", so the closure operator on sets of ordinals can in fact looked at as a Skolem hull operator

@ZuhairAl-Johar I have shown that $ZFC+Def$ is conservative to $ZFC+Definability\ rule$, if the former proves $\phi$ a statement in FOST and the latter does not, let $M$ be a witness (a model of the latter + $\lnot\phi$), then the $M_0$ I defined will be a model of $ZFC+Def+\lnot\phi$, which is a contradiction

@JosephVanName considering that Berkeley cardinals are known to be inconsistent (with AC), the assertion "If Berkeley cardinals are found to be inconsistent, not much would change" is as true as you can get