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I've got this property: For all $S\in 2^A$ and all partitions $P$ of $S$, $f(S) = f\left(\left\{f(M)\mid M \in P\right\}\right)$, i.e. $f$ maps a set of values to a single value and gives the same result whether we map the set "at once" or whether we map subsets first and then map the set of the results. Examples for $f$ would be sums of finite sets of numbers, or the least upper bound of subsets of a complete lattice.

So, if $A = \mathbb{N}\cup\left\{\infty\right\}$ and $f(S) = \sum_{s \in S} s$, the property is fulfilled, as, for example $f(\{1,2,3\}) = f(\{f(\{1,2\}),f(\{3\})\}) = 3+3 = 6$

Is that a well-known property? If so, what's it called so I can read up more on it?

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    $\begingroup$ The lack of explicit quantifiers makes your property hard to understand. Does your property say the following? For every $S \in 2^A$ and for every partition $P$ of $S$, $f(S) = f\left(\left\{f(M)\mid M \in P\right\}\right)$ $\endgroup$
    – Lee Mosher
    Commented Jul 4, 2023 at 15:46
  • $\begingroup$ Yes, exactly. I'll amend the question. $\endgroup$
    – Sacchan
    Commented Jul 4, 2023 at 16:22
  • $\begingroup$ Use \sum if you are talking about a sum. Use \Sigma if you are writing in Greek. $\endgroup$
    – jjagmath
    Commented Jul 4, 2023 at 16:30

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