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For two sets of numbers $A,B$, write $A<B$ iff $\max A<\min B$. For a sequence of integers $a_0,\cdots,a_{n-1}>0$, let $Prop(a_0,\cdots,a_{n-1})$ denote the following proposition:

Given $n$ sets of integers $A_0<\cdots <A_{n-1}$ with $|A_i| = a_i, i\leq n-1$, let $A = \bigcup\limits_{i\leq n-1}A_i$, there exists a function $f:\mathcal{P}(A)\rightarrow\{0,1\}$ such that for all $k\leq n-1$, any $a_k+1$ many mutually disjoint subsets $B_0,\cdots,B_{a_k}$ with $B_i\cap A_k = \{the\ i^{th}\ large\ element\ in\ A_k\}= \{\min B_i\}$, $0<i\leq a_k$, there exists $I,J\subseteq \{1,\cdots,a_k\}$ such that: $$f(B_0\cup (\bigcup\limits_{i\in I} B_i))\ne f(B_0\cup (\bigcup\limits_{i\in J}B_i)).$$

To get an intuition, this post describes $Prop(2,2,2,2)$, and an answer shows its falseness (the "proof" is given by computer verification).

The question is: does there exists a sequence of integers $a_0,\cdots,a_n,\cdots$ such that for all $m$, $Prop(a_0,\cdots,a_m)$ is true? For a starter, is $Prop(2,2,2,3)$ or $Prop(3,3,3,3)$ true?

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  • $\begingroup$ @PeterHeinig hm, why?! $\endgroup$ – Fedor Petrov Feb 22 '18 at 16:20

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