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Let $\mathcal{P}(\{0,\cdots,7\})$ denote all subsets of $\{0,\cdots,7\}$.

Is the following true?

For any function $f: \mathcal{P}(\{0,\cdots,7\})\rightarrow\{0,1\}$, there exists $0\leq k\leq 3$, three mutually disjoint sets $A,B,C\subseteq \{0,\cdots,7\}$ such that $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C)?$$

Let $\mathcal{P}(\{0,\dotsc,7\})$ denote the power set of $\{0,\dotsc,7\}$.

Is the following true?

For any function $f: \mathcal{P}(\{0,\dotsc,7\})\rightarrow\{0,1\}$ there exists $0\leq k\leq 3$ and three mutually disjoint sets $A,B,C\subseteq \{0,\dotsc,7\}$ such that $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C)?$$

Let $\mathcal{P}(\{0,\cdots,7\})$ denote all subsets of $\{0,\cdots,7\}$.

Is the following correct?

For any function $f: \mathcal{P}(\{0,\cdots,7\})\rightarrow\{0,1\}$, there exists $0\leq k\leq 3$, three mutually disjoint sets $A,B,C\subseteq \{0,\cdots,7\}$ such that $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C).$$

I can't find a counter example by far.

Let $\mathcal{P}(\{0,\cdots,7\})$ denote all subsets of $\{0,\cdots,7\}$.

Is the following true?

For any function $f: \mathcal{P}(\{0,\cdots,7\})\rightarrow\{0,1\}$, there exists $0\leq k\leq 3$, three mutually disjoint sets $A,B,C\subseteq \{0,\cdots,7\}$ such that $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C)?$$

Let $\mathcal{P}(\{0,\cdots,7\})$ denote all subsets of $\{0,\cdots,7\}$.

For any function $f: \mathcal{P}(\{0,\cdots,7\})\rightarrow\{0,1\}$, there exists $0\leq k\leq 3$, three mutually disjoint sets $A,B,C\subseteq \{0,\cdots,7\}$ such that $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C).$$

Let $\mathcal{P}(\{0,\cdots,7\})$ denote all subsets of $\{0,\cdots,7\}$.

Is the following correct?

For any function $f: \mathcal{P}(\{0,\cdots,7\})\rightarrow\{0,1\}$, there exists $0\leq k\leq 3$, three mutually disjoint sets $A,B,C\subseteq \{0,\cdots,7\}$ such that $\min A = 2k, \min B = 2k+1,$ and $$f(C) = f(A\cup C) = f(B\cup C) = f(A\cup B\cup C).$$

I can't find a counter example by far.