MHB What is the Identity for Unions and Intersections of Sets?

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The discussion focuses on proving the identity $(A\cup B)\cap (B\cup C)\cap (C\cup A) = (A\cap B)\cup (A\cap C)\cup (B\cap C)$. The proof involves demonstrating both subset relationships: first, showing that the left side is contained within the right side, and then vice versa. The argument starts with an element x in the left side, leading to the conclusion that x must belong to at least one of the intersections on the right side. The proof highlights the logical connections between the unions and intersections of the sets involved. Ultimately, the identity is established through careful reasoning about the relationships among the sets.
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$(A\cup B)\cap (B\cup C)\cap (C\cup A) = (A\cap B)\cup (A\cap C)\cup (B\cap C)$

For the identity, we will show $(A\cup B)\cap (B\cup C)\cap (C\cup A) \subseteq (A\cap B)\cup (A\cap C)\cup (B\cap C)$ and $(A\cup B)\cap (B\cup C)\cap (C\cup A) \supseteq (A\cap B)\cup (A\cap C)\cup (B\cap C)$.
Let $x\in (A\cup B)\cap (B\cup C)\cap (C\cup A)$.
Then $x\in A\cup B$ and $x\in B\cup C$ and $x\in C\cup A$.
So $x\in A$ or $x\in B$ and $x\in B$ or $x\in C$ and $x\in C$ or $x\in A$.

So I am stuck at this point.
 
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If ($x \in A$ or $x \in B$) and ($x \in B$ or $x \in C$) and ($x \in C$ or $x \in A$), then ($x \in A$ and $x \in B$) or ($x \in A$ and $x \in C$) or ($x \in B$ and $x \in C$). From there, $x \in (A \cap B) \cup (A \cap C) \cup (B \cap C)$.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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