Theorem or axiom: $x\in A \Leftrightarrow x\in A\wedge(x\in B\vee x\not\in B)$

Question

Part of my proof I had to use the following statement - Let $A,B$ be sets. So: $$ x\in A \Leftrightarrow x\in A\wedge(x\in B\vee x\not\in B) $$ Although I'm not sure if its a real theorem or an axiom. I'm trying to be more formal so I want explain why this statement is true. Is it an axiom? Maybe the "Axiom of regularity"? If it is not an axiom, what is the right way to prove this theorem?

Answer 1

$$x\in A \Leftrightarrow x\in A\wedge(x\in B\vee x\not\in B)$$

is a theorem.

You may use the fact that $$(x\in B\vee x\not\in B)=1$$

is a tautology and $$ x\in A\wedge 1 \iff x\in A $$