MHB Proving A∪B=A∩B iff A=B with Boolean Algebra

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To prove that A∪B=A∩B if and only if A=B using Boolean algebra, one must start with the definitions of union and intersection. The union A∪B includes all elements from both sets A and B, while the intersection A∩B includes only the elements common to both sets. If A∪B equals A∩B, it implies that every element of A is in B and vice versa, leading to the conclusion that A and B must be identical. This relationship is supported by the Axiom of Extensionality in set theory, which states that two sets are equal if they contain the same elements. Thus, the proof confirms that A∪B=A∩B holds true if and only if A=B.
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I want to prove:

$$A\cup B=A\cap B\Longleftrightarrow A=B$$ Forall A,B sets

By using the axioms and theorems of the Boolean Algebra.

Any hints ??
 
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solakis said:
I want to prove:

$$A\cup B=A\cap B\Longleftrightarrow A=B$$ Forall A,B sets

By using the axioms and theorems of the Boolean Algebra.

Any hints ??

This answer is an answer based on Zermelo Fraenkel set theory axioms.
So, then consider the definition of "union" .
Given A and B , then $$A \cup B$$ is a set C which contains all the elements that belong to A and all the elements that belong to B. In am more mathematical way C is a set such that $$d \in C$$ implies and is implied by $$d \in A $$ or $$d \in B$$
Now,
$$A\cup B=A\cap B$$

Considering the definition of intersection we get that every element of A is an element of B and every element of B is an element of A and thus from Axiom of Extensonality we get that A=B.
 
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