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

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SUMMARY

The discussion focuses on proving the equivalence \(A \cup B = A \cap B \Longleftrightarrow A = B\) using Boolean Algebra. The proof relies on the definitions of union and intersection, where \(A \cup B\) is defined as the set containing all elements from both sets A and B. By applying the Axiom of Extensionality from Zermelo-Fraenkel set theory, it is established that if the union and intersection of two sets are equal, then the sets themselves must be equal.

PREREQUISITES
  • Understanding of Boolean Algebra
  • Familiarity with set theory, specifically Zermelo-Fraenkel axioms
  • Knowledge of set operations: union and intersection
  • Basic mathematical logic and proofs
NEXT STEPS
  • Study the Axiom of Extensionality in Zermelo-Fraenkel set theory
  • Explore the properties of union and intersection in set theory
  • Learn about Boolean Algebra theorems and their applications
  • Practice proving set equivalences using formal logic
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Mathematicians, computer scientists, and students studying set theory and Boolean Algebra, particularly those interested in formal proofs and logical reasoning.

solakis1
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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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