Theorem

Let A, B and C be objects. Then we have A∩(B∪C)=A∩B∪A∩C.

Background

The following background is necessary to understand this theorem.

(.) Let x and y be objects. Then x=y is a proposition.

(.) Let A and B be objects. Then "A∪B" is an object.

(.) Let A and B be objects. Then "A∩B" is an object.

Background for Proof

The following background is necessary for the proof.

(.) Let A and x be objects. Then x∈A is a proposition.

(.) Let A and B be objects. Then "A⊆B" is a proposition.

(.) Let A and B be objects. Assume for all a ∈ A a∈B. Then we know A⊆B.

(.) Let A and B be objects. Assume A⊆B and B⊆A. Then we know A=B.

(.) Let A, B and x be objects. Assume x∈A. Then we know x∈A∪B.

(.) Let A, B and x be objects. Assume x∈B. Then we know x∈A∪B.

(.) Let A, B and x be objects. Let φ be a proposition. Assume x∈A∪B, x∈A implies φ and x∈B implies φ. Then we know φ.

(.) Let A, B and x be objects. Assume x∈A and x∈B. Then we know x∈A∩B.

(.) Let A, B and x be objects. Assume x∈A∩B. Then we know x∈A.

(.) Let A, B and x be objects. Assume x∈A∩B. Then we know x∈B.

Proof

Claim: A∩(B∪C)⊆A∩B∪A∩C.

Proof of Claim: It is enough to show for all x ∈ A∩(B∪C) x∈A∩B∪A∩C. Let x be a member of A∩(B∪C). We know x∈B∪C.

Claim: x∈B implies x∈A∩B∪A∩C.

Proof of Claim: Assume x∈B. We will show x∈A∩B. Clearly, x∈A. Using this and x∈B, we are done. This completes the proof of the claim.

Claim: x∈A∩B implies x∈A∩(B∪C).

Proof of Claim: Assume x∈A∩B. Thus x∈A. Using this, it is enough to show x∈B∪C. Since x∈A∩B, we have x∈B. Using this, we are done. This completes the proof of the claim.

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