Theorem

Let A and B be objects. Assume A∩B=B. Then we have B⊆A.

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 a proposition.

(.) 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 φ and ψ be propositions. Then φ≡ψ is a proposition.

(.) Let φ and ψ be propositions. Assume φ≡ψ and φ. Then we know ψ.

(.) Let x be an object. Then we know x=x.

(.) Let x and y be objects. Assume x=y. Then we know y=x.

(.) Let A and B be objects. Assume A=B. Let x and y be objects. Assume x=y. Then we know x∈A≡y∈B.

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

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

Proof

It is enough to show for all b ∈ B b∈A. Let b be a member of B. We will show b∈A∩B. Since A∩B=B, we know b∈B≡b∈A∩B. Using this, we are done.

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