Theorem

Let A and B be objects. Let a be a member of A. Let b be a member of B. Then we have {a,b}∈℘(A∪B).

Background

The following background is necessary to understand this theorem.

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

(.) ∅ is an object.

(.) Let x and y be objects. Then {x}∪y is an object.

(.) Let A be an object. Then ℘(A) is an object.

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

(.) We use {x1,...,xn} as shorthand notation for the finite set with elements x1, ..., xn.

Background for Proof

The following background is necessary for the proof.

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

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

(.) Let A and B be objects. Let a be a member of A. Let b be a member of B. Then we know {a,b}⊆A∪B.

Proof

Clearly, {a,b}⊆A∪B. Using this, we conclude {a,b}∈℘(A∪B).

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