Theorem

Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then we have {a∈A|φ(a)}⊆A.

Background

The following background is necessary to understand this theorem.

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

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then {a∈A|φ(a)} is an object.

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

Background for Proof

The following background is necessary for the proof.

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

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

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then we know {a∈A|φ(a)}∈℘(A).

Proof

We know {a∈A|φ(a)}∈℘(A). Using this, we conclude {a∈A|φ(a)}⊆A.

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