Theorem

Let A be an object. Then we have A⊆U({A}).

Background

The following background is necessary to understand this theorem.

(.) ∅ is an object.

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

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

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

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

(.) Let x and y be objects. Then we know x∈{x}∪y.

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

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

Proof

It is enough to show for all x x∈A implies x∈U({A}). Let x be an object. Assume x∈A. We have A∈{A}. Using this and x∈A, we conclude x∈U({A}).

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