Theorem

Let A be an object. Let a be a member of A. Then we have {a}⊆A.

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 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 x and y be objects. Then x=y is a proposition.

(.) Let x and y be objects. Let φ(x) be a proposition depending on an object x. Assume x=y and φ(y). Then we know φ(x).

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

(.) 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∈A. Let x be an object. Assume x∈{a}. Then x=a. Using this, we conclude x∈A.

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