Theorem

Let A be an object. Let X be a member of ℘(A). Let a be a member of A. Assume a∈X. Then we have a∈A \ (A \ X).

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. Then ℘(A) is an object.

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

Background for Proof

The following background is necessary for the proof.

(.) Let φ be a proposition. Then ¬φ is a proposition.

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

(.) "false" is a proposition.

(.) Let φ be a proposition. Assume φ implies false. Then we know ¬φ.

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

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

Proof

We will show ¬a∈A \ X. It is enough to show a∈A \ X implies false. Assume a∈A \ X. Hence ¬a∈X. Using this and a∈X, we conclude false.

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