Theorem

Let A be an object. Let X and Y be members of ℘(A). Assume X⊆A \ Y. Then we have Y⊆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 a proposition.

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

The following background is necessary for the proof.

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

(.) Let A be an object. Let X and Y be members of ℘(A). Assume for all a ∈ A a∈X implies a∈Y. Then we know X⊆Y.

(.) Let A be an object. Let X and Y be members of ℘(A). Assume X⊆A \ Y. Let a be a member of A. Assume a∈Y. Then we know a∈A \ X.

Since X⊆A \ Y, we have for all a ∈ A a∈Y implies a∈A \ X. Using this, we conclude Y⊆A \ X.

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