Theorem

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

Background for Proof

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

Proof

We know for all a ∈ A a∈A \ (A \ X) implies a∈X. Using this, we conclude A \ (A \ X)⊆X.

Click here to modify variable names (JavaScript Must Be Enabled).

Hide Background.

Test yourself on this item.