Theorem

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

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 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 A⊆B and x∈A. Then we know x∈B.

(.) 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, 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

It is enough to show for all a ∈ A a∈X implies a∈Y. Let a be a member of A. Assume a∈X. We will show ¬a∈Y implies false. Assume ¬a∈Y. Using a∈X, it is enough to show ¬a∈X. It is enough to show a∈A \ X. Since ¬a∈Y, we have a∈A \ Y. Using this and A \ Y⊆A \ X, we are done.

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