Theorem

Let A and B be objects. Assume ¬A⊆B. Then we have ¬A=B.

Background

The following background is necessary to understand this theorem.

(.) Let x and y be objects. Then x=y is a proposition.

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

(.) Let A and B be objects. Then "A⊆B" is a proposition.

Background for Proof

The following background is necessary for the proof.

(.) Let A and x be objects. Then x∈A 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(x). Then we know φ(y).

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

(.) Let A and B be objects. Assume for all a ∈ A a∈B. Then we know A⊆B.

Proof

Using ¬A⊆B, it is enough to show A=B implies A⊆B. Assume A=B. We have A⊆A. Using this and A=B, we conclude A⊆B.

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