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.

(.) ∅ 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 and x be objects. Then x∈A is a proposition.

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

(.) Let φ and ψ be propositions. Then φ≡ψ is a proposition.

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

(.) "false" is a proposition.

(.) Let x be an object. Assume x∈∅. Let φ be a proposition. Then we know φ.

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

(.) Let x be an object. Then we know x=x.

(.) Let A and B be objects. Assume A=B. Let x and y be objects. Assume x=y. Then we know x∈A≡y∈B.

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

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

Proof

It is enough to show for all x x∈A implies x∈B. Let x be an object. Assume x∈A. We will show ¬x∈B implies false. Assume ¬x∈B. It is enough to show x∈∅. Since A \ B=∅, we know x∈A \ B≡x∈∅. Since ¬x∈B and x∈A, we have x∈A \ B. Using this and x∈A \ B≡x∈∅, we are done.

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