Theorem

Let A, B and x be objects. Assume A⊆B and x∈A. Then we have x∈B.

Background

The following background is necessary to understand this theorem.

(.) Let A and x be objects. Then x∈A 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 be an object. Let φ(a) be a proposition depending on a member a of A. Then "∀a∈A⋅φ(a)" is a proposition.

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Assume ∀a∈A⋅φ(a). Let a be a member of A. Then we know φ(a).

(.) Let A and B be objects. Then A⊆B is the proposition given by ∀a∈A⋅a∈B.

Proof

Since A⊆B, we have ∀a∈A⋅a∈B. Using this, we conclude x∈B.

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