Theorem

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

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

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

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

Proof

We will show for all x ∈ X x∈Y∩Z. Let x be a member of X. Since X⊆Y, we have x∈Y. Since X⊆Z, we know x∈Z. Using this and x∈Y, we conclude x∈Y∩Z.

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