Theorem

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

Background

The following background is necessary to understand this theorem.

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

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

(.) Let x and y be objects. Then "x⊕y" is an object.

Background for Proof

The following background is necessary for the proof.

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

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then {a∈A|φ(a)} is an object.

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

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

(.) Let A and B be objects. Then "A∪B" is an object.

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

(.) Let A and B be objects. Then A⊕B is the object given by {x∈A∪B|¬x∈A∨¬x∈B}.

Proof

We will show x∈{x∈A∪B|¬x∈A∨¬x∈B}. Since ¬x∈A, we have ¬x∈A∨¬x∈B. Using this, we are done.

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