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. Assume φ and ¬φ. Then we know ψ.

(.) "false" is a proposition.

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

(.) Let A, B and x be objects. Assume x∈A⊕B. Let φ be a proposition. Assume x∈A implies ¬x∈B implies φ and ¬x∈A implies x∈B implies φ. Then we know φ.

Proof

It is enough to show x∈A⊕B implies false. Assume x∈A⊕B. Since x∈B, we know x∈A implies ¬x∈B implies false. Since x∈A, we have ¬x∈A implies x∈B implies false. Using this, x∈A implies ¬x∈B implies false and x∈A⊕B, we conclude false.

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