Theorem

Let x, y, z and u be objects. Assume {{x},{x,y}}={{z},{z,u}}. Then we have y=u.

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 x and y be objects. Then {x}∪y is an object.

(.) We use {x1,...,xn} as shorthand notation for the finite set with elements x1, ..., xn.

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 and ψ be propositions. Then φ∨ψ is a proposition.

(.) Let x and y be objects. Let φ(x) be a proposition depending on an object x. Assume x=y and φ(x). Then we know φ(y).

(.) Let φ and ψ be propositions. Assume φ∨ψ. Let θ be a proposition. Assume φ implies θ and ψ implies θ. Then we know θ.

(.) Let x and y be objects. Assume x=y. Then we know y=x.

(.) Let x and y be objects. Let φ(x) be a proposition depending on an object x. Assume x=y and φ(y). Then we know φ(x).

(.) Let x and y be objects. Then we know y∈{x,y}.

(.) Let x, y and z be objects. Assume z∈{x,y}. Then we know z=x∨z=y.

(.) Let x, y, z and u be objects. Assume {{x},{x,y}}={{z},{z,u}}. Then we know x=z.

(.) Let x, y, z and u be objects. Assume {{x},{x,y}}={{z},{z,u}} and z=u. Then we know y=u.

(.) Let x, y and z be objects. Assume {x,y}={z}. Then we know x=y.

Proof

Claim: {z,u}={x}∨{z,u}={x,y}.

Proof of Claim: We will show {z,u}∈{{x},{x,y}}. We have {z,u}∈{{z},{z,u}}. Using this and {{x},{x,y}}={{z},{z,u}}, we are done. This completes the proof of the claim.

Consider the case in which {z,u}={x}. Thus z=u. Using this and {{x},{x,y}}={{z},{z,u}}, we conclude y=u.

Consider the case in which {z,u}={x,y}.

Claim: u=x∨u=y.

Proof of Claim: It is enough to show u∈{x,y}. We know u∈{z,u}. Using this and {z,u}={x,y}, we are done. This completes the proof of the claim.

Consider the case in which u=x. Using {{x},{x,y}}={{z},{z,u}}, it is enough to show z=u. We will show u=z. Since {{x},{x,y}}={{z},{z,u}}, we have x=z. Using this and u=x, we are done.

Consider the case in which u=y. Using this, we conclude y=u.

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