**Proof**: *(related to "Justification of Set Union")*

- Let $A$ and $B$ be sets.

- By the axiom of pairing there is a set $Z$ containing $A$ and $B$ as elements:

- By the axiom of separation there is a subset $Z’\subseteq Z$ containing exactly $A$ and $B$ as elements:

- By the axiom of union there is a set $Z^*$ containing the elements of $A$ or
^{1}the elements of $B.$

- By the axiom of separation there is a subset $Z^\dagger \subseteq Z^*$ containing exactly the elements of $A$ or
^{1}the elements of $B.$, i.e. $Z^\dagger =\{z\mid z\in A\vee z\in B\}.$

- This is the required set union $A\cup B:=Z^\dagger .$
- The axiom of extensionality ensures the uniqueness of such a set.
- Altogether, we have shown:

^{1} We mean the logical or operation, in the natural English language “and/or”.

q.e.d

| | | | created: 2019-01-12 22:41:54 | modified: 2019-01-12 22:44:47 | by: *bookofproofs* | references: [983]

(none)

[983] **Ebbinghaus, H.-D.**: “Einführung in die Mengenlehre”, BI Wisschenschaftsverlag, 1994, 3

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