Bourbaki

Example

Tag 03TO, permanent. Cite it as [Bourbaki, Tag 03TO].

Theory of Sets, chapter I, § 1. Terms and relations

If we replace xx by □\square wherever xx occurs in the assembly ∨∈xy=xx\vee\in xy=xx, we obtain the assembly ∨∈□y=□□\vee\in\square y=\square\square.

If AA is an assembly and we are interested particularly in a letter xx, or two distinct letters xx and yy (which may or may not appear in AA), we shall often write A{x}A\{x\} or A{x,y}A\{x,y\}. In this case we write A{B}A\{B\} instead of (B∣x) A(B\mathbin{|}x)\,A. We denote by A{B,C}A\{B,C\} the assembly obtained by simultaneously replacing xx by BB and yy by CC wherever they occur in AA (note that xx and yy may appear in BB and in CC); if x′x' and y′y' are distinct letters, other than xx and yy, which appear in neither AA, BB, nor CC, then A{B,C}A\{B,C\} is the same as (B∣x′)(C∣y′)(x′∣x)(y′∣y)A(B\mathbin{|}x')(C\mathbin{|}y')(x'\mathbin{|}x)(y'\mathbin{|}y)A.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2004, Springer. Not translated.