Bourbaki

Theorem 1

Tag 03P3, permanent. Cite it as [Bourbaki, Tag 03P3].

Theory of Sets, chapter I, § 5. Equalitarian theories

x=xx = x.

Let SS denote the relation x=xx = x in T0\mathscr{T}_0. By C27 (§4, no. 1), for every relation RR in T0\mathscr{T}_0, (∀x)(R⇔R)(\forall x)(R \Leftrightarrow R) is a theorem in T0\mathscr{T}_0, and therefore, by S7, τx(R)=τx(R)\tau_x(R) = \tau_x(R), that is to say (τx(R)∣x)S(\tau_x(R)|x)S, is a theorem in T0\mathscr{T}_0. Taking RR to be the relation "not SS" and considering C26 (§4, no. 1), we see that (∀x)S(\forall x)S is a theorem in T0\mathscr{T}_0. By C30 (§4, no. 3), SS is therefore a theorem in T0\mathscr{T}_0.

The relation (∀x)(x=x)(\forall x)(x = x) is also a theorem in T0\mathscr{T}_0; and if TT is a term in T0\mathscr{T}_0, then T=TT = T is a theorem in T0\mathscr{T}_0 (cf. §4, no. 3). It is possible to transform later theorems in the same way into theorems in which no letter appears or into metamathematical criteria. From now on we shall not explicitly perform these transformations, but we shall often implicitly make use of them.

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