Bourbaki

Exercise 6, § 5

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

Theory of Sets, I, 59

Show that if the scheme (∃x)R⇒R(\exists x)R \Rightarrow R provides implicit axioms in T\mathscr{T}, then x=yx = y is a theorem in T\mathscr{T} (cf. Exercise 1).

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