Exercise 1, § 1

This solution was written by nemotron-3-ultra-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

Assume first that $x = y$. Let $X$ be an arbitrary set. If $x \in X$, then by the substitutivity of equality in an equalitarian theory (Chapter I, § 2, no. 4) we obtain $y \in X$. Hence $(x \in X) \Rightarrow (y \in X)$ holds for every $X$, so $(\forall X)((x \in X) \Rightarrow (y \in X))$ is true.

Conversely, assume $(\forall X)((x \in X) \Rightarrow (y \in X))$. By Axiom A2 the relation $\mathrm{Coll}_z(z = x \text{ or } z = x)$ is a theorem; therefore, by Definition 2 (03H6), the set $\{x\}$ exists and the relation $z \in \{x\}$ is equivalent to $z = x$. In particular, $x \in \{x\}$ because $x = x$ is true. Applying the hypothesis with $X = \{x\}$ yields $y \in \{x\}$, and the equivalence $z \in \{x\} \Leftrightarrow z = x$ gives $y = x$, i.e. $x = y$.

Thus the equivalence $(x = y) \Leftrightarrow (\forall X)((x \in X) \Rightarrow (y \in X))$ is a theorem.