Exercise 1, § 2

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.

Let $R\{x,y\}$ be a relation with $x$ and $y$ distinct letters, and let $z$ be a letter distinct from $x$ and $y$ which does not appear in $R\{x,y\}$.

By the definition of ordered pairs in §2, no. 1, the relation $z = (x,y)$ is equivalent to “$z$ is an ordered pair and $x = \operatorname{pr}_1 z$ and $y = \operatorname{pr}_2 z$”. Consider the relation $S\{z\}$ defined by

Now $(\exists z)S\{z\}$ is exactly $(\exists z)(\exists x)(\exists y)\bigl(z = (x,y) \text{ and } R\{x,y\}\bigr)$. Since the letter $z$ does not occur in $R\{x,y\}$, we may interchange the quantifiers to obtain $(\exists x)(\exists y)(\exists z)\bigl(z = (x,y) \text{ and } R\{x,y\}\bigr)$. For fixed $x$ and $y$, the relation $(\exists z)\bigl(z = (x,y) \text{ and } R\{x,y\}\bigr)$ is equivalent to $R\{x,y\}$ by criterion C47 (Chapter I, §5, no. 3), because $z$ is not free in $R\{x,y\}$ and there exists a $z$ (namely $(x,y)$) such that $z = (x,y)$. Hence $(\exists x)(\exists y)(\exists z)\bigl(z = (x,y) \text{ and } R\{x,y\}\bigr)$ is equivalent to $(\exists x)(\exists y)R\{x,y\}$. Therefore

For the second equivalence, apply the first equivalence to the relation $\neg R\{x,y\}$. We obtain