Bourbaki theory of set question

Question

In Chapter 2 section 1 subsection 5 of Bourbaki's "Theory of sets" there is this remark

$$\exists z(z\in \left\{x,y\right\} \wedge R\left\{z\right\})\Leftrightarrow R\left\{x\right\}\vee R\left\{y\right\}$$

Whereas this equivalence is perfectly clear, I could not deduced it from previous theorems in the book. Can someone provide a clear proof with theorems strictly from the Bourbaki book?

Note: I believe the original Boubaki text is "incorrect"; the equivalence it actually refers to is

$$\exists z(z\in \left\{x,y\right\} \wedge R\left\{z\right\})\Leftrightarrow (R\left\{x\right\}\vee R\left\{y\right\}) \wedge z\in \left\{x,y\right\}$$

or

$$z\in \left\{x,y\right\} \Rightarrow \exists zR\left\{z\right\}\Leftrightarrow R\left\{x\right\}\vee R\left\{y\right\} $$

Answer 1

The logical theorems used are [see page 69] :

C32. $\exists x \ (R \lor S) \Leftrightarrow (\exists x R \lor \exists x S), \ \ \forall x \ (R \land S) \Leftrightarrow (\forall x R \land \forall x S)$

C33. $\forall x \ (R \lor S)\Leftrightarrow (R \lor \forall x \ S), \ \ \forall x \ (R \lor S) \Leftrightarrow (R \lor \forall x S)$, if $x$ is not free in $R$,

and the properties of equality [page 44] :

S6. $(t=u) \to (R(t) \Leftrightarrow R(u))$,

and [page 46] : $x=x$.

From S6 we have, obviously : $((t=u) \land R(t)) \Rightarrow R(u)$.

Thus, from :

$\exists z \ [(z \in \{ x, y \}) \land R(z)]$

replacing $z \in \{ x, y \}$ with $z=x \lor z=y$, we get, using C32 :

$\exists z \ [(z=x \lor z=y) \land R(z)] \Leftrightarrow \exists z \ (z=x \land R(z)) \lor \exists z \ (z=y \land R(z)).$

By equality, this amounts to :

$\exists z \ R(x) \lor \exists z \ R(y).$

But $z$ is not free in $R(x)$, and thus it amounts to :

$R(x) \lor R(y).$

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Similar for : $\forall z \ [(z \in \{ x, y \}) \to R(z)]$.

Replacing $z \in \{ x, y \}$ with $z=x \lor z=y$ and distributing, we get, using C32 :

$\forall z \ (z=x \to Rz) \land \forall z \ (z=y \to Rz)$.

Instantiating both universal quantifiers (and some tautological transformations) we have :

$(x=x \to Rx) \land (y=y \to Ry)$.

By equality : $x=x$ and $y=y$ we finally get :

$Rx \land Ry$.