Proof of Characteristic Property of the Ordered Pair

Two sets are equal exactly when they have the same elements, and $\{u\}=\{v\}$ therefore holds exactly when $u=v$, while $\{u,v\}=\{u\}$ holds exactly when $v=u$.

If $x=x'$ and $y=y'$, then $\{x\}=\{x'\}$ and $\{x,y\}=\{x',y'\}$, so the two sets $(x,y)$ and $(x',y')$ of Ordered Pair have the same elements and are equal.

Conversely, suppose $(x,y)=(x',y')$, that is

$$\bigl\{\{x\},\{x,y\}\bigr\}=\bigl\{\{x'\},\{x',y'\}\bigr\}. \tag{$*$}$$

Case $x=y$. Then $\{x,y\}=\{x\}$, so the left-hand side of $(*)$ is $\bigl\{\{x\}\bigr\}$, whose only element is $\{x\}$. Since $\{x'\}$ and $\{x',y'\}$ are elements of the right-hand side, both equal $\{x\}$. From $\{x'\}=\{x\}$ we get $x'=x$. From $\{x',y'\}=\{x\}$ we get $y'=x$, and since $x=y$ this gives $y'=y$.

Case $x\ne y$. Then $\{x,y\}\ne\{x\}$, since $y$ belongs to the first set and not to the second.

The element $\{x\}$ of the left-hand side of $(*)$ belongs to the right-hand side, so $\{x\}=\{x'\}$ or $\{x\}=\{x',y'\}$. Suppose the latter held. Then $x'$ and $y'$ both belong to $\{x\}$, so $x'=x$ and $y'=x$, whence $\{x'\}=\{x\}=\{x',y'\}$ and the right-hand side of $(*)$ is $\bigl\{\{x\}\bigr\}$. Its only element being $\{x\}$, the element $\{x,y\}$ of the left-hand side would satisfy $\{x,y\}=\{x\}$, contradicting $x\ne y$. Hence $\{x\}=\{x'\}$ and therefore $x=x'$.

The element $\{x,y\}$ of the left-hand side of $(*)$ also belongs to the right-hand side, so $\{x,y\}=\{x'\}$ or $\{x,y\}=\{x',y'\}$. The first is impossible: $\{x'\}=\{x\}$ and $\{x,y\}\ne\{x\}$. Hence $\{x,y\}=\{x',y'\}=\{x,y'\}$, using $x=x'$. Now $y$ belongs to $\{x,y\}$, hence to $\{x,y'\}$, so $y=x$ or $y=y'$; the first is excluded, so $y=y'$.

In both cases $x=x'$ and $y=y'$, which proves the equivalence. The final assertion follows: if a pair $p$ is written both as $(x,y)$ and as $(x',y')$, then $x=x'$ and $y=y'$, so the first and second components of $p$ are determined by $p$ alone.