Prove/disprove: $f(x,y)=\{\{x\},\{y,\emptyset\}\}$ is an ordered pair map. (ZFC)
Progress:
We need to prove $f(x,y)=f(u,v)\iff x=u\quad \operatorname{and}\quad y=v$
proof.
$\impliedby$ is done by plugging.
$\implies$ we assume $f(x,y)=f(u,v)\iff \{\{x\},\{y,\emptyset\}\}=\{\{u\},\{v,\emptyset\}\}$
Case I: $(x=y)$
We have $\{\{x\},\{x,\emptyset\}\}=\{\{u\},\{v,\emptyset\}\}$.$\quad$
If $\{x\}=\{u\}$ $$x=u\implies\{\{x\},\{x,\emptyset\}\}=\{\{x\},\{v,\emptyset\}\}\implies x=y=u=v$$If $\{x\}=\{v,\emptyset\}$ $$x=v=\emptyset\implies\{\{x\}\}=\{\{u\},\{x\}\}\implies x=y=u=v$$Case II: $(x\neq y)$
If $\{x\}=\{v,\emptyset\}$ $$x=v=\emptyset\implies\{\{x\},\{y,x\}\}=\{\{u\},\{x\}\}\implies\{u\}=\{y,x\}\implies u=x=y \Rightarrow\!\Leftarrow$$Otherwise, $\{x\}=\{u\}$ $$x=u\implies \{\{x\},\{y,\emptyset\}\}=\{\{x\},\{v,\emptyset\}\}\implies y=v$$ comment:
I'm not sure about this proof. Furthermore, am I over-checking something in Case I? (and do I even need Case I?)