$(a,b)=(c,d)$ if and only if $a=c$ and $b=d$

Following is a proof that the ordered pairs $(a,b)$ and $(c,d)$ are equal if and only if $a=c$ and $b=d$.

Proof.

If $a=c$ and $b=d$, then $(a,b)=\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}=(c,d)$.

Assume that $(a,b)=(c,d)$ and $a=b$. Then $\{\{c\},\{c,d\}\}=(c,d)=(a,b)=\{\{a\},\{a,b\}\}=\{\{a\},\{a,a\}\}=\{\{a\},\{a\}\}=\{\{a\}\}$.Thus, $\{c,d\}\in \{\{a\}\}$. Therefore, $\{c,d\}=\{a\}$. Hence, $a=c$ and $a=d$. Since it was also assumed that $a=b$, it follows that $a=c$ and $b=d$.

Finally, assume that $(a,b)=(c,d)$ and $a\ne b$. Then $\{a\}\ne \{a,b\}$. Note that $\{\{a\},\{a,b\}\}=(a,b)=(c,d)=\{\{c\},\{c,d\}\}$. Thus, $\{c\}\in \{\{a\},\{a,b\}\}$. It cannot be the case that $\{c\}=\{a,b\}$ (lest $a=c=b$). Thus, $\{c\}=\{a\}$. Therefore, $a=c$. Hence, $\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}=\{\{a\},\{a,d\}\}$. Note that $\{a,b\}\in \{\{a\},\{a,d\}\}$. Since $\{a\}\ne \{a,b\}$, it must be the case that $\{a,b\}=\{a,d\}$. Thus, $b\in \{a,d\}$. Since $a\ne b$, it must be the case that $b=d$. It follows that $a=c$ and $b=d$.∎