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if and only if 
and 
Following is a proof that the ordered pairs $(a,b)$ and $(c,d)$ are equal if and only if $a=c$ and $b=d$ .
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 $a \neq b$ . Then $\{a\} \neq \{a,b\}$ . 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 $\{a\} \neq \{a,b\}$ , it must be the case that $\{a,b\}=\{a,d\}$ . Thus, $b \in \{a,d\}$ . Since $a \neq b$ , it must be the case that $b=d$ . It follows that $a=c$ and $b=d$ . 