How do I prove that a definition of an ordered pair does not satisfy the required property? It is to my knowledge that any definition of an ordered pair, (a, b), must satisfy this condition:
$\quad (a,b) = (c, d) \iff a = c \land b = d$
However, I chanced upon this definition of an ordered pair:
$\quad (a, b) = \{a, \{b\}\}$
which I know does not satisfy the condition. However, how do I prove this formally?
 A: I'm rather new to proofs, so correct me if something goes wrong somewhere and tell me if anything needs fixing.
$$\quad (a, b) = (c, d) \iff a = c \land b = d$$
$$\quad assume \space (a, b) = (c, d)$$
$$\quad (a, b) = (c, d) \implies \{a, \{b\}\} = \{c,\{d\}\}$$
$$\quad suppose \space a = \{d\} \space \land \space c = \{b\}$$
$$\quad \{\{d\},\{b\}\} = \{\{b\},\{d\}\}$$
$$\quad let \space b = 3 \space and \space d = 5, \space then \space b \neq d \space and \space a \neq c \space but \space \{\{5\}, \{3\}\} = \{\{3\},\{5\}\}$$
$$which \space does \space not \space satisfy \space the \space property.$$
A: In order to prove false a statement of the form “for all $a,b$, $P(a,b)$”, you just need to provide a counterexample.
In your case the statement is

for all $a,b$, if $(a,b)=(c,d)$, then $a=c$ and $b=d$

so you need to find $a,b,c,d$ such that $\{a,\{b\}\}=\{c,\{d\}\}$, but with $a\ne c$ or $b\ne d$.
If we want the statement to be false, we can try to make $b\ne d$, say $b=0$ and $d=1$. Now we want to find $a$ and $c$ so that
$$
\{a,\{0\}\}=\{c,\{1\}\}
$$
and taking $a=\{1\}$, $c=\{0\}$ ends the search.
