Using an alternative definition of ordered pairs, choose two different sets $\displaystyle \triangle$ and $\displaystyle \square$, define $\displaystyle \langle a, b \rangle = \lbrace \lbrace a, \square \rbrace, \lbrace b, \triangle \rbrace \rbrace$

State and prove an analogue of Theorem 1.2

so Theorem 1.2 states (a, b) = (a' , b') if and only if a = a' and b = b'.

now the only part I'm not sure about is $\displaystyle \triangle$ and $\displaystyle \square$, so would the statement be:

$\displaystyle \langle a, b \rangle = \langle a', b' \rangle

$ if and only if a = a' b = b', $\displaystyle \triangle = \triangle '$ and $\displaystyle \square = \square '$

or simply $\displaystyle \langle a, b \rangle = \langle a', b' \rangle

$ if and only if a = a' b = b'. where the triangle and square are sort of fixed?