$\forall$$A$:Type, $B$:($A$$\rightarrow$Type), $f$:$a$:$A$ fp$\rightarrow$ $B$($a$), ${\it eq}$:EqDecider($A$).  $\oplus$ $f$ $\sim$ $\langle$1of($f$)$,\,$$\lambda$$a$.2of($f$)($a$)$\rangle$