$\forall$$A$:Type, $B$:($A$$\rightarrow$Type), $f$:$a$:$A$ fp$\rightarrow$ $B$($a$), ${\it eq}$:EqDecider($A$), $x$:$A$, $z$:$B$($x$). $f$($x$)?$z$ $\in$ $B$($x$)