$\forall$$A$:Type, $B$:($A$$\rightarrow$Type), $f$:fpf($A$; $a$.$B$($a$)), ${\it eq}$:EqDecider($A$), $x$:$A$, $z$:$B$($x$).
\\[0ex]fpf{-}cap($f$; ${\it eq}$; $x$; $z$) $\in$ $B$($x$)