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