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