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