$\forall$$A$:Type, $B$:($A$$\rightarrow$Type), ${\it eq}$:EqDecider($A$), $f$,$g$:fpf($A$; $a$.$B$($a$)).
\\[0ex]($f$ = $g$) $\Rightarrow$ fpf{-}sub($A$; $a$.$B$($a$); ${\it eq}$; $f$; $g$)