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