$\forall$$A$:Type, $B_{1}$,$B_{2}$:($A$$\rightarrow$Type), ${\it eq}$:EqDecider($A$), $f$:fpf($A$; $a$.$B_{1}$($a$)), $g$:fpf($A$; $a$.$B_{2}$($a$)).
\\[0ex]fpf{-}sub($A$; $a$.$B_{1}$($a$); ${\it eq}$; $f$; fpf{-}join(${\it eq}$; $f$; $g$))