$\forall$$A$:Type, $B$,$C$:($A$$\rightarrow$Type), ${\it eq}$:EqDecider($A$), $f$:fpf($A$; $a$.$B$($a$)), $g$:fpf($A$; $a$.$C$($a$)), $x$:$A$.
\\[0ex]($\uparrow$fpf{-}dom(${\it eq}$; $x$; $f$)) $\Rightarrow$ (fpf{-}ap(fpf{-}join(${\it eq}$; $f$; $g$); ${\it eq}$; $x$) = fpf{-}ap($f$; ${\it eq}$; $x$) $\in$ $B$($x$))