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