$\forall$$A$:Type, ${\it eq}$:EqDecider($A$), $B$:($A$$\rightarrow$Type), $f$, $g$:$a$:$A$ fp$\rightarrow$ $B$($a$).
\\[0ex]l\_disjoint($A$;1of($f$);1of($g$)) $\Rightarrow$ $f$ $\parallel$ $g$