$\forall$$A$:Type, $B_{1}$, $B_{2}$:($A$$\rightarrow$Type), ${\it eq}$:EqDecider($A$), $f$:$a$:$A$ fp$\rightarrow$ $B_{1}$($a$), $g$:$a$:$A$ fp$\rightarrow$ $B_{2}$($a$). $f$ $\subseteq$ $f$ $\oplus$ $g$