$\forall$$A$:Type, $B$:($A$$\rightarrow$Type), ${\it eq}_{1}$, ${\it eq}_{2}$:EqDecider($A$), $f$:$a$:$A$ fp$\rightarrow$ $B$($a$), $x$:$A$.
\\[0ex]$x$ $\in$ dom($f$) $=$ $x$ $\in$ dom($f$) $\in$ $\mathbb{B}$