$\forall$$T$:Type, $f$:($T$$\rightarrow$$T$), $L$:($T$ List), $x$, $y$, $z$:$T$.
\\[0ex]$z$=$f$$\ast$($x$) via $L$ $\Rightarrow$ ($y$ $\in$ $L$) $\Rightarrow$ ($f$($y$) = $y$) $\Rightarrow$ ($y$ = $z$)