In this case, $f(\lambda) = ()$ and $g(f(\lambda)) = g(())$. Since $\lambda = ()$, we have $g(()) = 1$.
Now, let's consider the case when $\lambda$ is not empty, i.e., $\lambda = \operatorname{cons}(a, \lambda')$ for some element $a$ and list $\lambda'$. In this
Show more…