Every Implementable Algorithm is Extensionally a Lookup Table

Proof. Let $f:I\to O$ with $I$ finite. The function $f$ is completely specified by the finite set of pairs

$$T_f=\left\{\right(i,f\left(i\right)):i\in I\}.$$

This set $T_f$ is a lookup table representation of $f$. Hence $f$ and $T_f$ are extensionally equivalent.   ∎

Proof. By definition, a stochastic algorithm specifies for each $i\in I$ a distribution $P(·\mid i)$ over $O$. Thus it is represented by the finite table

$$T_P=\left\{\right(i,P(·\mid i)):i\in I\}.$$

Equivalently, if the algorithm is implemented as a deterministic map $h:I\times S\to O$ with random seed $s~\mu$, then the induced conditional distribution is

$$P(o\mid i)=\sum _{s\in S}\mathbf{1}\left\{h\right(i,s)=o\}\mu \left(s\right).$$

Since $I\times S$ is finite, $h$ is representable by the finite table

$$T_h=\left\{\right((i,s),h(i,s)):i\in I,s\in S\}.$$

Thus both formulations reduce to finite tables.   ∎

Proof. For the deterministic case,

$$(gˆf)\left(i\right)=g\left(f\right(i\left)\right)\forall i\in I.$$

Thus $gˆf$ is specified by the finite table

$$T_{gˆf}=\left\{\right(i,g\left(f\right(i\left)\right)):i\in I\}.$$

For stochastic maps, let $f\left(i\right)$ induce a distribution $P(·\mid i)$ on $O$, and let $g\left(o\right)$ induce $Q(·\mid o)$ on $O\text{'}$. Then the composition induces

$$R(o\text{'}\mid i)=\sum _{o\in O}Q(o\text{'}\mid o)P(o\mid i),$$

which is again a finite table of distributions. Hence closure holds.   ∎