Combination of Recursive Functions

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Theorem

Let $f: \N^k \to \N$ and $g: \N^k \to \N$ be recursive functions (not necessarily total), where $k \ge 1$.

Let $\RR$ be a $k$-ary relation such that:

if $\map \RR {n_1, n_2, \ldots, n_k}$ holds, then $\map f {n_1, n_2, \ldots, n_k}$ is defined

if $\map \RR {n_1, n_2, \ldots, n_k}$ does not hold, then $\map g {n_1, n_2, \ldots, n_k}$ is defined.

Let $h: \N^k \to \N$ be the function defined as:

$\map h {n_1, n_2, \ldots, n_k} = \begin {cases} \map f {n_1, n_2, \ldots, n_k} & : \text{if } \map \RR {n_1, n_2, \ldots, n_k} \text { holds} \\ \map g {n_1, n_2, \ldots, n_k} & : \text{otherwise} \end {cases}$

so that $h$ is total.

Then $h$ is recursive.

This article contains statements that are justified by handwavery. In particular: It seems to me that we need to assume something about $\RR$ to make it computable which of the two cases holds at any given tuple You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Handwaving}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Proof

Let $P_f, P_g, P_\RR$ be the URM programs computing, respectively, the functions $f$ and $g$ and the characteristic function $\chi_\RR$.

From Recursive Function is URM Computable, these programs are guaranteed to exist.

An informal algorithm for computing $h$ is as follows.

1. Input $\tuple {n_1, n_2, \ldots, n_k}$.
2. Use $P_\RR$ to determine whether $\map \RR {n_1, n_2, \ldots, n_k}$ holds.
3. If it holds, use $P_f$ to compute $\map f {n_1, n_2, \ldots, n_k}$.
4. If it does not hold, use $P_g$ to compute $\map g {n_1, n_2, \ldots, n_k}$.

The result follows from URM Computable Function is Recursive.

$\blacksquare$

Fallacious Proof

We might attempt to show that $h$ is recursive by using the equation:

\(\ds \map h {n_1, n_2, \ldots, n_k}\)

\(=\)

\(\ds \map f {n_1, n_2, \ldots, n_k} \map {\chi_\RR} {n_1, n_2, \ldots, n_k}\)

\(\ds \)

\(+\)

\(\ds \map g {n_1, n_2, \ldots, n_k} \map {\overline \sgn} {\map {\chi_\RR} {n_1, n_2, \ldots, n_k} }\)

where $\overline \sgn$ is the signum bar function, known to be primitive recursive.

Then the right hand side is obtained by substitution from known recursive functions, so is recursive, and hence $h$ is recursive.

However, this fails as follows.

Although $\map h {n_1, n_2, \ldots, n_k}$ is always defined,

\(\text {(1)}: \quad\)

\(\ds \)

\(\)

\(\ds \map f {n_1, n_2, \ldots, n_k} \map {\chi_\RR} {n_1, n_2, \ldots, n_k}\)

\(\ds \)

\(+\)

\(\ds \map g {n_1, n_2, \ldots, n_k} \map {\overline \sgn} {\map {\chi_\RR} {n_1, n_2, \ldots, n_k} }\)

might not be.

$f$ and $g$ might not be total, in which case there may be some $\tuple {n_1, n_2, \ldots, n_k}$ for which at least one of $\map f {n_1, n_2, \ldots, n_k}$ and $\map g {n_1, n_2, \ldots, n_k}$ is not defined.

This means $(1)$ is not defined for those values of $\tuple {n_1, n_2, \ldots, n_k}$.

Note that $(1)$ does define a recursive function, but its domain consists only of those $\tuple {n_1, n_2, \ldots, n_k}$ for which $\map f {n_1, n_2, \ldots, n_k}$ and $\map g {n_1, n_2, \ldots, n_k}$ are both defined.

Thus we establish the result by means of the indirect approach requiring the use of the results concerning the equivalence of recursive functions and URM computable functions.