1
$\begingroup$

I am trying to solve this problem of my textbook related to recursive primitive functions and PRC. The problem is the following:

Let $C_i$ be the class of initial functions, i.e., the class that contains:$$n(x)=0, s(x)=x+1, {u_i}^n(x_1,...,x_n)=x_i, \text{for each natural n and some fixed i between 1 and n} $$

and let $C_c$ be the minimal class that contains $C_i$ and is closed by composition, that means, if $f,g_1,...,g_m$ are in $C_c$ then $$h(x_1,...,x_n)=f(g_1(x_1,...,x_n),...,g_m(x_1,...,x_n))$$ is also in the class.

i) Show that for all $f:\mathbb N \to \mathbb N$, $f$ is in $C_c$ if and only if there is $k >=0$ such that $f(x_1,...,x_n)=k$ of $f(x_1,...,x_n)=x_i+k$ for some fixed $i$.

ii) Show that there is a primitive recursive function that is not contained in $C_c$.

I could only show that for any natural $k$, if a function $f$ is of the form $f(x_1,...,x_n)=k$ or $f(x_1,...,x_n)=x_i+k$, then $f$ is in $C_c$. I've proved this by induction, and using the fact that the initial functions are in $C_c$ and that this class is closed by composition. This proof was easy but I couldn't show the other way round, which is, if a function $f$ is in the class, then $f$ has to be of the form mentioned above.

I also can't think of an example of a primitive recursive function which is not in this class.

I would really appreciate help in the part of point i) where I got stuck and in finding some example for ii). Thanks in advance.

1 Answers 1

1

When you have a class of objects (like terms, functions, formulas etc) that is generated from some initial set of objects by using some operations (or formation rules), the induction on the complexity (or on the construction) of such an object is often used.

$C_c$ being the smallest class containing the initial functions and closed under composition means that any $f \in C_c$ is obtained from the initial functions using finite number of compositions.

To prove that all $f \in C_c$ have some property reduces to proving that

  1. All the initial functions have this property.

  2. Any function obtained via composition from the functions having this property also has this property.

The proof then is by induction on the complexity of $f \in C_c$. It reduces to the induction on natural numbers by letting $r(f)$ (for any $f \in C_c$) to be a number of the composition operations needed to construct $f$ from the initial functions.

To prove i) observe that all the initial functions (the functions with $r(f) = 0$) have this form. Now consider $$h(x_1, \dots, x_n) = f(g_1(x_1, \dots, x_n), \dots, g_m(x_1, \dots, x_n)),$$ and assume the property holds for all $g \in C_c$ with $r(g) < r(h)$. Note that $$r(h) > \max(r(f), r(g_1), \dots, r(g_m))$$ (that is, $f, g_1, \dots, g_m$ are of the desired form). We consider several possibilities:

  1. If $f(x_1, \dots, x_m) = k$ then so is $h$.

  2. If $f(x_1, \dots, x_m) = x_i + k$ then $h(x_1, \dots, x_n) = g_i(x_1, \dots, x_n) + k$. Now we have same possibilities for $g_i$ and we conclude that $h$ has the desired form in both cases.

For ii) note that all the functions of the form $k$ or $x_i + k$ essentially depend at most on one variable. Can you think of some (basic arithmetic) function that depeneds essentially on two variables?