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We want to show that the following function $F$ is primitive recursive (using the primitive recursion scheme) :

For all $x,p,a\in \mathbb N$,

$F(p,a,0)=g(\eta^{p}(a))$
$F(p,a,x+1)=h(\eta^{(p-(x+1))}(a),x,F(p,a,x))$

where $\eta, g : \mathbb{N}\to \mathbb{N}$ and $h: \mathbb{N}^3 \to \mathbb{N}$ are primitive recursive functions.

We want to apply the primitive recursion scheme which is defined as follows : if $g',h'$ are primitive recursive functions then the function $f$ defined by : $f(\bar y,0)=g'(\bar y)$ and $f(\bar y,x+1)=h'(\bar y,x,f(\bar y,x))$ is also primitive recursive.

Then I have to find the functions $f,g',h'$.

To build $g'$ let us consider that $\eta^{p}(a)=k(p,a)$ hence $g'(a,p)=g \circ k(p,a)$ which is a primitive recursive function by hypothesis.

To build $h'$ it seems that the difficulty is to get rid of the term in $\eta^{(p-(x+1))}(a)$ which depends of $x$ in particular. In fact we want $h'(p,a,x,F(p,a,x)) = h(-,-,-)$ but I am stuck. Any hints for this step would be very helpful

Thanks in advance !

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1 Answer 1

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Note that you can show that $k(p,a) = \eta^{p}(a)$ is primitive recursive. Also addition and subtraction are primitive recursive, thus definiing $h'(p,a,x,...) := h(k(p-(x+1),a),...) = h(\eta^{p-(x+1)}(a),...)$ is not a problem.

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  • $\begingroup$ Thank you for the precisions. Indeed all the operations here are primitive recursive. Maybe it could be better to write $k(p-(x+1),a)=k'(p,a,x)$ because this function $\eta^{p-(x+1)}(a)$ depends on $3$ parameters no ? $\endgroup$
    – Maman
    Sep 26, 2022 at 13:20
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    $\begingroup$ Yes if you want to be rigorous. But note that composition of prim. rec. function is also prim. rec. So that $k(sub(p,add(x,1)),a)$ is prim. rec. directly follows from the definition that (finite) composition of primitive recursive functions is also primitive recursive. $\endgroup$
    – plshelp
    Sep 27, 2022 at 9:35

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