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I know, loosely speaking, if we can define a function $f$ in term of \begin{align} &f(0,\vec{x})=g(\vec{x})\\ &f(n+1,\vec{x})=h(f(n),n,\vec{x}) \end{align} where functions $g,h$ are primitive recursive. Then $f$ is primitive recursive.

However, what it means to show a function is primitive recursive by induction?

I had read above explaination on page 93 on book $\textit{Computability}$ by Epstein and Carnielli, but still I'm not sure if I got the idea. Could someone provide some examples about how a inductive definition shows a function is primitive recursive?

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  • $\begingroup$ Your first question --how to show that a function is $PR$ by induction-- is not exactly what the quoted text is talking about. Let me explain the role of induction in the text and then maybe you can confirm if you still have a question. In the text they are defining a set, $PR$. They define $PR$ to be the union of sets $PR_0,PR_1,...,PR_n,...$ that are defined for all natural numbers $n$. The set $PR_0$ is defined as all basic functions. For each natural number $n$ the set $PR_{n+1}$ is defined as the functions that can be obtained from $PR_n$ by applying the basic operations one or ... $\endgroup$ – plop Sep 29 '20 at 11:11
  • $\begingroup$ ... zero times. The role of induction is to be able, from those two statements to make the claim that -- for all natural numbers $n$ we have defined a set $PR_n$ --. After that the definition of $PR$ ends by saying that $PR$ is the union of all $PR_n$ for all natural numbers $n$. $\endgroup$ – plop Sep 29 '20 at 11:12
  • $\begingroup$ Does the book contain no examples? $\endgroup$ – Yuval Filmus Sep 29 '20 at 11:15
  • $\begingroup$ The thing is that the definition is being based on two propositions: [(1) The definition of $PR_0$ and (2) the definition of $PR_{n+1}$ in terms of $PR_n$, for each natural number $n$.] But to end the definition one needs the proposition that [(3) For all natural numbers $n$, the set $PR_n$ is defined.] A logical deduction is needed to get from ((1) and (2)) to (3). Induction is that logical deduction. $\endgroup$ – plop Sep 29 '20 at 11:17
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Wikipedia gives such an example. You haven't stated the definition of basic functions or basic operations, so I will assume the Wikipedia definition.

The successor function $S(x) = x+1$ is a basic function, and so is assigned 0. Projection functions such as $P_1^1(x) = x$, $P_2^3(x,y,z) = y$ and $P_3^3(x,y,z) = z$ are also assigned 0. The constant function $z(x) = 0$ is also assigned 0.

The function $g(x,y,z) = S(P_2^3(x,y,z)) = y + 1$ is a composition of two functions assigned 0, and so is assigned 1.

Using these functions, we can define a function $h(x,y)$ by primitive recursion: $h(0,y) = P_1^1(y) = y$, and $h(S(x),y) = g(x,h(x,y),y) = h(x,y) + 1$. This function is assigned 2, and you can check that $g(x,y) = x + y$.

The function $r(x,y,z) = h(P_2^3(x,y,z),P_3^3(x,y,z)) = g(y,z) = y + z$ is assigned 3.

We can define a function $k(x,y)$ by primitive recursion: $k(0,y) = z(y) = 0$, and $k(S(x),y) = r(x,k(x,y),y) = k(x,y)+y$. This function is assigned 4, and you can check that $k(x,y) = x\cdot y$.

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