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Let $R$ be a equivalence relation over $\mathbb{N}^k$. We say that $R$ is primitive recursive (p.r) if its indicator function is primitive recursive.

Let $R$ be p.r and $f_R$ its indicator function.

Well I already proved that the equivalence classes defined by $R$ are also primitive recursive: if $[\vec{n}]$ in an equivalence class we define $f_{[\vec{n}]}(\vec{m})=f_R(\vec{n},\vec{m})$ and it works as its indicator function.

But I don't know how to prove the if the converse is true or not.

Thanks in advance

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The answer is no - the equivalence classes of $R$ can be extremely simple, while $R$ itself is extremely complicated.

For instance, let $X$ be an arbitrary set of natural numbers, and define $R_X$ as follows:

  • $R_X(k,k)$ always holds (so $R_X$ is reflexive).

  • $R_X(2k, 2k+1)$ and $R_X(2k+1, 2k)$ hold iff $k\in X$.

  • Those are the only times $R_X$ holds.

Then:

  • Each equivalence class of $R$ is p.r. - in particular, each equivalence class of $R$ consists of either one or two elements. Can't get much simpler than that!

  • On the other hand, $R_X$ is as complicated as $X$. Pick a really complicated $X$ - say, $X$ not computable - and this yields an $R_X$ which is also really complicated (in particular, not p.r.).