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In other words, I'm looking for a binary relation $P(x,y)$, being arithmetical, i.e being expressed in the first order language of Peano arithmetic (hence non-recursively, i.e. using addition and multiplication only - without using the very symbol $P$), which satisfies:

  1. For every $x$, there exists a unique $y$, satisfying $P(x,y)$.
  2. $P(0,1)$.
  3. Every $x,y$, satisfying $P(x,y)$, satisfy $P(x+1,2y)$.

If such an arithmetical function (as defined above) does not exist, I will be glad to read the proof of this fact.

(No Gödel's coding please).

Eli
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    Please clarify when a binary relation is "arimethical". For example, $2^n$ can be defined inductively using multiplication (by $2$). So I presume you do not allow induction – Ewan Delanoy Jun 06 '19 at 12:10
  • @Ewan Delanoy: Are you sure you can define the relation $P(x,y)$ (as defined above), by means of induction, using addition and multiplication only, i.e. without using the very symbol $P$ itself? – Eli Jun 06 '19 at 12:22
  • @Ewan Delanoy: My question has already answered your question: A binary relation $P(x,y) is arithmetical, if and only if its definition uses addition and multiplication only, without using the very symbol P itself. – Eli Jun 06 '19 at 12:32
  • Perhaps you just mean that $P$ should be a polynomial in $x,y$ ? (if you know what that is) – Ewan Delanoy Jun 06 '19 at 12:36
  • @Ewan Delanoy: I'm looking for a definition like "Every $x,y$ satisfy $2^{x}=y$ if and only if $P(x,y)$", while the term $P(x,y)$ is expressed in the first order language of Peano arithmeic, i.e, without the symbol $P$. – Eli Jun 06 '19 at 12:40

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