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I here define the hierarchy of basic mathematical operators and their respective "inverse" operation (see hyperoperation).

$$ \begin{array}{c|c|c|} & \text{Operator} & \text{"Inverse"} \\ \hline \text{Incrementation} & a+1 & a-1 \\ \hline \text{Addition} & a+b & a-b \\ \hline \text{Multiplication} & ab & \frac{a}{b} \\ \hline \text{Exponentiation} & a^b & \sqrt[b]{a} \\ \hline \text{Tetration} & ^ba & \sqrt[b]{a}_s \\ \hline \text{} \vdots & \vdots & \vdots \\ \hline \end{array} $$

Now, it is clear that each "level" is simply the previous one except the process is done several times. Adding two integers is like incrementing one integer many times. Multiplication is addition of the same number many times.

My question is: does anything precede incrementation?

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  • $\begingroup$ How could anything precede the successor operation? By definition, there are no natural numbers strictly between $n$ and $n + 1$... (Also, if you're interested in higher levels, you might want to look at how the Ackermann function is constructed.) $\endgroup$ – Zhen Lin Jul 13 '12 at 16:31
  • $\begingroup$ No, because the successor operation is no longer a binary operation. I really don't see how you plan on breaking down a unary operation as "repeating", though I'm more than willing to be proven wrong. $\endgroup$ – Robert Mastragostino Jul 13 '12 at 16:46
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No, because the successor operation enters the stage already in the definition of natural numbers.

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We can notice that

$a^{S(n)}=mul_a(a^n)=a\times (a^n)$

$a \times S(n)=add_a(a\times n)=a+(a\times n)$

$a+S(n)=S(a+n)$

here the seqence breaks, since successor is not a binary operation. but we can continue finding a function $f$ that is the "$(-1)$ step of the sequence (if the Successor is the step $0$).

$S(S(n))=f(S(n))$ that becomes

$n+2=f(n+1)$ and for $n=m-1$ we get

$(m-1)+2=f(m)=m+1$

so $S(n)=f(n)$: "successor precedes" the successsor in the sequence

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To expand on MphLee's answer:

Successor can also be viewed as a binary function $H_0(a,b) = S(b)$. Then we have that $H_0(a, Sb) = H_{-1}(a,H_0(a,b)) = H_{-1}(a,Sb)$ so $H_{-1}$ agrees with $H_0$ when the second argument is greater than $0$. However, if we stop there, $H_{-1}(a,0)$ can be specified freely.

If we continue further, we can show that $H_{-k}(a, b) = S(b)$ for $b \ge k$:

If $b \ge k$, then $SSb = H_{-k}(a, Sb) = H_{-k-1}(a, H_{-k}(a,b)) = H_{-k-1}(a, Sb)$.

However, then $2 = H_{-1}(a, 1) = H_{-2}(a, H_{-1}(a,0))$, so if $f(a) := H_{-1}(a,0) \ge 2$, we get that $2 = f(a) + 1$, so $f(a) = 1$, a contradiction. Therefore, $f(a) \le 1$.

If we continue further it gets more complicated. I think if you extend it infinitely, they all have to be successor, even though they don't in a finite extension.

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  • $\begingroup$ Define $H_n(b,x)=H_n(x)$ (for a fixed $b$) as recursive functions $H_n:\Bbb N\rightarrow D\subset\Bbb N$. We want them to satisfie $H_n(H_{n+1}(x))=H_{n+1}(S(x))$ for all the ranks $n\in \Bbb Z$. You are right that if we stay on the naturals we can't actually conclude $H_{k\le0}=S$ we can just prove that $S\cap H_{k\le 0}\neq\varnothing$ (negative-rank Hos. agrees with the successor for some values). This is true because if we stay on the naturals we have to define 0-th rank Ho. to be $b[0]x=b+1+(x\dot -b)$ where $\dot - $is the cutoff subtraction. continue $\endgroup$ – MphLee Mar 28 '15 at 18:51
  • $\begingroup$ I haven't proved it yet but it seems that the intersections $S\cap H_{k}$ really get "smaller" as $k$ grows (as you say). Anyways the requirement of invertibility for $H_1={\rm addition}$ is enough to prove $H_{k\le 0}=S$ $$H_0\circ H_{1}=H_{1}\circ S\implies H_0=H_1\circ S\circ H_1^{-1}$$ Since addition commutes with successor we have the equality then all the $H_{k\le 0}$ coincides with the successor on $\Bbb Z$. $\endgroup$ – MphLee Mar 28 '15 at 18:58

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