Does anything precede incrementation in the operator "hierarchy?" 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?
 A: 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
A: 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.
A: No, because the successor operation enters the stage already in the definition of natural numbers.
