Peano Addition is recursive?

Question

Looking at this Wikipedia article on Peano's Axioms, specifically on addition, I see $(2)$ saying that

$a + S(b) = S(a + b)$

It is called recursive, but I'm not clear why, in and of itself, this is necessarily recursive -- other than by immediately jumping into the example where adding successive numbers to $a$ are built up from $(2)$ starting from $(1)$ where $a + 0 = a$.

But if you don't start "at the beginning," this would imply $a + b = a + S(k)$ where $b = S(k)$ -- which might imply $k = b - 1$, so

$a + b = a + b - 1 = S(a + b - 1) = a + b - 1 + 1$

assuming $S(a + b - 1) = a + b - 1 + 1$ ... which seems circular and pointless. It seems like I've jumped to the conclusion that $1$ (either adding or subtracting) is what addition within the context of succession is based on. Can, must I do this? Seen in a programming language, you decrement $b$, then add the decremented $b$ to $a$, then increment the sum. What have I gained for this exercise? Yes, building up from $0$ "inductively" is interesting, however, starting with any old $a + b$ seems to stretch things badly. Taken with this quote (above in the article):

Axioms 1, 6, 7, 8 define a unary representation of the intuitive notion of natural numbers: the number 1 can be defined as S(0), 2 as S(S(0)), etc. However, considering the notion of natural numbers as can be derived from the axioms, axioms 1, 6, 7, 8 do not imply that the successor function generates all the natural numbers different than 0. Put differently, they do not guarantee that every natural number other than zero must succeed some other natural number.

I'm duly confused. This is something that has always disturbed me about Peano, i.e., that we can't really assume adding $1$ to something is synonymous with its successor, and, likewise subtracting $1$ gets you from $S(a)$ "back to" $a$. "Next" on a integer line sequence is $a + 1$ doesn't really seem supported by the actual definition of successor.

Answer 1

The presentation of Peano's Axioms at Wikipedia is unusual. More commonly used these days is that presented at Wolfram Math World. In the notation of set theory, they are:

1. $0\in N$
2. $S: N\to N$
3. $S$ is injective
4. For all $x\in N$, we have $S(x)\ne 0$
5. For all $P\subset N$, if $0\in P$ and for all $x\in P$ we also have $S(x)\in P$, then $P=N$

Notice that there is no mention of addition. Using only these axioms, the rules of logic and set theory, we can actually prove the existence of a binary function $+$ on $N$ such that:

1. For all $x\in N$, we have $x+0=x$
2. For all $x,y\in N$, we have $x+S(y)=S(x+y)$

This function perfectly models the addition operation that we have all known since childhood. From these two rules, we can determine the sum of any two natural numbers.

Example

To prove $2+3=5$, we start by applying (1) for $x=2$ to obtain:

$2+0=2$.

Then we apply (2) three times to obtain:

$2+1=2+S(0)=S(2+0)= S(2)=3$

$2+2= 2+S(1) = S(2+1)=S(3)=4$

$2+3= 2+S(2) = S(2+2)=S(4)=5$

Answer 2

Very short answer to why

we can't really assume adding 1 to something is synonymous with its successor

That's because in Peano there is no such thing as adding 1, or addition. Those operations are defined recursively (as your question states) using the postulated properties of the successor operation.

Then that statement you wish were an assumption becomes a theorem.

It's the successor that's in the axioms, so you don't define it in terms of addition or next on the number line.