Full disclaimer: I have edited the question to make it simpler, and therefore some of the comments may no longer make much sense.
Assume we have defined the set $\mathbb{N}$ using the first nine Peano axioms. For brevity, $1$ is defined as $S(0),\ 2\ $ is defined as $S(1),\ $ etc.
Define the function, $+': (\mathbb{N}\cup\{0\}) \times (\mathbb{N}\cup\{0\})\to (\mathbb{N}\cup\{0\}) $ recursively by:
- $0\ +'\ a=a\qquad (1)$
- $S(a)\ +'\ S(b) = S(a\ +'(\ b\ +'\ 1)\ )\qquad (2)$
So for example,
$1\ +'\ 1\ \overset{def}{=}\ S(0)\ +'\ S(0) \overset{(2)}{=}\ S( 0\ +'\ (0\ +'\ 1)\ ) \overset{(1)}{=}\ S(0\ +'\ 1) \overset{(1)}{=}\ S(1) \overset{def}{=} 2. $
And I believe we can prove all other additions match our expectations using similar reasoning.
Edit: I don't see how we show that $\ 1\ +'\ 0 = 1.$
I think this is an alternative way of defining addition.
Is this correct? Does my definition of addition give us the expected results and therefore also match wikipedia's definition?
Given the two axioms in the question, we can deduce that $a\ +'\ b = a+b\ $ for all $a$ and all $b\geq 1.$
However, I don't see a reason why $1\ +'\ 0$ cannot equal any natural number (or zero).
Same for $a\ +'\ 0$ for any $a:\ $ it can be equal to any natural number and this doesn't lead to contradictions, e.g. by the pigeonhole principle.
Obviously there are many ways a third axiom would give us $a\ +'\ 0 = a\quad \forall a,\ $ for example associativity:
$ a\ +'\ (b\ +'\ c) = (a\ +'\ b)\ +'\ c\quad \forall\ a,b,c $
This would work because, for example:
$S(1)\ +'\ S(0) = S(1\ +'\ (0\ +'\ 1)) = S(2) = S((1\ +'\ 0)\ +'\ 1)\implies\ 1\ +'\ 0 = 1,\ $ because $+': (\mathbb{N}\cup\{0\}) \times (\mathbb{N}\cup\{0\})\to (\mathbb{N}\cup\{0\}) $ is a function and the only value of $k$ for which $k\ +'\ 1 = 2\ $ is $\ k=1.$
Commutativity is an alternative third axiom that would make the usual addition work for all $a,b$ because along with axiom $(1)$ we would have: $a\ +'\ 0=0\quad \forall a.$