Can we use Peano's axioms to prove that integer = prime + integer? Every integer greater than 2 can be expressed as sum of some prime number greater than 2 and some nonegative integer....$n=p+m$. Since 3=3+0; 4=3+1; 5=3+2 or 5=5+0...etc it is obvious that statement is true.My question is: Can we use Peano's axioms to prove this statement (especially sixth axiom which states "For every natural number $n$, $S(n)$ is a natural number.")?
 A: Note: My usual definition of $\mathbb{N}$ includes $0$, as for example in Halmos's Naive Set Theory. This does not matter here. The proposed result asks for an expression of natural numbers greater than $2$ (which means it doesn't matter if you consider $0$ a natural number or not) as a sum of a prime greater than 2 and a nonnegative integer; the proof below shows that every nonnegative integer (in particular, every "natural number if you don't include $0$") is the sum of a prime greater than two and a "natural-number-including-zero" (that is, a "nonnegative integer"). 
Let $S=\{n\in\mathbb{N}\mid n\leq 2\text{ or there exists a prime }p\gt 2\text{ and a natural }m\text{ such that }n=p+m\}$.
Note that $0$, $1$, and $2$ lies in $S$.
Assume that $k\in S$; we want to prove that $s(k)$, the successor of $k$, lies in $S$. 


*

*If $k\lt 2$, then $s(k)\leq 2$, so $s(k)\in S$. 

*If $k=2$, then $s(k) = k+1 = 2+1 = 3+0$ is the sum of a prime, $3$, and a natural number, $0$, so $s(k)\in S$.

*If $k=p+m$ for some prime $p$ and some natural number $m$, then
$$s(k) = s(p+m) = p+s(m),$$
and by Peano's postulates, since $m\in\mathbb{N}$ then $s(m)\in\mathbb{N}$. So $s(k)$ is the sum of a prime and a natural number, hence $s(k)\in S$.


By Peano's Fifth Postulate (Induction), $S=\mathbb{N}$. This proves that every natural number greater than $2$ is the sum of a prime and natural number. $\Box$

If you don't consider the natural numbers to include zero, it is very simple to translate the proof above: the definition of $S$ just has to replace the word "natural" with "nonnegative integer":
$$S = \{n\in\mathbb{N}\mid n\leq 2\text{ or there exists a prime }p\gt 2\text{ and a nonnegative integer }m\text{ such that }n=p+m\}.$$
Then the first line of the proof can omit the observation that $0\in S$, and where it says "natural number" afterwards, just replace with "nonnegative integer", and the observation that if $m$ is a nonnegative integer, then it is either a natural-greater-than-1, so $s(m)$ is a natural number (under either definition), or else $m=0$ in which case "$s(m)$" denotes $1$, which is a natural number (under either definition). 
A: HINT $\ $ For nonempty $\rm\ S\subset \mathbb N\ $ show $\rm\ S\:+\:\mathbb N\ =\: \min(S)\: +\: \mathbb N\:.\ $ Your case is $\rm\:S =\:$ odd primes.
A: Yes we can use the Peano's axiom to prove that integer = prime + integer. Think of $0 + a = a$. 
