Assuming the following definition of the addition of natural numbers, how do I prove that $\forall a:0+a=a$

Question

Natural numbers and the succesor function S are defined according to the Peano axioms.

Addition is defined recursively (DIFFERENTLY from the traditional Peano definition, though I am trying to prove that they are equivalent):

$\forall a:a+0=a$

$\forall a \forall b:a+S(b)=S(a)+b$

I tried and failed to prove this by induction on a.

Answer 1

Consider the set of numbers that verify every property. Let's set: $M=\{n\in \mathbb{N}:n+0=0\}$. $0$ is trivially in $M$. We suppose that there is a natural number $n$ that verifies the given property. Since the sum is defined recursively we have: $$ S(n)+0=n+S(0)=n+1=S(n) $$ The other statement is proved similarly.

Answer 2

I showed how you can derive the 'normal' $a + S(b) = S(a+b)$ from what you have in an earlier post, and using that, the inductive proof for $0 + a = a$ is trivial.