Proofs with axioms for natural numbers

Question

I just started a course in discrete mathematics and i am totally stuck on something basic.

Construct proofs of the following rules using only axioms for natural numbers:

$1 * n = n$ and $(a+b)*c = ac + bc$

I don't need the solutions. Is there any specific methodology to solve that kind of problems?

EDIT: Axioms list:

1) a + b

2) a x b

3) a + b = b + a

4) (a + b) + c = a + (b + c)

5) a x b = b x a

6) (a x b) x c = a x (b x c)

7) if m x z = n x z then m = n

8) There is a special element of N, denoted by 1, which has the property that n x 1 for all n as natural numbers.

9) a * (b + c) = (a * b) + (a * c)

Answer 1

You certainly can't prove the first because you are not given that there is an element called $1$ or one that functions as one unless that is part of the definition of multiplication. Note that the even naturals satisfy all your axioms but there is no $1$. Similarly, you have nothing to link addition and multiplication, so you can't prove the distributive law. We can just interchange addition and multiplication with your axioms and the distributive law fails.

If you write the added axioms correctly, you are all set. $8$ must be $n*1=n$, so you just apply $5$ to that to get $1*n=n$. For the second, you start with $9$ and apply $5$, then rename $a,b,c$.

Answer 2

Axiom 5 tells you that $1*n =n*1$ and Axiom 8 tells you $n*1 = n$

And Axiom 5 tells yoou that

$(a+b)*c = c*(a+b)$ and axiom 9 tells you $c*(a+b) = c*a +c*b$ and axiom 5 tells you $c*a + c*b = a*c + b*c$.