<s>[INST] Proving distributive law of natural numbers Is my proof correct?
If we define multiplication for natural numbers as
$a \times S(b) = (a \times b) + a$
$a \times 0 = 0$
And addition as
$a + 0 = a$
$a + S(b) = S(a+b)$
Where $S(n)$ is the successor function of $n$ (and assume we've already proven commutative property, etc).
Then I wish to prove that $a \times (b + c) = (a \times b) + (a \times c)$ via induction on $c$.
Base case: Let $c=0$. Then $a \times (b + 0) = a \times b$ and $(a \times b) + (a \times 0) = (a \times b) + 0 = a \times b$. Therefore $a \times (b + 0) = (a \times b) + (a \times 0)$.
Inductive Step: Suppose $a \times (b + c) = (a \times b) + (a \times c)$ for some $c$. We wish to show that $a \times (b + S(c)) = (a \times b) + (a \times S(c))$.
$a \times (b + S(c)) = a \times (S(b + c)) = a \times (b+c) + a = (a \times b) + (a \times c) + a$
and
$(a \times b) + (a \times S(c)) = (a \times b) + (a \times c) + a$
Since both sides can be shown to equal the same quantity $(a \times b) + (a \times c) + a$, we close the induction.
 [/INST] Perfect ... though you may want to note that you need associativity of addition, since:
$a \times (b + S(c)) = a \times (S(b + c)) = (a \times (b+c)) + a = \color{red}{(}(a \times b) + (a \times c)\color{red}) + a$
while:
$(a \times b) + (a \times S(c)) = (a \times b) + \color{red}((a \times c) + a\color{red})$
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