I'm not very fluent in mathematical proofs. High School has, sadly, not taught me any kind of proof-theory. That's why I would like your help with my proof of
$$m \equiv S_m \pmod 3.$$
where $S_m$ is the digit sum of $m$ and $\wedge$ is a logical-AND.
I'm going to use the very little that I know: Mathematical induction.
Basis:
$$A(1): S_m = 1 \wedge m = 1$$
$$m \equiv S_m \pmod 3\quad \text{because}\quad 1 \equiv 1 \pmod 3$$
Inductive Step:
$$A(n)\colon m = n \land S_m = \sum_{k=0}^K m_k = (m_0 + m_1 + \ldots + m_K)$$
$$m \equiv S_m \pmod 3.$$
So then
$$A(n+1)\colon m = n+1 \wedge S_m = \sum\limits_{k=0}^{K'} m_k = (m_0 + m_1 + \ldots + m_{K'})$$
$$m \equiv S_m \pmod 3.$$
I'm having trouble at the step of $A(n+1)$ because I don't know how to actually prove that it's valid? Please note that $m_0, \ldots, m_M$ refers to the summation of the digits of the number $m$. I just didn't find any good documentation that showed how to depict that in mathematical terms. Feel free to correct!