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If $m = s + t$ then $m \pmod k \equiv (s \pmod k + t \pmod k) \pmod k$

Can someone provide a proof of the above statement. I can understand it for the case of $s \equiv 0 \pmod k$ or $t \equiv 0 \pmod k$, or when both are zero, but not when neither are $0$.

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    Could you use this: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference? – 2017-02-16
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    Please edit the body of your Question to include a self-contained problem statement. Mere implied reference to the title short-changes your Readers by omitting a fuller set-up and explanation. – 2017-02-16

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$s \equiv r_1 \pmod k$ is the same as $s = nk + r$ for a multiple $n\in\mathbb{Z}$. This is that $s$ is to a distance of $r$ units of a multiple $n$ of $k$, where $r_1$ is the remainder

Same for $t \equiv r_2 \pmod k$. Then you know that $s$ is to a distance of $r_1$ and $t$ to a distance of $r_2$.

If you add $t$ and $s$ then $m=s+t$. What will happen if you perform modulo operation to $m$? Exactly, the remainder of $m \pmod k$ will be $r_1+r_2 \pmod k$ since both distances of $s$ and $t$ will sum up.

Then we can say that $m \equiv s+t \equiv r_1+r_2 \pmod k$

Finally take a look at Modular arithmetic since this is one of the congruence relations properties.