Prove that if $a\equiv b \pmod m $ , then $a \bmod m = b \bmod m$ Question

Prove that if $a\equiv b \pmod  m$ , then
$a \bmod  m = b \bmod m$

Approach
Given,

$a\equiv b \pmod m$

$\implies m\mid (a-b)$
$\implies (a-b)=m\cdot k$
$\implies a=b+m\cdot k$
Now,

$a \bmod  m$   can be written as
$\implies b+m\cdot k  \pmod m$

No idea how to move forward to get
$$b \pmod m \impliedby b +m\cdot k  \pmod m$$
Please help me out!!!
Thanks!
Hint $\, $ Show  $\ c\equiv  d\pmod{\!m}\,$  and $\,|c-d|&lt;m\,\Rightarrow\,  c= d$

Apply that to $\,\ \underbrace{a\bmod m}_{\Large c}\!\equiv a\equiv b\equiv \underbrace{b\bmod m}_{\Large d}\,\ \pmod{\!m}$

Remark $ $ Any sequence of $m$ consecutive integers forms a complete system $S$ of representatives (normal forms) of all equivalence classes mod $m,\,$ so the above implies that congruence reduces to equality on these normal forms, i.e.  $\ a\equiv b\iff \bar a = \bar b,\,$ where $\bar n =$ unique normal-form rep in $S$ congruent to $n\,$ (which is analogous to: $ $ fractions are equivalent iff their least form rep's are equal, i.e.  least terms fractions are normal forms for equivalence of fractions).

An alternative convenient choice of normal forms for modular arithmetic arises by choosing the least magnitude reps (signed or balanced residue system), e.g. $\,\{-2,\color{#c00}{-1},0,1,2\}\pmod{\! 5}.\,$ In these signed systems the  normal form of $\,m\!-\!1\,$ is $\,-1,\,$ e.g. above $\,4\equiv\color{#c00}{-1}\pmod{\!5},\,$ simplifying common calculations such as $\,(m\!-\!1)^k\equiv (-1)^k,\,$  e.g. when casting out elevens.
I think it's obvious:

You have that $a \equiv b+mk \pmod m$.

But, $mk \equiv 0 \pmod m$ , so $a \equiv b+mk \equiv b\pmod m$.

Done.