# Congruence arithmetic proof of $2i \equiv 2j \pmod{2m}$ implies $i \equiv j \pmod{2m}$ or $i\equiv j + m \pmod{2m}$.

If $2i \equiv 2j \pmod{2m}$, then either $i \equiv j \pmod{2m}$ or $i + m \equiv j \pmod{2m}$. This is easy to show by using the definition, i.e. if $2m$ divides $2(i-j)$, then either $2m$ divides $(i-j)$ (the first congruence), or not, in which case we have $km = i - j$ with $k$ odd. But then $(k+1)m = i - j + m$ and $k+1$ is even, hence $2m$ divides $i - j + m$.

But is there a way to show this just using the congruence relations listed for example here or here, i.e. without explicitly unfolding the definition but just using the "congruence calculus"?

By no. 12 on the Wolfram page, you have $i \equiv j \pmod{m}$, so $j-i \equiv 0 \pmod{m}$. Thus the entire problem reduces to showing that $k \equiv 0 \pmod{m}$ implies $k \equiv 0,m \pmod{2m}$. At this point I would depart from the lists of properties you linked to and say that, since $m | 2m$, there is a well-defined mapping from $Z_{2m}$ to $Z_m$ that takes an integer from its congruence class modulo $2m$ to its congruence class modulo $m$. We are interested in finding the inverse image of $0 \pmod{m}$ under this mapping. So we need only observe that the multiples of $m$ among $0,1,\dots 2m-1$ are precisely $0$ and $m$.
$$2i\equiv 2j\pmod{2m}\iff i\equiv j\pmod{m}$$
$$i=mk+j\iff \begin{cases}i=m(2t)+j\\\text{or}\\i=m(2t+1)+j\end{cases}$$
$$\iff \begin{cases}i=2mt+j\\\text{or}\\i=2mt+m+j\end{cases}\iff \begin{cases}i\equiv j\pmod{2m}\\\text{or}\\i\equiv m+j\pmod{2m}\end{cases}$$