I know that $a \equiv b \pmod n$, is like saying that $n \mid b-a$, that is like saying $\operatorname{rem}(a, n) = \operatorname{rem} (b, n)$, where $\operatorname{rem}(h, k)$ is the remainder of the division $a/n$.
If $a \equiv a'$ and $b \equiv b'$, then $a + a' \equiv b + b' \pmod d$.
From this derive that if $l \equiv m \pmod g$, and $k$ is a positive integer, then $l + k \equiv m + k \pmod g$; so if $a \equiv b \pmod d$, then $a-b \equiv 0 \pmod d$.
This seems strange to me: when we say that $a \equiv b$ and $a-b$ is congruent to zero, we're saying that $d \mid b-a-0$, and so, $\operatorname{rem}(a-b, d) = \operatorname{rem}(0, d)$; that's impossible, since the remainder on the division by $d$ of $a-b$ must be zero, for the hypothesis that they're congruent.
How can explain this behavior?
p.s. I'm sorry for the bad english, isn't my native language.
$rem(0,d)$ is the remainder when $0$ is divided by $d$. So $rem(0,d)=0$.