Is $((a \mod n) + (b \mod n) ) = (a + b) \mod n$? As we know $(a + b) \mod n = ((a\, \bmod\, n) + (b\, \bmod\, n))\, \bmod\, n$ 
Is their reversal also true like this $((a \mod n) + (b \mod n)) = (a + b) \mod n$. If not then what could be its alternative?
 A: You can add, subtract and multiply congruences all day as long as you mind the "wrap-around."
Look at an analog clock or watch, let's do some arithmetic modulo 12. Suppose you work an 8-hour shift at the factory starting at 1300 (1 p.m.) Indeed $(13 \mod 12) + (8 \mod 12) = 9$ and $13 + 8 \equiv 9 \mod 12$. Suppose you're asked to work 4 hours overtime immediately after your regular shift. We have $9 + 4 \equiv 1 \mod 12$ and likewise $(21 \mod 12) + (4 \mod 12) = 1$.
Now, I don't know what your lunch arrangements would be in this hypothetical scenario, but the math can easily be changed to accommodate it.
You might complain that I'm being cagey in not exactly copying your usage of the equal sign. Truth is, I think it could be misleading. Maybe you manage to punch in at exactly 1300 one day and punch out exactly at 0100 the next day. The clock's hands are in the exact same position, and indeed $1 = 1$, but in some ways, the 1 p.m. that you started your regular shift at is very different from the 1 a.m. you ended your overtime shift at.
But if we were to declare that we're going to be working in, say, $\mathbb{Z}_{12}$, a finite ring which consists of precisely 12 integers, I'd have no problem at all with statements like $9 + 4 = 1$ and $1 = 9 + 4$. Equality is unequivocally commutative. Equivalence is commutative with caveats.
