By popular demand, I’ll prove the theorem (and correct a typo).
a≡b(modm) → a=q1m+r, b=q2m+r → (a-b)=qm
c≡d(modm) → (c-d)=q’m
* So I don’t leave with a guilty conscience,
a-b=qm, a=q1m+r1, b=q2m+r2 → r1=r2→ a≡b(modm).
The point of the theorem was to show it is not a definiton of addition.
Now that you understand congruences and modular arithmetic (definitions and operations), don’t look back at any problems- remember Lot's wife.
EDIT: For a rainy day, if m is prime there is a cancellation law. Didn't want to clutter up the structure.