By popular demand, I’ll prove the theorem (and correct a typo).

a≡b(modm) → a=q_{1}m+r, b=q_{2}m+r → (a-b)=qm

c≡d(modm) → (c-d)=q’m

(a+c)-(b+d)=q’’m

(a+c)≡(b+d)(modm)*

* So I don’t leave with a guilty conscience,

a-b=qm, a=q_{1}m+r_{1}, b=q_{2}m+r_{2}→ r_{1}=r_{2}→ a≡b(modm).

Notes

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.