Correct another congruence/mod proof, please!

Can somebody please check my work on this proof? I would greatly appreciate it!

Prove a ≡ b(mod m) if and only if there exists an integer k such that a = km+b.

(1)

If a ≡ b(mod m), then a = km+b, where k is an existing integer.

Proof: (⇒)Suppose a ≡ b(mod m), then m|a-b.

Thus a-b = km where k is an existing integer.

Therefore, by addition, a = km+b.(⇐)

(2)

If a =km+b and k is an existing integer, then a ≡ b(mod m).

Proof: Suppose a=km+b for an existing integer k.

Then a-b=km. Thus m|a-b.

Therefore a ≡ b(mod m).

Hence, a ≡ b(mod m) ⇐⇒ a=km+b for an existing integer k.

Re: Correct another congruence/mod proof, please!