For all integers a, b, c, m with m > 0 and c > 0, if ac ≡ bc (mod m), then a ≡ b (mod m)

The statement is saying that for all integers a, b, c, m where m is greater than 0 and c is greater than 0, if ac is congruent to bc modulo m, then a is congruent to b modulo m.

Here's a step-by-step proof:

1. Start with the given congruence: ac ≡ bc (mod m). This means that m divides (ac - bc).

2. Factor out c from the difference: ac - bc = c(a - b). So, m divides c(a - b).

3. According to the definition of divisibility, there exists an integer k such that c(a - b) = mk.

4. If c > 0, then we can divide both sides of the equation by c to get a - b = mk/c.

5. Since m, k, and c are all integers, and m and c are positive, mk/c is also an integer.

6. Therefore, m divides (a - b), which means a ≡ b (mod m).

So, the statement is true.