Note that assuming that there is another solution y besides x would not work in the final proof because the existence of this y is precisely what you need to prove. In the initial attempt, however, it is fine to assume that such y exists in order to see what properties it must have. So you concluded that b(x-y) = 0 must hold. Can it in fact hold since b is a zero divisor?I guess that if there is a solution to bx = c, there must also be a y such that by = c, where x != y. I could set the two equal to each other: bx = by, and then simplify to b(x-y) = 0.