Our disagreement seems to derive from my use of the words "different flawed step" and your use of "same flaw". Eliezer suggested substituting 1 for x and y in :

(x+y)(x-y) = y(x-y)

x+y = y

yielding

(1+1)(1-1) = 1(1-1) (true)

1+1 = 1 (false)

Thus, since a true equation was transformed into a false one, the step must have been flawed.

Under my suggestion, we have:

(0+0)(0-0) = 0(0-0) (true)

0+0 = 0 (true)

So, under Eliezer's suggested criterion (turning true to false) this is not a flawed step, though if you look carefully enough, you can still notice the flaw - a division by zero.