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Emeritus
PF Gold
P: 16,101

## Another proof I have a question about

Well, let's think how we would "normally" attack this problem:

$$\begin{equation*}\begin{split} x^2 = x \\ x^2 - x = 0 \\ x(x-1) = 0 \\ x = 0 \vee x = 1 \end{split}\end{equation*}$$

So our first thought should be if we can write this derivation in terms of the axioms. The only difficulty here is going from the third to the fourth step, and I'll give you a hint for it:

Can you prove that:

$$a b = 0 \Leftrightarrow a = 0 \vee b = 0$$

?

P.S. you have a typo in your cancellation law

P.P.S. this thing I asked you to prove, the "no zero divisors law" (I don't think that's a standard name), is an important one to remember; whether or not it is true has a big impact on the algebra of a system. For example, when you're working over the integers mod 6, this law fails (2 * 3 = 0), and your equation has four solutions (0, 1, 3, and 4).