Prove If a,b are both non-zero, then ab is not equal to 0

Question

I cannot figure out how to prove the above theorem (theorem 3 in image) using only the axioms and theorems one and two shown in the image above, please help if you can! (https://i.stack.imgur.com/nwepP.jpg)

Answer 1

Proof by Contradiction:

1. Let $a,b \in \mathbb{R}$ such that $a \neq 0$ and $b \neq 0$. In other words, it is NOT the case that $a=0$ or $b=0$.
2. Assume $ab=0$.
3. Theorem 2 states that if $ab=0$, then $a=0$ or $b=0$. We have already assumed that $ab=0$, so we may conclude that $a=0$ or $b=0$.
4. In line $1$ we stated $a \neq 0$ and $b \neq 0$, but in line $3$ we deduced that $a=0$ or $b=0$. Hence, we have a contradiction.
5. Therefore, our assumption that $ab=0$ is false, and it must be true that $ab \neq 0$