Applying axioms to inequalities

Question

For example I'm supposed to prove that if $xy>0$, then $x>0$ and $y>0$, or $x<0$ and $y<0$. I'm supposed to do this using only these axioms:

1. $(x+y)+z=(x+z)+y$

2. $x+y=y+x$

3. $x+0=x$

4. There exists an $x'$ such that $x'+x=0$.

5. $(x*y)*z=(x*z)*y$

6. $x*y=y*x$

7. There exists element $1$ such that $1\neq 0$ and $x*1=x$

8. There exists an $x'$ such that $x'*x=1$

9. $x*(y+z)=x*y+x*z$

10. $x>0, x<0, or x=0$

My first idea was to just use axiom $10$ and apply that to both $x$ and $y$ and keep showing that contradictions exist, but in that case it seems like I'm only using axiom $10$ at the beginning and just using logic from there instead of using ONLY axioms. Any ideas on how I'd apply these to inequality proofs like this one?

Answer 1

I suspect you are missing some axioms, because without more axioms about the behavior of the order relation, here's a simple counterexample.

Start with the real numbers $\mathbb{R}$ equipped with their ordinary addition and multiplication operations, which satisfy axioms 1--9. Define the relation $<$ on $\mathbb{R}$ as follows: $x>0$ for all $x \ne 0$ except for $x=-1$; define $-1<0$. Extend over all other pairs which do not involve $0$ in any arbitrary way you please. Axiom 10 is satisfied. But we have $-1 \cdot 2 > 0$, but $-1 < 0$ and $2>0$, which violates what you are supposed to prove.

Answer 2

Suppose there is an element $t$ other than $1$ and $0$, and suppose $t>0$ and $1<0$. Then $1t=t$ by axiom 7, but now $1t=t>0$ while $1<0$ and $t>0$ which contradicts what you are trying to prove.