Taken from Problem 2.6 in Intermediate Real Analysis by Emanuel Fischer
Prove: (a) $x+z<y+z$ implies $x<y$, and (b) $x+z=y+z$ implies $x=y$.
The thing is, the author has not introduced the symbol $0$ nor the additive inverse yet (till the next page). How would you answer to this problem based on the previous axioms the author stated (p. 4-6)?
a) Suppose that $x > y$. By Axiom $0_4$ this would implies $x+z>y+z$ a contradiction with Axiom $0'_2$. If $x=y$ then we would have $x+z=y+z$ [1], again a contradiction. From Axiom $0'_2$, $x<y$.
b) Suppose $x < y$. By Axiom $0_4$ this would implies $x+z<y+z$ a contradiction with Axiom $0'_2$. Using the same argument $x>y$ yields a contradiction. Hence by Axiom $0'_2$, $x=y$.
[1] This holds because the operation $+:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is well defined. In fact, given $(x,y) \in \mathbb{R} \times \mathbb{R}$, from Axiom $A_1$, there is an unique $r \in \mathbb{R}$ such that $r=+(x,y)=x+y$. You can find discussions about it in these links Algebra: What allows us to do the same thing to both sides of an equation? Why do we have to do the same things to both sides of an equation?
