Replacing the "if $x ≤ y$, then $x + z ≤ y + z$" axiom in Reals.

Question

How can I prove that we cannot (or maybe can) replace preservation of order under addition i.e. "If $x \leq y$, then $x + z \leq y + z$ with "if $0<x$ and $0<y$ , then $0<x+y$" in axioms of the real numbers $(R,+,.,<)$?

Edit: Axiom of the reals:

1. The set $\Bbb{R}$ is a field.

2. The field $\Bbb{R}$ is ordered.

   if $a ≤ b$ then $ a + c ≤ b + c$

   if $0 ≤ a$ and $0 ≤ b$ then $0 ≤ a×b$

3. The order is Dedekind-complete; that is, every non-empty subset S of $\Bbb{R}$ with an upper bound in $\Bbb{R}$ has a least upper bound (also called supremum) in $\Bbb{R}$.

Answer 1

Replacing that axiom would make certain structures into models that were originally not models. For example take standard $\mathbb{R}$ but define every number in the interval $[-10,-5)$ to be greater than every number in the interval $[-5,-4)$. (Additionally we must specify that $[-4,\infty)$ comes after $[-10,-5)$ and $(-\infty, -10)$ comes before $[-5,4)$... essentially we can muck about with the order of any entirely negative half closed intervals).

We see that $<$ is still asymmetric, dense, and D-complete. Addition is unchanged. The statement that "$x \le y \rightarrow x + z \le y + z$" is false: example $-7 \le -6$ but $-7 + 1 \ge -6 + 1$, but the new axiom is still true.