Ordering on Cuts is Compatible with Addition of Cuts

Theorem
Let $\alpha$, $\beta$ and $\gamma$ be cuts.

Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.

Let $\beta < \gamma$ denotes the strict ordering on cuts defined as:
 * $\beta < \gamma \iff \exists p \in \Q: p \in \beta, p \notin \gamma$

Then:
 * $\beta < \gamma \implies \alpha + \beta < \alpha + \gamma$

Proof
By definition of the strict ordering on cuts and addition of cuts:


 * $\alpha + \beta \le \alpha + \gamma$

Suppose $\alpha + \beta = \alpha + \gamma$.

Then:

Thus:
 * $\beta = \gamma$

and the result follows.