Proposition 3.1.5. Let $G$ be a commutative ordered semigroup, and $\seq{g_n}\subset G$ such that for each $n \in \natp$, $g_{n+1}+ g_{n+1}\le g_{n}$. For each $x \in \mathbb{D}\cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_{n}$, then
For any $x, y \in \mathbb{D}\cap [0, 1)$ such that $x + y \in [0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.
For any $x, y \in \mathbb{D}\cap [0, 1)$ with $x \le y$, $\phi(x) \le \phi(y)$.
Proof. Assume without loss of generality that $x, y \in (0, 1)$. First suppose that $\text{rk}(x) = 2$ and $\text{rk}(y) \le 2$. In which case, since $G$ is commutative, it is sufficient to consider two cases: $(x, y) = (1/4, 1/4)$ and $(x, y) \in (1/2, 1/4)$. In the first case,
In the second, $\phi(x) + \phi(y) = \phi(x + y)$.
Now assume inductively that the proposition holds for all $x, y \in \mathbb{D}\cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1}\setminus \mathbb{D}_{n}$ and $y \in \mathbb{D}_{n+1}$. By Lemma 3.1.3, there exists $x_{0} \in \mathbb{D}_{n}$ such that $x = x_{0} + 2^{-n-1}$. If $y \in \mathbb{D}_{n}$, then
by the inductive assumption. Otherwise, there exists $y_{0} \in \mathbb{D}_{n}$ such that $y = y_{0} + 2^{-n-1}$, so
by the inductive assumption.$\square$