3. Monotonicity for purely infinite series

Fix some $x,y\in\mathbb{U}$ with $x>y>\mathbb{R}$. We prove by induction on $\mathrm{ER}(\gamma)$ that $\gamma\circ x>\gamma\circ y$. Since by construction $0<\log(x)-\log(y)<x-y$, by iterating $\log$, we find that for all $k>n>0$

Therefore, the statement is true for $\mathrm{ER}(\gamma)=0$, and even for $\gamma=\delta-\eta$ when $\mathrm{ER}(\delta)=\mathrm{ER}(\eta)=0$ and $\delta>\eta$.

Now let $\gamma\in\mathbb{J}^{>0}$ have rank $\mathrm{ER}(\gamma)>0$. We assume by induction that the conclusion holds for any $\delta$ such that $\mathrm{ER}(\delta)<\mathrm{ER}(\gamma)$. Let $re^{\delta}$ be the leading term of $\gamma$. Recall that $\mathrm{ER}(\delta)<\mathrm{ER}(\gamma)$.

We claim that $\gamma\circ x-\gamma\circ y\sim re^{\delta\circ x}-re^{\delta\circ y}$. Let $e^{\eta}$ be any monomial in the support of $\gamma$ distinct from $e^{\delta}$ (if there is no such monomial, then the conclusion is obvious as $\gamma=re^{\delta}$). Again $\mathrm{ER}(\eta)<\mathrm{ER}(\gamma)$, and moreover $\delta>\eta>0$, since $\gamma\in\mathbb{J}$. We have one of $\mathrm{ER}(\delta)=\mathrm{ER}(\eta)=0$ or $\mathrm{ER}(\delta-\eta)\leq\max\{\mathrm{ER}(\delta),\mathrm{ER}(\eta)\}<\mathrm{ER}(\gamma)$, hence by inductive hypothesis

Moreover, $\eta\circ x-\eta\circ y>0$, again by inductive hypothesis. Therefore,

since $\exp$ is an increasing function and $e^{a}>1$ for $a>0$.

Since $(\delta-\eta)\circ y>\mathbb{R}$, we have $e^{(\delta-\eta)\circ y}>\mathbb{R}$, or in other words, $e^{\delta\circ y}\succ e^{\eta\circ y}$. Combining the inequalities together,

It follows at once that $\gamma\circ x-\gamma\circ y\sim re^{\delta\circ x}-re^{\delta\circ y}$.

Granted the claim, by inductive hypothesis, $\delta\circ x>\delta\circ y$. Recall moreover that $r>0$, since $\gamma>0$. It follows that $re^{\delta\circ x}-re^{\delta\circ y}>0$, hence $\gamma\circ x-\gamma\circ y>0$, as desired. ∎