Geometric series are invertible in the completed character ring

For $k\ge 1$ every exponent of $u^k$ lies in $-Q_{+}\setminus \{0\}$ and has height at most $-k$: each exponent is the negative of a sum of $k$ nonzero elements of $Q_{+}$, a nonzero element of $Q_{+}$ has height at least $1$ by [F2], and heights add; hence for a fixed $\eta$ the coefficient of $e^{\eta }$ in ${\sum }_{k\ge 0}(-u{)}^k$ vanishes for $k>-\mathrm{ht}(\eta )$ and for $\eta \mathrm{\notin }Q$ or $\eta \mathrm{\notin }-Q_{+}$, whereas for each single $k$ the coefficient is a finite integer by [F1], so the sum defines an element $v\in \mathcal{R}$; multiplying out with [F1] and [F3] gives $(1+u)v={\sum }_{k\ge 0}(-u{)}^k+{\sum }_{k\ge 0}(-1{)}^k{u}^{k+1}={\sum }_{k\ge 0}(-u{)}^k-{\sum }_{j\ge 1}(-u{)}^j=1$, so $v$ is the inverse of $1+u$.

Claim (ii): expanding the finite product gives ${\prod }_{\alpha \in {\mathrm{\Phi }}^{+}}(1-e^{-\alpha })=1+w$ with $w={\sum }_{\mathrm{\varnothing }\mathrm{\ne }S\subseteq {\mathrm{\Phi }}^{+}}(-1{)}^{\mathrm{\mid }S\mathrm{\mid }}{e}^{-{\sum }_{\alpha \in S}\alpha }$, and each exponent $-{\sum }_{\alpha \in S}\alpha$ with $S$ nonempty lies in $-Q_{+}\setminus \{0\}$ because every $\alpha \in {\mathrm{\Phi }}^{+}$ is a nonzero element of $Q_{+}$ by [F2], so $w$ is supported in $-Q_{+}\setminus \{0\}$ and step 1.1 makes the product invertible; likewise each geometric series ${\sum }_{k\ge 0}{e}^{-k\alpha }$ is the inverse of $1-e^{-\alpha }$, since $e^{-\alpha }$ is supported in $-Q_{+}\setminus \{0\}$ and step 1.1 applies, so the coefficientwise product $G={\prod }_{\alpha \in {\mathrm{\Phi }}^{+}}{\sum }_{k\ge 0}{e}^{-k\alpha }$ is the inverse of the product (a finite product of inverses is the inverse of the product in a commutative ring), and $G$ lies in $\mathcal{R}$ because at a fixed exponent only finitely many tuples $(k_{\alpha })$ can sum to it by [F2]. Claim (i) is immediate from [F3].

Claim (iv): grouping the finite defining sum of $A(\rho )$ by $w=1$ and $w\mathrm{\ne }1$ gives $A(\rho )=e^{\rho }+{\sum }_{w\mathrm{\ne }1}(-1{)}^{\mathrm{\ell }(w)}{e}^{w\rho }=e^{\rho }(1+u^{\mathrm{\prime }})$ with $u^{\mathrm{\prime }}={\sum }_{w\mathrm{\ne }1}(-1{)}^{\mathrm{\ell }(w)}{e}^{w\rho -\rho }$; for $w\mathrm{\ne }1$ the exponent $w\rho -\rho =-(\rho -w\rho )$ lies in $-Q_{+}\setminus \{0\}$ by [F4], so $u^{\mathrm{\prime }}$ is supported in $-Q_{+}\setminus \{0\}$ and step 1.1 shows that $1+u^{\mathrm{\prime }}$ is invertible; hence $A(\rho )=e^{\rho }(1+u^{\mathrm{\prime }})$ is a product of the invertible elements $e^{\rho }$ and $1+u^{\mathrm{\prime }}$, with inverse $e^{-\rho }(1+u^{\mathrm{\prime }}{)}^{-1}$ by [F3] and multiplicativity of inversion. Claim (iii) is the same multiplicativity applied to $e^{\rho }$ and the invertible product of claim (ii).