A regular weight has a unique dominant dot translate

Since $\mu$ is regular it is not fixed by any reflection, so it lies in exactly one open chamber $D$ of [F1]. By [F2] the orbit $W\mu$ meets the closed chamber $\bar{C}$ in exactly one point, necessarily regular and hence in $C$; this gives existence and uniqueness of $w$ with $w\mu$ dominant. For that $w$ and each $i$, $⟨w\cdot \lambda ,{\alpha }_i^{\vee }⟩=⟨w\mu -\rho ,{\alpha }_i^{\vee }⟩=⟨w\mu ,{\alpha }_i^{\vee }⟩-⟨\rho ,{\alpha }_i^{\vee }⟩$; here $⟨w\mu ,{\alpha }_i^{\vee }⟩=⟨\lambda ,(w^{-1}{\alpha }_i{)}^{\vee }⟩+⟨\rho ,(w^{-1}{\alpha }_i{)}^{\vee }⟩$ is a nonzero integer by [F4], [F6] and [F7], and $⟨\rho ,{\alpha }_i^{\vee }⟩=1$ by [F7]. Hence $w\cdot \lambda$ is dominant exactly when $⟨w\mu ,{\alpha }_i^{\vee }⟩$ is a nonnegative integer for all $i$, that is exactly when $w\mu$ is dominant; uniqueness transfers as well.

Let $\alpha$ be simple. By [F4] the map $\beta \mapsto s_{\alpha }\beta$ is an involution of ${\mathrm{\Phi }}^{+}\setminus \{\alpha \}$ and sends $\alpha$ to $-\alpha$. For $\beta \in {\mathrm{\Phi }}^{+}\setminus \{\alpha \}$ the $W$-invariance [F5] gives $⟨s_{\alpha }\mu ,s_{\alpha }{\beta }^{\vee }⟩=⟨\mu ,{\beta }^{\vee }⟩$, while $⟨s_{\alpha }\mu ,{\alpha }^{\vee }⟩=-⟨\mu ,{\alpha }^{\vee }⟩$. Since $s_{\alpha }$ permutes ${\mathrm{\Phi }}^{+}\setminus \{\alpha \}$, summing the defining conditions of $N$ over ${\mathrm{\Phi }}^{+}$ gives $N(s_{\alpha }\mu )=\mathrm{\#}\{\beta \in {\mathrm{\Phi }}^{+}\setminus \{\alpha \}:⟨\mu ,{\beta }^{\vee }⟩<0\}+[⟨\mu ,{\alpha }^{\vee }⟩>0]$, that is $N(s_{\alpha }\mu )=N(\mu )-[⟨\mu ,{\alpha }^{\vee }⟩<0]+[⟨\mu ,{\alpha }^{\vee }⟩>0]$; regularity of $\mu$ makes exactly one bracket equal to $1$, which gives the two asserted values.

For $\alpha \in {\mathrm{\Phi }}^{+}$ one has $⟨\mu ,{\alpha }^{\vee }⟩<0$ if and only if $⟨w\mu ,(w\alpha {)}^{\vee }⟩<0$ by $W$-invariance [F5]. As $\alpha$ runs over ${\mathrm{\Phi }}^{+}$, $w\alpha$ runs over $w{\mathrm{\Phi }}^{+}$; since $w\mu$ is strictly dominant, it pairs negatively with $w\alpha$ exactly when $w\alpha \in {\mathrm{\Phi }}^{-}$. Thus $N(\mu )=\mathrm{\#}\{\alpha \in {\mathrm{\Phi }}^{+}:w\alpha \in {\mathrm{\Phi }}^{-}\}=\mathrm{\mid }\mathrm{Inv}(w)\mathrm{\mid }=\mathrm{\ell }(w)$ by [F3].

Write $M=N(\mu )$ and let $w$ be as in step 1.1, so $\mathrm{\ell }(w)=M$ by step 2.1. Assume $w_j=s_{i_j}\cdots s_{i_1}$ has been constructed with $N(w_j\mu )=M-j$ and $j<M$. Then $w_j\mu$ is regular, because $W$-invariance [F5] shows $⟨w_j\mu ,{\gamma }^{\vee }⟩=0$ only if $⟨\mu ,(w_j^{-1}\gamma {)}^{\vee }⟩=0$, and $w_j^{-1}\gamma$ runs over all roots. Also $N(w_j\mu )=M-j>0$, so $w_j\mu$ is not dominant: if it were dominant then every positive root would pair nonnegatively with it by [F1], contradicting $N(w_j\mu )>0$. As dominance is tested on the simple coroots [F6] and $w_j\mu$ is regular, some simple $\alpha$ has $⟨w_j\mu ,{\alpha }^{\vee }⟩<0$; put $s_{i_{j+1}}=s_{\alpha }$, so step 1.2 gives $N(w_{j+1}\mu )=M-j-1$. Starting at $j=0$ this produces $w_M$ with $N(w_M\mu )=0$, so $w_M\mu$ is strictly dominant and hence $w_M=w$ by the uniqueness in step 1.1. To see the word is reduced, apply step 1.1 to the regular weight $w_j\mu$: the unique element $u_j$ with $u_j{w}_j\mu$ dominant satisfies $\mathrm{\ell }(u_j)=N(w_j\mu )=M-j$ by step 2.1 and $u_j{w}_j\mu =w\mu$, so $u_j{w}_j=w$ by uniqueness; hence $\mathrm{\ell }(w{w}_j^{-1})=M-j$, while $\mathrm{\ell }(w)=M$ and subadditivity give $M=\mathrm{\ell }(w{w}_j^{-1}{w}_j)\le \mathrm{\ell }(w{w}_j^{-1})+\mathrm{\ell }(w_j)=M-j+\mathrm{\ell }(w_j)$, that is $\mathrm{\ell }(w_j)\ge j$. Since $w_j$ is a product of $j$ simple reflections, $\mathrm{\ell }(w_j)\le j$ by [F3], so $\mathrm{\ell }(w_j)=j$ and the expression is reduced. This proves the first part of (iii) with $\mathrm{\ell }=M$.

Now suppose in addition that $\mu$ is dominant, so $N(\mu )=0$ and $\mu$ is strictly dominant by regularity. Repeat the construction of step 3.1 with the opposite choice: given $w_j$ with $N(w_j\mu )=j<N$, the regular weight $w_j\mu$ is not strictly antidominant, and strict antidominance is tested on the simple coroots [F6]; hence some simple $\alpha$ has $⟨w_j\mu ,{\alpha }^{\vee }⟩>0$, and step 1.2 increases $N$ by one. Starting at $j=0$ produces $w_N$ with $N(w_N\mu )=N=\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }$: every positive root pairs negatively with $w_N\mu$ by [F1], so $w_N\mu$ lies in the chamber of $w_0\mu$ by [F2]; regularity gives $w_N=w_0$, and $w_N\mu =w_0\mu$ is strictly antidominant. The word has $N$ factors and $w_N=w_0$ has length $N$ by [F2], so it is reduced.

Let $w\in W$ and $\alpha \in {\mathrm{\Phi }}^{+}$. Since $w_0$ sends positive roots to negative roots and negative roots to positive roots [F2], the root $w_{0}w\alpha$ is negative exactly when $w\alpha$ is positive. Counting positive roots gives $\mathrm{\ell }(w_{0}w)=\mathrm{\#}\{\alpha \in {\mathrm{\Phi }}^{+}:w_{0}w\alpha \in {\mathrm{\Phi }}^{-}\}=\mathrm{\#}\{\alpha \in {\mathrm{\Phi }}^{+}:w\alpha \in {\mathrm{\Phi }}^{+}\}=\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }-\mathrm{\ell }(w)$, which is (iv). All the cited inputs are choice-free, and no step invokes a choice principle.