[proofplan] We compare the Weyl group with the abstract group defined by the stated generators and relations. The defining relations give a natural surjective homomorphism from the abstract Coxeter group to the Weyl group. To prove injectivity, we construct an inverse map from $W$ to the abstract group by choosing reduced expressions; Matsumoto's theorem makes this choice independent of the reduced expression, and the Coxeter relations give exactly the required braid moves. The inverse property is then checked by multiplying reduced expressions by one simple reflection at a time. [/proofplan]

[step:Construct the abstract presentation group and map it onto the Weyl group] Let $G$ denote the abstract group \begin{align*} G := \langle \sigma_1,\dots,\sigma_r : (\sigma_i\sigma_j)^{m_{ij}}=e_G \text{ for all } i,j \rangle, \end{align*} where $e_G$ is the identity element of $G$. Define a map on generators by \begin{align*} \pi_0: \{\sigma_1,\dots,\sigma_r\} &\to W, \\ \sigma_i &\mapsto s_i. \end{align*} For each pair $i,j$, the definition of $m_{ij}$ gives $(s_i s_j)^{m_{ij}}=e_W$, where $e_W$ is the identity element of $W$. Therefore the universal property of a group given by generators and relations induces a group homomorphism \begin{align*} \pi: G &\to W, \\ \sigma_i &\mapsto s_i. \end{align*} Since the simple reflections $s_1,\dots,s_r$ generate the Weyl group $W$, the homomorphism $\pi$ is surjective. [/step]

[step:Use the Weyl group exchange theorem to control word length]Let $\Phi$ be the given reduced crystallographic root system, let $\Delta=\{\alpha_1,\dots,\alpha_r\}$ be the simple root system determined by the chosen Weyl chamber, and let $s_i=s_{\alpha_i}$ be the reflection in the hyperplane perpendicular to $\alpha_i$. Define the simple-reflection length function \begin{align*} \ell: W &\to \mathbb{N}\cup\{0\}, \\ w &\mapsto \min\{k\in\mathbb{N}\cup\{0\}: w=s_{i_1}\cdots s_{i_k}\text{ for some }i_1,\dots,i_k\in\{1,\dots,r\}\}. \end{align*} A word $s_{i_1}\cdots s_{i_k}$ is called reduced when $k=\ell(s_{i_1}\cdots s_{i_k})$. We use the [Exchange Theorem for Weyl Groups](/page/Exchange%20Theorem%20For%20Weyl%20Groups), applied to the Coxeter generating set $\{s_1,\dots,s_r\}$ associated to the simple system $\Delta$: if \begin{align*} w=s_{i_1}\cdots s_{i_k} \end{align*} is not reduced, then there exist indices $1\le a<b\le k$ such that \begin{align*} s_{i_1}\cdots s_{i_k}=s_{i_1}\cdots \widehat{s_{i_a}}\cdots \widehat{s_{i_b}}\cdots s_{i_k}, \end{align*} where the hats mean that the indicated factors are omitted. The hypotheses needed for the exchange theorem are satisfied because $W$ is the Weyl group of the reduced root system $\Phi$, and $s_1,\dots,s_r$ are precisely the simple reflections associated to $\Delta$. In particular, every word in the alphabet $s_1,\dots,s_r$ can be transformed, without changing its element of $W$, to a reduced expression by repeated exchange deletions; each deletion strictly lowers the simple-reflection word length.[/step]

[guided]The purpose of this step is to isolate the one structural fact about Weyl groups that controls word reduction. Let $\Phi$ be the reduced crystallographic root system in the theorem, let $\Delta=\{\alpha_1,\dots,\alpha_r\}$ be its simple root system, and let $s_i=s_{\alpha_i}$ be the simple reflection attached to $\alpha_i$. We invoke the exchange theorem for Weyl groups. Its hypotheses require a Weyl group generated by the reflections attached to a chosen simple root system. These hypotheses are exactly the present setting: $W$ is the Weyl group of $\Phi$, and $s_1,\dots,s_r$ are the corresponding simple reflections. The theorem says that if a product \begin{align*} w=s_{i_1}\cdots s_{i_k} \end{align*} is not reduced, then two letters can be deleted without changing the Weyl group element: \begin{align*} s_{i_1}\cdots s_{i_k}=s_{i_1}\cdots \widehat{s_{i_a}}\cdots \widehat{s_{i_b}}\cdots s_{i_k} \end{align*} for some $1\le a<b\le k$. This is the mechanism that replaces cancellation in a free group. A word in simple reflections may have no adjacent equal letters, but it can still be non-reduced because a more global exchange is possible. Repeating the exchange deletion strictly lowers the length of the word, so the process terminates at a reduced expression representing the same element of $W$.[/guided]

[step:Identify the only relations between reduced expressions]For $i\ne j$ and $m=m_{ij}$, let $A_m(\sigma_i,\sigma_j)$ denote the alternating word of length $m$ beginning with $\sigma_i$, and let $A_m(\sigma_j,\sigma_i)$ denote the alternating word of length $m$ beginning with $\sigma_j$. [claim:The Coxeter power relation gives the rank-two braid relation] For every $i\ne j$, the relation $(\sigma_i\sigma_j)^{m_{ij}}=e_G$ implies \begin{align*} A_{m_{ij}}(\sigma_i,\sigma_j)=A_{m_{ij}}(\sigma_j,\sigma_i) \end{align*} in $G$. [/claim] [proof] The relations with equal indices give $\sigma_i^2=e_G$ and $\sigma_j^2=e_G$, because $m_{ii}=m_{jj}=1$. Set $a=\sigma_i$, $b=\sigma_j$, and $m=m_{ij}$. Then $a^{-1}=a$, $b^{-1}=b$, and $(ab)^m=e_G$. If $m=2q$ is even, then $A_m(a,b)=(ab)^q$ and $A_m(b,a)=(ba)^q$. Since $(ab)^{-1}=ba$ and $(ab)^{2q}=e_G$, we have \begin{align*} (ab)^q=(ab)^{-q}=(ba)^q. \end{align*} If $m=2q+1$ is odd, then $A_m(a,b)=(ab)^q a$ and $A_m(b,a)=(ba)^q b$. Their quotient is \begin{align*} \big((ab)^q a\big)^{-1}(ba)^q b &=a(ba)^q(ba)^q b \\ &=a(ba)^{2q}b \\ &=(ab)^{2q+1} \\ &=e_G, \end{align*} so $(ab)^q a=(ba)^q b$. This proves the braid relation in both parity cases. [/proof] We use [Matsumoto's Theorem for Weyl Groups](/page/Matsumoto%27s%20Theorem%20For%20Weyl%20Groups), applied to the Coxeter generating set $\{s_1,\dots,s_r\}$ associated to the simple system $\Delta$: if \begin{align*} s_{i_1}\cdots s_{i_k}=s_{j_1}\cdots s_{j_k} \end{align*} are two reduced expressions for the same element of $W$, then one expression is obtained from the other by a finite sequence of rank-two braid moves. By the claim, every such braid move is valid in $G$. Hence two reduced words in the $\sigma_i$ whose images under $\pi$ are the same element of $W$ represent the same element of $G$.[/step]

[guided]We now identify exactly which relations can occur between reduced expressions. First we check that the defining power relation in $G$ really contains the usual braid relation. For $i\ne j$ and $m=m_{ij}$, let $A_m(\sigma_i,\sigma_j)$ be the alternating word of length $m$ beginning with $\sigma_i$, and let $A_m(\sigma_j,\sigma_i)$ be the alternating word of length $m$ beginning with $\sigma_j$. The equal-index relations are important: since $m_{ii}=m_{jj}=1$, the defining relations give $\sigma_i^2=e_G$ and $\sigma_j^2=e_G$. Set $a=\sigma_i$ and $b=\sigma_j$. Then $a^{-1}=a$, $b^{-1}=b$, and $(ab)^m=e_G$. If $m=2q$, the two alternating words are $(ab)^q$ and $(ba)^q$. Since $(ab)^{-1}=ba$ and $(ab)^{2q}=e_G$, \begin{align*} (ab)^q=(ab)^{-q}=(ba)^q. \end{align*} If $m=2q+1$, the two alternating words are $(ab)^q a$ and $(ba)^q b$. We compute the quotient: \begin{align*} \big((ab)^q a\big)^{-1}(ba)^q b &=a(ba)^q(ba)^q b \\ &=a(ba)^{2q}b \\ &=(ab)^{2q+1} \\ &=e_G. \end{align*} Therefore $(ab)^q a=(ba)^q b$. Thus the relation $(\sigma_i\sigma_j)^{m_{ij}}=e_G$ gives the rank-two braid relation in both the even and odd cases. Now apply [Matsumoto's Theorem for Weyl Groups](/page/Matsumoto%27s%20Theorem%20For%20Weyl%20Groups). Its hypotheses require the Weyl group together with the simple reflections attached to a chosen simple root system. These hypotheses hold here because $W$ is the Weyl group of $\Phi$ and $s_1,\dots,s_r$ are the simple reflections associated to $\Delta$. If \begin{align*} s_{i_1}\cdots s_{i_k}=s_{j_1}\cdots s_{j_k} \end{align*} are reduced expressions for the same Weyl group element, Matsumoto's theorem connects them by finitely many braid moves inside rank-two subgroups generated by pairs $s_i,s_j$. The computation above shows that each corresponding move is already valid among the generators $\sigma_i,\sigma_j$ in $G$. Hence reduced words in the $\sigma_i$ with the same image in $W$ are equal in $G$.[/guided]

[step:Build an inverse map from reduced expressions] Define a map \begin{align*} \rho: W &\to G \end{align*} as follows. For $w\in W$, choose a reduced expression $w=s_{i_1}\cdots s_{i_k}$ and set \begin{align*} \rho(w):=\sigma_{i_1}\cdots\sigma_{i_k}. \end{align*} This definition is independent of the chosen reduced expression by the previous step, because any two reduced expressions for $w$ give the same element of $G$. We prove that $\rho\circ\pi=\operatorname{id}_G$. First note the multiplication rule: for every $w\in W$ and every simple reflection $s_i$, one has \begin{align*} \rho(ws_i)=\rho(w)\sigma_i. \end{align*} Indeed, if $\ell(ws_i)=\ell(w)+1$, then appending $s_i$ to a reduced expression for $w$ gives a reduced expression for $ws_i$, so the equality follows from the definition of $\rho$. If $\ell(ws_i)=\ell(w)-1$, then appending $s_i$ to a reduced expression for $ws_i$ gives a reduced expression for $w$; hence $\rho(w)=\rho(ws_i)\sigma_i$, and using $\sigma_i^2=e_G$ gives $\rho(w)\sigma_i=\rho(ws_i)$. Let $g\in G$ and choose a word \begin{align*} g=\sigma_{i_1}\cdots\sigma_{i_k}. \end{align*} Set $w_0=e_W$ and, for $1\le t\le k$, define $w_t=s_{i_1}\cdots s_{i_t}\in W$. Applying the multiplication rule successively gives \begin{align*} \rho(\pi(g)) =\rho(w_k) =\rho(w_0)\sigma_{i_1}\cdots\sigma_{i_k} =e_G\sigma_{i_1}\cdots\sigma_{i_k} =g. \end{align*} Thus $\rho\circ\pi=\operatorname{id}_G$, so $\pi$ is injective. Since $\pi$ is also surjective, it is an isomorphism. Under this isomorphism $\sigma_i$ corresponds to $s_i$, and the stated presentation is the presentation of $W$. [/step]