[proofplan]
We compare $N(w)$ with $N(ws_i)$ for a single right multiplication by a simple reflection. The key root-system fact is that $s_i$ sends $\alpha_i$ to $-\alpha_i$ and permutes the positive roots different from $\alpha_i$. This gives an exact formula: right multiplication by $s_i$ either adds the one root $\alpha_i$ to the inversion set or removes it, according to the sign of $w\alpha_i$. Applying this formula along a reduced expression gives $|N(w)|=\ell(w)$, and the same one-step formula then gives the two length-change criteria.
[/proofplan]
[step:Record the one simple reflection property of positive roots]
For each $i\in I$, the simple reflection $s_i$ satisfies
\begin{align*}
s_i\alpha_i=-\alpha_i,
\end{align*}
and the restricted map
\begin{align*}
s_i:\Phi^+\setminus\{\alpha_i\} &\to \Phi^+\setminus\{\alpha_i\} \\
\beta &\mapsto s_i\beta
\end{align*}
is a bijection. We use here the standard simple-root positivity lemma for Coxeter root systems: a simple reflection sends its own simple root to its negative and preserves the positive system on all other positive roots.
Consequently, when testing membership in $N(ws_i)$, the root $\alpha_i$ is the only positive root whose sign is reversed by the final application of $s_i$ itself.
[/step]
[step:Compute how the inversion set changes under right multiplication]
Fix $w\in W$ and $i\in I$. We compare
\begin{align*}
N(ws_i)=\{\beta\in\Phi^+ : ws_i\beta\in -\Phi^+\}
\end{align*}
with
\begin{align*}
N(w)=\{\gamma\in\Phi^+ : w\gamma\in -\Phi^+\}.
\end{align*}
First consider the root $\beta=\alpha_i$. Since $s_i\alpha_i=-\alpha_i$, we have
\begin{align*}
ws_i\alpha_i=-w\alpha_i.
\end{align*}
Therefore
\begin{align*}
\alpha_i\in N(ws_i)
\quad\Longleftrightarrow\quad
-w\alpha_i\in -\Phi^+
\quad\Longleftrightarrow\quad
w\alpha_i\in \Phi^+.
\end{align*}
Now let $\beta\in\Phi^+\setminus\{\alpha_i\}$, and define
\begin{align*}
\gamma:=s_i\beta.
\end{align*}
By the previous step, $\gamma\in\Phi^+\setminus\{\alpha_i\}$, and since $s_i^2=1$, the assignment $\beta\mapsto \gamma=s_i\beta$ is a bijection of $\Phi^+\setminus\{\alpha_i\}$. For such $\beta$,
\begin{align*}
\beta\in N(ws_i)
&\Longleftrightarrow ws_i\beta\in -\Phi^+ \\
&\Longleftrightarrow w\gamma\in -\Phi^+ \\
&\Longleftrightarrow \gamma\in N(w).
\end{align*}
Hence
\begin{align*}
N(ws_i)=
\begin{cases}
s_iN(w)\cup\{\alpha_i\}, & \text{if } w\alpha_i\in\Phi^+,\\
s_i\bigl(N(w)\setminus\{\alpha_i\}\bigr), & \text{if } w\alpha_i\in-\Phi^+.
\end{cases}
\end{align*}
In the first case, $\alpha_i\notin N(w)$, because $w\alpha_i\in\Phi^+$. In the second case, $\alpha_i\in N(w)$, because $w\alpha_i\in-\Phi^+$. Thus the displayed formula is a bijective description of the change in inversion sets; whenever $N(w)$ is finite it also gives
\begin{align*}
|N(ws_i)|=
\begin{cases}
|N(w)|+1, & \text{if } w\alpha_i\in\Phi^+,\\
|N(w)|-1, & \text{if } w\alpha_i\in-\Phi^+.
\end{cases}
\end{align*}
[guided]
Fix $w\in W$ and $i\in I$. We want to understand exactly which positive roots become negative after applying $ws_i$. By definition,
\begin{align*}
N(ws_i)=\{\beta\in\Phi^+ : ws_i\beta\in -\Phi^+\}.
\end{align*}
There is one exceptional positive root for $s_i$, namely $\alpha_i$. Since $s_i\alpha_i=-\alpha_i$, we compute
\begin{align*}
ws_i\alpha_i=-w\alpha_i.
\end{align*}
Thus $\alpha_i$ is an inversion of $ws_i$ precisely when $-w\alpha_i$ is negative, which is the same as saying that $w\alpha_i$ is positive:
\begin{align*}
\alpha_i\in N(ws_i)
\quad\Longleftrightarrow\quad
w\alpha_i\in \Phi^+.
\end{align*}
For every other positive root, the simple reflection $s_i$ does not leave the positive system. Let $\beta\in\Phi^+\setminus\{\alpha_i\}$ and define $\gamma:=s_i\beta$. Then $\gamma\in\Phi^+\setminus\{\alpha_i\}$, and $s_i^2=1$ gives a bijective correspondence between such $\beta$ and such $\gamma$. Under this correspondence,
\begin{align*}
\beta\in N(ws_i)
&\Longleftrightarrow ws_i\beta\in -\Phi^+ \\
&\Longleftrightarrow w\gamma\in -\Phi^+ \\
&\Longleftrightarrow \gamma\in N(w).
\end{align*}
So all non-exceptional inversions of $ws_i$ are exactly the $s_i$-translates of inversions of $w$, except that the root $\alpha_i$ must be handled separately. If $w\alpha_i\in\Phi^+$, then $\alpha_i\notin N(w)$ and right multiplication by $s_i$ adds the new inversion $\alpha_i$:
\begin{align*}
N(ws_i)=s_iN(w)\cup\{\alpha_i\}.
\end{align*}
If $w\alpha_i\in-\Phi^+$, then $\alpha_i\in N(w)$ and right multiplication by $s_i$ removes that inversion:
\begin{align*}
N(ws_i)=s_i\bigl(N(w)\setminus\{\alpha_i\}\bigr).
\end{align*}
Since $s_i$ is a bijection on $\Phi$, these formulas give the corresponding cardinality relation whenever $N(w)$ is finite:
\begin{align*}
|N(ws_i)|=
\begin{cases}
|N(w)|+1, & \text{if } w\alpha_i\in\Phi^+,\\
|N(w)|-1, & \text{if } w\alpha_i\in-\Phi^+.
\end{cases}
\end{align*}
[/guided]
[/step]
[step:Apply the one-step formula along a reduced expression]
Let $w\in W$, and choose a reduced expression
\begin{align*}
w=s_{i_1}s_{i_2}\cdots s_{i_k},
\end{align*}
where $k=\ell(w)$. For each $m\in\{0,1,\dots,k\}$, define
\begin{align*}
w_m:=s_{i_1}s_{i_2}\cdots s_{i_m},
\end{align*}
with $w_0=e$, the identity element of $W$.
The expression for $w_m$ is reduced for every $m$, so
\begin{align*}
\ell(w_m)=m.
\end{align*}
We now use the standard root-exchange lemma for reduced words, stated independently of inversion cardinalities: if $s_{j_1}\cdots s_{j_r}$ is reduced, then for every $q\in\{1,\dots,r\}$ the root
\begin{align*}
s_{j_1}\cdots s_{j_{q-1}}\alpha_{j_q}
\end{align*}
belongs to $\Phi^+$. Applying this lemma to the reduced word $w_m=w_{m-1}s_{i_m}$ with $q=m$ gives
\begin{align*}
w_{m-1}\alpha_{i_m}\in\Phi^+.
\end{align*}
The one-step formula from the previous step therefore gives
\begin{align*}
|N(w_m)|=|N(w_{m-1})|+1
\end{align*}
for every $m\in\{1,\dots,k\}$. Since $N(e)=\varnothing$, induction on $m$ gives
\begin{align*}
|N(w_m)|=m
\end{align*}
for every $m\in\{0,1,\dots,k\}$. Taking $m=k$ yields
\begin{align*}
|N(w)|=k=\ell(w).
\end{align*}
[guided]
Choose a reduced expression
\begin{align*}
w=s_{i_1}s_{i_2}\cdots s_{i_k},
\end{align*}
so that $k=\ell(w)$. Define the partial products
\begin{align*}
w_m:=s_{i_1}s_{i_2}\cdots s_{i_m}
\end{align*}
for $m\in\{0,1,\dots,k\}$, with $w_0=e$.
Because the full expression is reduced, every initial segment is reduced. Hence
\begin{align*}
\ell(w_m)=m.
\end{align*}
At the $m$-th stage, we pass from $w_{m-1}$ to $w_m=w_{m-1}s_{i_m}$. We use the standard root-exchange lemma for reduced words: if $s_{j_1}\cdots s_{j_r}$ is reduced, then each root
\begin{align*}
s_{j_1}\cdots s_{j_{q-1}}\alpha_{j_q}
\end{align*}
with $q\in\{1,\dots,r\}$ lies in $\Phi^+$. This lemma is a statement about the root sequence attached to a reduced word, not about the cardinality of inversion sets. Applying it to the reduced word $w_m=w_{m-1}s_{i_m}$ at the last position gives
\begin{align*}
w_{m-1}\alpha_{i_m}\in\Phi^+.
\end{align*}
This is the precise point where reducedness is used: it says that the next simple reflection crosses a positive root hyperplane not already crossed before.
Now the one-step inversion formula applies with $w=w_{m-1}$ and $i=i_m$. Since $w_{m-1}\alpha_{i_m}$ is positive, the formula says that exactly one new inversion is added:
\begin{align*}
|N(w_m)|=|N(w_{m-1})|+1.
\end{align*}
The base case is the identity element. Since $e\alpha=\alpha$ for every $\alpha\in\Phi^+$, no positive root is sent to a negative root, so
\begin{align*}
N(e)=\varnothing.
\end{align*}
Therefore $|N(w_0)|=0$, and induction gives
\begin{align*}
|N(w_m)|=m
\end{align*}
for all $m\in\{0,1,\dots,k\}$. In particular,
\begin{align*}
|N(w)|=|N(w_k)|=k=\ell(w).
\end{align*}
[/guided]
[/step]
[step:Derive the two right descent criteria]
Fix $w\in W$ and $i\in I$. Since $s_i$ is a generator of order two, right multiplication by $s_i$ changes Coxeter length by exactly one:
\begin{align*}
\ell(ws_i)\in\{\ell(w)+1,\ell(w)-1\}.
\end{align*}
Using the equality already proved for both $w$ and $ws_i$, the one-step inversion formula gives
\begin{align*}
\ell(ws_i)=|N(ws_i)|=
\begin{cases}
|N(w)|+1, & \text{if } w\alpha_i\in\Phi^+,\\
|N(w)|-1, & \text{if } w\alpha_i\in-\Phi^+.
\end{cases}
\end{align*}
Since $|N(w)|=\ell(w)$, this becomes
\begin{align*}
\ell(ws_i)=
\begin{cases}
\ell(w)+1, & \text{if } w\alpha_i\in\Phi^+,\\
\ell(w)-1, & \text{if } w\alpha_i\in-\Phi^+.
\end{cases}
\end{align*}
Because every root lies in exactly one of $\Phi^+$ and $-\Phi^+$, the two equivalences follow:
\begin{align*}
\ell(ws_i)=\ell(w)+1 \quad \Longleftrightarrow \quad w\alpha_i\in\Phi^+,
\end{align*}
and
\begin{align*}
\ell(ws_i)=\ell(w)-1 \quad \Longleftrightarrow \quad w\alpha_i\in-\Phi^+.
\end{align*}
This completes the proof.
[/step]
