Let $G$ be a group generated by two subgroups $B$ and $N$, with $T := B \cap N$ normal in $N$. Let $W = T/N$, and suppose $S \subseteq W$ is a set consisting of elements of order two which generates $W$. Also suppose that for $\sigma \in W, \rho \in S$,
$$\rho B \sigma \subseteq B\sigma B \space \cup B\rho \sigma B $$
and that $\rho B \rho \neq B$. Then the data $(G,B,N,S)$ is called a Tits system.
The length $\ell$ of an element in $W$ is its shortest expression as a product of elements in $S$. If $I \subseteq S$, $W_I$ denotes the subgroup of $W$ generated by $I$, and $P_I$ denotes the product set $BW_IB$. Using the axioms above, you can show that $P_I$ is actually a subgroup of $G$.
I'm trying to understand the proof of this lemma in Humphreys, Linear Algebraic Groups (p. 176):
It is clear that the subgroup generated by $B$ and $\sigma B \sigma^{-1}$ is contained in the one generated by $B$ and $\sigma$, and that both of these groups are contained in $P_I$. The lemma asserts that these three groups are actually equal.
What I don't understand is the claim: "Combining these two steps, we conclude that $P_I = BW_IB$ is generated by $B$ and $\sigma B \sigma^{-1}$."
Since $\rho_1 \in \langle B, \sigma B \sigma^{-1} \rangle$ by part (b) of Lemma A, we see that $$\langle B, \rho_1 \sigma B \sigma^{-1} \rho_1 \rangle \subseteq \langle B, \sigma B\sigma^{-1} \rangle$$
because a generator of the form $\rho_1 \sigma b \sigma^{-1} \rho_1$ is obtained by combining the elements $\rho_1, \sigma b \sigma^{-1}$ of the latter group. Now also $\rho_2, ... , \rho_k \in \langle B, \rho_1 \sigma B \sigma^{-1} \rho_1 \rangle \subseteq \langle B, \sigma B\sigma^{-1} \rangle$, so $$P_I \subseteq \langle B, \sigma B\sigma^{-1} \rangle \subseteq P_I$$
and everything is equal.