This is a minor detail in a paper I'm reading. Let $W$ be a Weyl group with simple roots $\Pi$, and suppose $J\subseteq\Pi$. Let $w_I$ denote the longest element of the parabolic subgroup $W_I$ for $I\subseteq\Pi$. For any $\alpha\in\Pi-J$, the following set containment is known to be true: $$ K:=w_{J\cup\{\alpha\}}w_J(J)\subseteq J\cup\{\alpha\}. $$
So you can pick $\beta\in\Pi$ such that $K\cup\{\beta\}=J\cup\{\alpha\}$. Why then does it follow that $w_{K\cup\{\beta\}}w_K=(w_{J\cup\{\alpha\}}w_J)^{-1}$?
My thinking is this. Let $w:=w_{K\cup\{\beta\}}=w_{J\cup\{\alpha\}}$. Then the question can be rephrased as why $ww_K=w_Jw$, or why $ww_Jw=w_K$. But $ww_Jw(K)=ww_Jw_J(J)=wJ$, and $wJ$ consists entirely of negative roots since $w$ is the longest element of $W_{J\cup\{\alpha\}}\supset W_J$. So $ww_Jw$ turns every simple root in $K$ negative, and hence every positive root in $\Phi_K$ negative. Hence if $ww_Jw\in W_K$, then $ww_Jw=w_K$, but I'm not sure if $ww_Jw\in W_K$ in the first place.