For $n\ge 3$. Let $s\_1\cdots s\_n$ be a reduced expression of $x$. Suppose $s\_1\cdots s\_{n-1}\le w$ and $s\_2\cdots s\_{n}\le w$. 



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> Does this imply $x\le w$?
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Reduced expression and Bruhat order
Let $W = S\_5$ (as a side note, we could let $W = S\_3 \times S\_2$). Let $w = (132)(45), x = (123)(45) \in W$; we write $x = (12)(45)(23)$ as a reduced expression. Then the subexpressions written in the question are $(12)(45)$ and $(45)(23) = (23)(45)$. The reduced expressions for $w$ are $(23)(12)(45), (23)(45)(12), (45)(23)(12)$; for each of those, the appearance of both $(12)(45) = (45)(12), (45)(23) = (23)(45)$ as outputs for subexpressions is clear, so $(12)(45), (23)(45) \leq w$, satisfying the hypothesis of the theorem, but $x \nleq w$, so the answer to the question is **no**.


