Partial ordering on $\mathfrak{h}^*$ and Bruhat ordering In section 5.2 (p.95) of Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$.
Let $\mu\le \lambda$ if $\lambda-\mu\in \Gamma$, where $\Gamma$ is the set of $\mathbb{Z}^{\ge 0}$-linear combinations of simple roots.
Let $\lambda$ be a regular, integral, antidominant weight.
It is easy to show $(s_\alpha w) \cdot\lambda < w \cdot \lambda \iff s_\alpha w < w$ in the Bruhat ordering.
According to Humphreys, we can show $w'\cdot\lambda<w\cdot\lambda\iff w'<w$ by iteration.


How to do iteration in order to show $w'\cdot\lambda<w\cdot\lambda\implies w'<w$?
Or is this a typo? Any conterexample if it is a typo?

 A: This question was answered correctly by Sam Hopkins. Let me just abuse this space for answers for pointing out that James Humphreys keeps up to date corrections on the AMS webpage of the book. Direct link to pdf of the errata.
(In this particular case he says that one should replace both occurences of $<$ symbol with  $\uparrow$ and $\leq$ respectively.)
A: I think the implication you are asking about might not in fact hold.
If my computations are correct, we can see this already for $S_4$.
Let me ignore the dot action part of the question (I don't think that matters) and try to simplify it as follows: let $\lambda$ be any regular anti-dominant weight, and define a partial order $\preceq$ on $W$ by $u \preceq v$ if and only if $u\lambda \leq v\lambda$ (this is the usual partial order on weights with $\mu \leq \lambda$ if and only if $\lambda-\mu=\sum_{i}c_i\alpha_i$ with $c_i\in\mathbb{Z}_{\geq 0}$). The question becomes: is $\preceq$ the same partial order as $\leq$, the usual (strong) Bruhat order?
My answer is that, no, they are not the same and this can be seen already for $S_4$. For instance, let us take $\lambda =(1,2,3,4)$ as our regular anti-dominant weight so that $w\lambda$ is just the one-line notation of the permutation $w \in S_4$. Then I claim that $(1,4,2,3) \preceq (2,3,4,1)$ but  $(1,4,2,3)\not \leq (2,3,4,1)$. That $(2,3,4,1)-(1,4,2,3)= 1*(1,-1,0,0) + 2*(0,0,1,-1)$ shows $(1,4,2,3) \preceq (2,3,4,1)$. And my computer tells me that $(1,4,2,3)\not \leq (2,3,4,1)$ (actually I think that this is also easy to see since we cannot ever move the $4$ rightward in $(1,4,2,3)$ when going up in Bruhat order).
EDIT:
Regarding the dot action aspect of the question, here is a larger portion of the text in question which shows the notions of regular and antidominant are meant relative to the dot action:

