Unclear relation in the number of permutations consistent with Hasse diagrams

I have been reading the paper 'Time Space Tradeoff for Sorting on Non-Oblivious Machines' by Borodin et al. (Link). Lemma 1 in that paper gives a relation between the number of permutations consistent with a Hasse diagram $$H$$ and the number of permutations that are $$i$$-consistent with $$H$$.

Let me explain for self containment. Let $$H$$ be a Hasse diagram on $$n$$ vertices. A permutation $$\pi$$ on $$(1,2,\cdots, n)$$ is said to be consistent with $$H$$ if $$\pi(i)>\pi(j)$$ whenever there is a path of positive length from $$i$$ to $$j$$ in $$H$$. Two elements $$j$$ and $$k$$ are said to be comparable if there exists some path from $$j$$ to $$k$$ or $$k$$ to $$j$$ in $$H$$. Now, define the following.

1. $$P(H)$$ is the set of all permutations that are consistent with $$H$$.
2. $$C(H,i)$$ is the set of all elements that are comparable to $$i$$ in $$H$$.
3. $$H-i$$ is the Hasse diagram on $$n-1$$ vertices after removing $$i$$ from $$H$$ and adding an edge from each parent of $$i$$ in $$H$$ to all children of $$i$$ in $$H$$.

Lemma 1 states that $$n\cdot |P(H-i)| \le |C(H,i)|\cdot |P(H)|$$ for any Hasse diagram $$H$$ on $$n$$ vertices and has a vertex labelled $$i$$. In order to understand the proof, I tried to take some examples.

Consider the Hasse diagram on $$3$$ vertices that has only two edges. The edges are $$1\rightarrow 2$$ and $$1\rightarrow 3$$. Say $$i=1$$. Now, we can get the following

1. $$P(H) = \{ \{312\}, \{321\} \}$$ where $$\{abc\}$$ denotes the permutation $$\pi$$ such that $$\pi(1)=a, \pi(2)=b, \pi(3)=c$$.
2. $$C(H,i) = \{2,3\}$$.
3. $$P(H-i) = \{\{123\},\{132\},\{213\},\{231\},\{312\},\{321\}\}$$.

Clearly, $$n\cdot |P(H-i)| = 3\cdot 6 > 2\cdot 2 = |C(H,i)|\cdot |P(H)|$$. I tried the same for a few other examples, and I ended up in a similar situation.

So, is the lemma incorrect or am I interpreting the lemma incorrectly?