Nodes of height h in an n-element heap

This is exercise 6.3-3 of CLRS version 2.

Question:Show that there are at most $\lceil n/2^{h+1} \rceil$ nodes of height $h$ in any $n$-element heap.

Solution: First some facts

1. According to exercise 6.1-7, an $n$-element heap has exactly $\lceil n/2 \rceil$ leaves.
2. Notice that the nodes with height $i$ become leaves after deleting all the nodes with height $0,\cdots, i-1$.

Let $y_i$ be the total number of elements of the new tree after deleting all the nodes with height $0,\cdots, i-1$, and let $x_i$ be the number of leaves of the new tree. Then we have $x_i = \lceil y_i/2 \rceil$ by fact #1 and $y_{i+1}=y_i - x_i$ by fact #2. Thus we have $y_{i+1} \leq y_i/2$, and this leads to
$$y_i \leq n/2^i, \quad \forall n=0,1,\cdots.$$
The final conclusion follows from the relation $x_i = \lceil y_i/2 \rceil$.