# In depth reduction of arithmetic formula why we get a $v$ st $\frac{s}3\leq |\Phi_v|\leq \frac{2s}{3}$

I am reading Depth Reduction of Arithmetic Formula form the survey of Ramprasad Saptharishi. Now in the proof of depth reduction due to Brent, 74 that

Let $$f$$ be an n-variate degree d polynomial computed by an arithmetic formula $$\Phi$$ of size s. Then, $$f$$ can also be computed by a formula $$\Phi^{\prime}$$ of size $$s^{\prime}=\operatorname{poly}(s, n, d)$$ and depth $$O(\log s)$$

In the proof they have written that

Consider the first node in this path $$V$$ such that the size of the formula rooted at $$v$$ is smaller than $$\frac{2s}{3}$$. Let $$\Phi_v$$ refer to the sub-formula rooted at $$V$$. By the choice of the path from the root, we have $$\frac{s}3\leq |\Phi_v|\leq \frac{2s}{3}$$

Now how are we getting the $$\frac{s}3\leq |\Phi_v|$$ inequality?

You do not cite the part of the survey that is actually relevant for getting the $$s/3$$ lower bound:
Starting from the root, walk down to the leaves by always taking the child with a larger sub-tree under it. Consider the first node in this path $$v$$ such that the size of the formula rooted at v is smaller than $$2s/3$$.
$$v$$ cannot be the root since the subtree rooted at the root has size $$s$$. Let $$w$$ be the predecessor of $$v$$. By definition, the subtree rooted at $$w$$ contains more than $$2s/3$$ nodes. Since it has most two children and $$v$$ is the child with the largest subtree under it, it has more than half the nodes under $$w$$, that is, more than $$s/3$$.
• See also Emil's linked thread since indeed, you may not only have a value precisely between $s/3$ and $2s/3$ but between $\lceil s/3 \rceil$ and $\lceil 2s/3 \rceil$.