I've been confused about the proof of this in the textbook. The proof so far is as follows:
"If $G$ is a forest, then $q = p - c \le p - 1$, and since $p \ge 3$, then $p - 1 \lt 3p - 6$ and the result holds.
Now assume that $G$ is not a forest and has a cycle of length at least 3. Consider an embedding of the plane of $G$ and call its $r$ regions
{$r_1, r_2, . . . , r_r$}.
Let $b_i$ be the number of edges incident to $r_i$, up to multiplicity. (If $r_i$ is on both sides of an edge, that edge is counted twice.) Since $G$ is not a forest, every region is bounded by a closed walk. Since $G$ is simple, we have $b_i ≥ 3$ for every $i$. Since each edge is counted twice, we have $\sum_{i=1}^r b_i = 2q$. It follows that $3r \le \sum_{i=1}^r b_i = 2q$. By Euler’s Formula, we have $r = p − q + r = q − p + c + 1 \ge q − p + 2$. Combining these, we get $3(q − p + 2) \le 3r \le 2q$ and thus $q \le 3p − 6$."
The part that confuses me the most is how we ended up getting $r=p-q+r$, because it then implies that $p=q$ which isn't necessarily true. Can someone please explain this to me?
Thank you in advance.
I think that is simply a typographical error. It is certainly not true that $r=p-q+r$. On the other hand, (I presume) you have learnt Euler's formula $p-q+r=c+1$ where there are $p$ vertices, $q$ edges, $r$ regions and $c$ components in a planar drawing. It follows that $r=q-p+c+1$.
So just omit the middle equality and proceed.