How is equation 1 simplified to equation 2 as shown below In CLRS (Intro to algorithms) on page 362, it says eqn(1) :

can be simplified to this equation(2):

I would like to know how this simplification was arrived at. It shouldn't necessarily be a formal proof but one that makes logical and intuitive sense. (General arguments or assumptions that lead to this simplification are also welcome). Any effort is much appreciated. (Thanks)
Update
I want to know how the second equation is valid if we are not considering the optimal revenues on both sides but only the optimal revenue of the right-hand piece as suggested by the second equation. I.e. in the second equation, we use the price of the left piece and then add it to the optimal revenue of the right-hand piece.
I feel like this is basically paraphrasing the explantion in CLRS but maybe it will help.
Let $j$ be the length of the leftmost rod piece in an optimal solution. This piece has a cost of $p_j$ and so it contributes exactly that much to the total revenue. It is easy to see that the remaining part of the rod (of length $n-j$) must be divided up in an optimal manner as well for the whole thing to be optimal
(and thus leads to a revenue of $r_{n-j}$).
Thus, we have $r_n = p_j +r_{n-j} $. Because $r_n$ is the optimal revenue, $r_n \geq \max_{1\leq i \leq n}\{p_i+r_{n-i}\}$. But we also have $r_n = p_j +r_{n-j} \leq \max_{1\leq i\leq n}\{p_i+r_{n-i}\}$.
Putting both ineqalities together gives $r_n = \max_{1\leq i \leq n}\{p_i+r_{n-i}\}$.