p. 66 — total idle time of machine 3 is $\max_{1 \le u \le v \le n}(H_v + K_u)$

Let $A_i, B_i, C_i > 0$, let $n \ge 1$, and let $\sigma$ be any ordering of the items. With $K_u$ and $H_v$ the quantities of $\sigma$ (positions $1, \dots, n$ in the paper's indexing), the total elapsed time of the as-soon-as-possible three-machine schedule of $\sigma$ is

Equivalently, since machine 3 is busy for $\sum_i C_i$ and idle for $\sum_i Y_i$, the total idle time of machine 3 is $\sum_{i=1}^{n} Y_i = \max_{1 \le u \le v \le n} (H_v + K_u)$, the display of p. 66. On p. 68 Johnson restates it: $\max_{u \le v \le n}(K_u + H_v + \sum_{i=1}^{n} C_i)$ "is the maximum sum of elements passed through on all 'walks' in the time matrix from the upper left-hand corner to the lower right-hand corner, taking steps to the right or downward. The problem is to find a scheduling of items which minimizes this maximum walk."

This turns the three-stage problem into the minimization of an explicit function of the ordering.

Formalization Note The statement is in elapsed-time form. The maximum over pairs is written as $\max_v \max_{u \le v}$ over 0-based positions. It holds for every ordering; the special-case hypothesis $\min A \ge \max B$ is not assumed.