This question relates to pages 89 and 96 the following text (pages 102 and 109 of the pdf):
The author gives the following variant of Farkas' lemma:
Let $A$ be a matrix and $b$ be a vector. Then the system of linear inequalities $Ax \le b$ has a solution $x$, if and only if $yb \ge 0$ for each row vector $y \ge 0$ with $yA = 0$.
He then proves that:
for a bounded and feasible linear program
$$\max \{ cx \mid Ax \le b \} = \min \{ yb \mid y \ge 0, yA = c \} $$
if the maximum problem has no optimum $x_0$ with $a_ix_0 \lt b_i$ then the minimum problem must have an optimum $y_0$ with a positive $\mathit{i}$th component.
The proof starts as follows:
It is assumed that there is no optimum solution $x_0$ for the maximum problem with $a_ix_0 \lt b_i$. Then, if $\delta$ is the optimum value of the maximum and minimum problems, $Ax \le b, cx \ge \delta$ implies $a_ix \ge \beta_i$. So, by Corollary 7.1e (Farkas' lemma as given above), $yA - \lambda c = -a_i$ and $yb - \lambda \delta \le -\beta_i$ for some $y, \lambda \ge 0$.
My question is:
How does the last sentence ("So") follow from what preceeds it?
Specifically, how can "$Ax \le b, cx \ge \delta$ implies $a_ix \ge \beta_i$" be cast in a form such that application of Farkas' lemma yields $yA - \lambda c = -a_i$ and $yb - \lambda \delta \le -\beta_i$ for some $y, \lambda \ge 0$?
Neither the positive form ("$Ax \le b, -cx \le -\delta, -a_ix \le -\beta_i$ has a solution") nor the negative form ("$Ax \le b, -cx \le -\delta, a_ix \le \beta'_i$, with $\beta'_i \lt \beta_i$, has no solution") yields exactly the desired inequalities when Farkas' lemma is applied. For example, signs or inequality directions differ from those sought.
Any help would be greatly appreciated.