How to prove strict complementary slackness by means of Farkas' lemma

This question relates to pages 89 and 96 the following text (pages 102 and 109 of the pdf):

https://promathmedia.files.wordpress.com/2013/10/alexander_schrijver_theory_of_linear_and_integerbookfi-org.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.