# Cannot Understand solution: Inconsistent systems of linear inequalities proof.

I'm trying to understand the solution to this question in Bertsimas 4.29: Question: Let $$a_1,....a_m$$ be some vectors in $$R^n$$ with $$m>n+1$$. Suppose that the system of inequalities, $$a_i'x \geq b_i,i=1,....m$$ does not have any solutions. Show that we can choose $$n+1$$ of these inequalities, so that the resulting system of inequalities has no solutions.

The solution given is: Questions:

Questions:

1. Why is the polyhedron, $$P$$ considered with the transposes $$A^Tp=0, b^Tp=1$$ as opposed to $$p^TA=0^T$$

2. Why does theorem 4.7 imply that if $$x$$ is a solution to the subsystem we must have $$0^T x=0 \leq -1$$? I also don't get how this is implied by $$\hat{p}^T A=0^T, \hat{p}^T b=1, p \geq 0$$. Theorem 4.7 is given as:

Where does the $$-1$$ come from in the inequality $$0 \leq -1$$ is my main question. Thank you.

Notice that $$p^TA = 0^T \iff A^Tp=0$$, these two conditions are equivalent, so it doesn't matter if we consider its transpose or not.
To use theorem $$4.7$$,
Notice that $$\hat{p}^T(-A)=0^T$$, here in theorem $$4.7$$, $$c=0$$. We also have $$p\ge 0$$, and $$\hat{p}^T(-b) =-1 \le -1$$, here $$d=-1$$. That is part $$(b)$$ of condition $$4.7$$ holds, then we conclude that result for part $$(a)$$ holds
If $$\hat{x}$$ is a solution to the subsystem, then $$(-A)x\le (-b)$$, hence we have $$c'x=0 \le d = -1$$ which is a contradiction.