Deducing Farkas' lemma from its convex hull formulation (full details inside) I want to prove the converse direction of this question: Linear algebra classic, Farkas lemma application. To quote explicitly:
Assume:

For every $x_1,x_2,\dots,x_m \in \mathbb{R}^n$, exactly one of the following holds:
(a) $0 \in$ convex.hull$\{ x_1,x_2,\dots,x_m \}$
(b) there is $y \in \mathbb{R}^n$ such that $\langle y,x_i \rangle > 0$ for all $i$.

I want to prove:

$A \in M_{m \times n}(\mathbb{R})$ and $b \in \mathbb{R}^m$.
  Exactly one of the following holds:
(c) there exists some $x \in \mathbb{R}^n$, $x \geq 0$, such that $Ax = b$
(d) there exists some vector $y \in \mathbb{R}^m$ such that $y^TA \geq 0$ and $y^Tb < 0$.


Approach
Define $x_1,...,x_{n+1}$ to be the columns of $A$ and $-b$. 
if (b) holds then we have $A^T y >0$ and $y^Tb<0$ so (d) holds, even with strict positiveness.
So assume (a), and we want to prove that (c) or (d) hold.
This suffices since easily we cannot have both (c) and (d) instantaneously.
We have $p_1,...,p_{n+1} \geq 0$ with sum $1$ such that $A \begin{pmatrix}
           p_{1} \\
           p_{2} \\
           \vdots \\
           p_{n}
         \end{pmatrix}=p_{n+1}b$. 
If $p_{n+1} \neq 0$ then we divide by it and get (c), so assume $p_{n+1}=0$.
Thus we have some probability vector $p \geq 0$, $p_1+...+p_n = 1$ with $Ap = 0$. Here I'm stuck: how can I deduce (c) or (d)?
 A: Notice that for $y=(y_1,\dots,y_n)$ we have that
$$A^Ty = (\langle x_1, y\rangle, \dots, \langle x_n, y\rangle).$$
Hence, in order for $(d)$ to hold we must show that there is some $y\in\mathbb R^n$ with $\langle b, y\rangle< 0$ and $\langle x_i, y\rangle \geq 0$ for all $i$.
Every equation $\langle x_i, y\rangle \geq 0$ defines a halfspace $H_i$ whose boundary is a hyperplane through the origin, given by $\langle x_i, y\rangle = 0$.
If $(b)$ does not hold, it means that the intersection of the open halfspaces $\text{int}(H_i)$ with the open halfspace given by $\langle b, y\rangle > 0$ is empty.
The intersection of $n$ closed halfspaces through the origin is either a half cone or a line.
If it's a half cone $C$, the existence of $y$ with $\langle b,y\rangle <0$ is clear from the fact that $\langle b, y\rangle > 0$ does not intersect $C$.
Finally, a line is only possible if at least two of the halfspaces share boundaries -- in other words, if $x_i=-\alpha\,x_j$ for some $i \neq j$ and $\alpha >0$.
But this would mean $(a)$ holds, which contradicts our hypotheses.
