I have the following relationship for the primal and dual problems listed below.
$Max: c^{T}x $
$s.t: Ax\leq-c $
$x\geq0$ where $c \in \mathbb{R}^{m}$ and $A^{T} = -A$.
I have formulated the dual as the following:
$Min: -c^{T}y $
$s.t: -Ay\geq c $
$y\geq0$
I am trying to show that there are only two possibilities to this system. Either the Primal and Dual are infeasible or the optimal value is 0. I have tried to use the following relationship: $Ax+c\leq 0$ (from primal) and $c-Ay\geq 0$ (from dual) then $Ax+c \leq c-Ay$ which gives $Ax\leq -Ay$. Suppose that $x >0$ then this would imply that $y<0$ which cannot be true since the dual defines $y\geq0$, and the same vice versa. If an optimal solution exists it must be 0. Is this a valid argument for the two possibilities?