# Prove if $x,y$ are vectors and $A$ is a matrix then $\exists x$ st $x \le 0$ and $Ax \le b$ xor $\exists y$ st $A^Ty\ge 0, y\ge 0$ and $by<0$

I'm supposed to use the fundamental theorem of LP, weak duality and strong duality to prove this.

Fundamental Theorem of LP:
1) If there's a feasible solution there's a basic feasible solution
2) If there's an optimal solution there's a basic optimal solution

Weak Duality:
If x is a solution to primal constraints:
$$Ax\le b$$ and $$x\ge 0$$
and y is a solution to the dual
$$A^Ty\ge c$$
$$y\ge 0$$
then $$cx\le by$$

Strong Duality:
Above conditions met and x and y are optimal solutions then
$$cx= by$$

I'm still not entirely sure what's the difference between a "feasible solution" is and a "basic feasible solution." So if someone can explain that, that'd be very helpful.

Using the above I have to prove $$x,y$$ are vectors and $$A$$ is a matrix then

$$\exists x$$ st $$x \le 0$$ and $$Ax \le b$$ or $$\exists y$$ st $$A^Ty\ge 0, y\ge 0$$ and $$by<0$$, but not both

This seems very similar to Farkas lemma, except $$Ax=b$$ is a requirement in Farkas so it's not exactly the same.