Consider the following system with the vector of unknowns $x\equiv (x_{y,v}: y\in \{0,1\}, v\in \mathcal{V})$, where $\mathcal{V}$ is finite and $|\mathcal{V}|\geq 2$. $$ \begin{aligned} & (1) \text{ }\sum_{v\in \mathcal{V}} x_{y,v}=q_y \text{ }\forall y\in \{0,1\} \\ & (2) \text{ }\sum_{v\in \mathcal{V}} x_{1,v}(z_{1,v}-z_{0,v})\geq 0 \\ & (3) \text{ }\sum_{v\in \mathcal{V}} x_{0,v}(z_{1,v}-z_{0,v})\leq 0\\ & (4) \text{ }\sum_{y\in \{0,1\}, v\in \mathcal{V}} x_{y,v}=1 \\ & (5) \text{ }0\leq x_{y,v}\leq 1 \text{ }\forall y\in \{0,1\}, \forall v \in \mathcal{V} \end{aligned} $$
We also know that $q_0+q_1=1$, $q_0\geq 0$, $q_1\geq 0$. I would like to show that the system admits a solution for any $q_0, q_1, (z_{1,v}, z_{0,v}: v\in \mathcal{V})$. Can you help?
How far I went: Observe that the upper bound in (5) is implied by (4) and the lower bound in (5). Also, (4) is implied by (2). Hence, the problem reduces to: $$ \begin{aligned} &x_0 1_{|\mathcal{V}|}=q_0 \\ &x_1 1_{|\mathcal{V}|}=q_1 \\ &x_1 z^\top \geq 0 \\ & x_0 z^\top \leq 0 \\ &x_0\geq 0 \\ & x_1\geq 0 \end{aligned} $$ where $$ \begin{aligned} &x_{y}\equiv (x_{y,v}: v \in \mathcal{V})\\ &z_{y}\equiv (z_{y,v}: v \in \mathcal{V})\\ &z\equiv z_1-z_0\\ &1_{|\mathcal{V}|}\text{ is a column vector of ones of size $|\mathcal{V}|\times 1$.} \end{aligned} $$
How can I proceed from here?