I would like your help to show that the system below has at least one solution.
Let $\mathcal{Y}\equiv \{0,1\}$. Let $\mathcal{V}$ be a finite set containing positive and negative numbers.
Consider the system of equations/inqualities below. The vector of unknowns is $(x_{y,v}: y\in \mathcal{Y}, v\in \mathcal{V})$. We know the vectors $(w_v: v\in \mathcal{V})$, $(q_y: y\in \mathcal{Y})$, and $(z_{y,v}: y\in \mathcal{Y}, v\in \mathcal{V})$, and the sets $\mathcal{V},\mathcal{Y}$. $$ (*) \quad \begin{cases} &(1) \quad \sum_{y\in \mathcal{Y}}x_{y,v} =w_v \quad \forall v \in \mathcal{V},\\ &(2) \quad \sum_{v\in \mathcal{V}} x_{y,v}=q_y\quad \forall y\in \mathcal{Y},\\ & -----------------------\\ &(3) \quad \sum_{v\in \mathcal{V}} x_{1,v} *z_{1,v} \geq \sum_{v\in \mathcal{V}} x_{1,v} *z_{0,v},\\ &(4) \quad\sum_{v\in \mathcal{V}} x_{0,v} *z_{0,v} \geq \sum_{v\in \mathcal{V}} x_{0,v} *z_{1,v} ,\\ &--------------------\\ &(5) \quad \sum_{y\in \mathcal{Y},v\in \mathcal{V}} x_{y,v}=1,\\ &(6) \quad 0\leq x_{y,v}\leq 1 \quad \forall y\in \mathcal{Y}, v\in \mathcal{V},\\ &(7) \quad \sum_{v\in \mathcal{V} } w_v=1,\\ &(8) \quad 0\leq w_v\leq 1 \quad \forall v\in \mathcal{V},\\ &(9) \quad \sum_{y\in \mathcal{Y} } q_y=1,\\ &(10) \quad 0\leq q_y\leq 1 \quad \forall y\in \mathcal{Y}.\\ \end{cases} $$
Question: Show that $(*)$ may not have a solution.
Note: I've posed a similar question here for the case where $z_{y,v}\equiv y*v$ (hence, constraints (3) and (4) look simpler there). Can the same counterexample be extended to this more general setting?