\omega} v_{\Lambda_0} \quad \textrm{and} \\
& \label{vakuum3} x_{\tau}(-1) x_{\delta}(-1)x_{\gamma}(-1) v_{\Lambda_0} = 0\end{aligned}$$ for $\gamma, \delta, \tau \in \Gamma.$

Let us show relation . We first analyse $\mathcal{Y}(e^{\gamma},z)v_{\Lambda_0}$. Since $\langle \gamma, \omega_0 \rangle = 0$, formula implies that minimal $j$ such that $x_{\gamma}(-j)v_{\Lambda_0} \neq 0$ is $j=1$. We get: $$\begin{aligned}
x_{\gamma}(-1) v_{\Lambda_0} = C \cdot e^{\gamma + \omega_0} = C \cdot e^{\gamma}.\end{aligned}$$ Now we observe $\mathcal{Y}(e^{\delta},z)e^{\gamma}$ for $\delta \neq \underline{\gamma}$. Since $\langle \delta, \gamma \rangle \geq 1$, minimal $j$ such that $x_{\delta}(-j)e^{\gamma} \neq 0$ is $j=2$. Therefore we have $x_{\delta}(-1)e^{\gamma} = 0$, i.e. $x_{\delta}(-1)x_{\gamma}(-1) v_{\Lambda_0} = 0 $.

If, on the other hand, $\delta = \underline{\gamma}$, then $\langle \underline{\gamma}, \gamma \rangle = 0$ and minimal $j$ such that $x_{\underline{\gamma}}(-j)e^{\gamma} \neq 0$ is $j=1$. We have $$x_{\underline{\gamma}}(-1)x_{\gamma}(-1) v_{\Lambda_0} = C \cdot x_{\underline{\gamma}}(-1)e^{\gamma} = C' \cdot e^{\gamma + \underline{\gamma}} = C' \cdot e^{2 \omega} v_{\Lambda_0}$$ i.e. we got the relation . Since $x_{\delta}(-1)x_{\gamma}(-1) v_{\Lambda_0} = 0$ if $\delta \neq \underline{\gamma}$, the only claim left to prove is the relation $$x_{\tau}(-1) x_{\underline{\gamma}}(-1) x_{\gamma}(-1) v_{\Lambda_0} = 0.$$ Formula implies again that minimal $j$ such that $x_{\tau}(-j) x_{\underline{\gamma}}(-1) x_{\gamma}(-1) v_{\Lambda_0} \neq 0$ must be $j=2$ because $\langle \tau, 2 \omega \rangle = 2$. This completes the 