Observe that $\langle v_{i},w_{j}\rangle=1,\ i,j=0,1$. Hence, if $n\in\mathbb{Z}^{d}_{v_{0}}$, then $n+\langle v_{1},n\rangle(w_{1}-w_{0})\in\mathbb{Z}^{d}_{v_{0}}$. Now put $n^{\prime}:=n-\langle v_{1},n\rangle w_{0}\in\mathbb{Z}^{d}_{v_{1}}$. Then $n^{\prime}\equiv n\mod w_{0}$ together with $n^{\prime}\equiv n+\langle v_{1},n\rangle(w_{1}-w_{0})\mod w_{1}$ imply the desired formula. ∎