Plucker relations imply Leibniz formula?

Question

Let $E$ be a finite set, $n \in \mathbb{N}$ and $f:E^n \rightarrow \mathbb{Q}$ be an alternating function such that $$\sum_{i=0}^n (-1)^i f(y_i,x_2, \dots, x_n)f(y_0, \dots , \hat{y_i}, \dots, y_n)=0, $$ for all $(x_2,\dots, x_n)\in E^{n-1}$ and $(y_0, \dots, y_n) \in E^{n+1}$. I claim that for all $(a_1,\dots, a_n)\in E^{n}$ and $(b_1, \dots, b_n) \in E^{n}$ the following formula holds: $$\sum_{\sigma \in \mathfrak{S}_n}(-1)^{\operatorname{sgn} \sigma} \prod_{i=1}^n f(a_1, \dots a_{i-1}, b_{\sigma(i)}, a_{i+1}, \dots, a_n)=f(a_1, \dots, a_n)^{n-1} f(b_1, \dots, b_n).$$ Is it true? can you prove it?