  Let $ E=\{ (1, b_1, \ldots , b_n)\in R^{n+1} \mid \; b_i= \pm 1 ,\; i=1,
\ldots, n \}$, $E^{\times n}_{\ne 0} := \{ W=(w_{i_1}, \ldots , w_{i_n}) \mid
w_{i_k}\in E, \, k=1, \ldots, n, \, dim \, span(w_{i_1}, \ldots , w_{i_n}) = n
\},$ and $q^W_l := |span(w_{i_{n-l+1}}, \ldots , w_{i_n}) \cap E|.$ Then for
any weights $p=(p_1, \ldots, p_{2^n})$, $p_i\in R$, $\sum_{i=1}^{2^n}{p_i} =1$
we have for the number of threshold functions $P(2,n)$ the following lower
bound $$P(2, n) \geq 2\sum_{W\in E^{\times n}_{\ne 0}}{\frac{1- p_{i_1}
-p_{i_2} - \cdots - p_{i_{q_n^W}}}{q_n^W\cdot q_{n-1}^W\cdots q_1^W}},$$ and
the right side of the inequality doesn't depend on the choice of $p$. Here the
indices used in the numerator correspond to vectors from $span(w_{i_1}, \ldots
, w_{i_n})\cap E = \left\{w_{i_1}, \ldots, w_{i_n}, \ldots
w_{i_{q_n^W}}\right\}$.
