Sums of truth-table values mod 2 range over all truth tables Let $A=\lbrace0,1\rbrace$. There are 16 distinct functions $f_i:A^2\to A$.
Choose a permutation $P=\left(a_1,\ldots,a_4\right)$ of the elements of $A^2$, and for each $i$ consider the ordered quadruple $\left(\sum_{j=1}^nf_i(a_j)\pmod2\right)_{n=1}^4\in A^4$. Clearly this quadruple is $f_k(P)$ for some $k$. I claim that as $i$ ranges over $\left(1,\ldots,16\right)$ we obtain all sixteen $f_k$ this way.
(My proof is by inspecting a single choice of $P$ — which I did manually — and handwavingly claiming that the choice of $P$ doesn't matter because everything's symmetric.)
Question: Why is this true? (Something (a proof not by inspection or an explanation) that generalizes would be most welcome.)

The space of $f_i$'s is a 4-dimensional vector space over $\mathbb F_2$. Each permutation P gives rise to a linear transformation of this vector space whose matrix is
$$\pmatrix{1&1&1&1\\0&1&1&1\\0&0&1&1\\0&0&0&1}$$
with respect to appropriate bases. The determinant of this matrix is clearly $1$, so it is invertible, and therefore the transformation is bijective.
