The question of how many linearly separable boolean functions exist on an $n$-dimensional hypercube is a long standing problem without a solution. However, I am looking for an explanation of why the following reasoning is true for the case $n=3$:
Let $C = \big\{ (x_1,x_2,x_3) : x_1 , x_2 , x_3 \in \{0,1\} \big\}$ be the three dimensional cube and $f : C \to \{0,1\}$ a boolean function. The following statement is clear:
- If there exists a plane through $4$ points on that cube such that these $4$ points are not linearly separable by a line on said plane, then $f$ is not linearly separable.
However, if we count all of these functions, we end up with the exact number of linearly separable functions on the $3$-dimensional cube (see wiki from above), implying that above statement (1.) is in fact an equivalence.
Unfortunately I have no idea on how the other direction of above statement works and would be happy about any hints or help.
We could try, for given $f$ where all above mentioned $2$D slices are $1$D-separable, to construct a surface linearly separating $f$. I don't know how, though.