Proof. Using the same set up as Oddtown Theorem 7.6.1 over $\bb{F}_{2}$ we have the hypothesis is equivalent to

1. (1)

   $\indi{A}\cd \indi{A}\congruent{0}{2}$

2. (2)

   $\indi{A}\cd\indi{B}\congruent{1}{2}$

We now define a new family $\cali{F}'\suf\cali{P}([n+1])$ by

which satisfies Oddtown Theorem 7.6.1 so $\abs{F}=\abs{F'}\leq n+1$. If there is a linear relation over $\bb{F}_{2}$ that is not the trivial relation then fix $B\in\F$ we have

Let $C\neq B\in\F$ then over $\bb{F}_{2}$

Hence $c_{B}=c_{C}$ for any $B,C\in\F$ and so $c_{A}=1$ for all $A\in\F$ and

Taking dot product with $\indi{B}$ again we see

showing that $\abs{\F}$ is odd. Observe that

so for each $A\in\F$ we have

where $V$ is the subspace of vectors with even number of $1$ in coordinates and $\dim V=n-1$ as the kernel of the linear functional $x\mto x\cd\indi{[n]}$. Using the trick of $\F'$ we consider

and using the hypothesis we have

1. (1)

   $\indi{A'}\cd \indi{A'}\congruent{1}{2}$

2. (2)

   $\indi{A'}\cd\indi{B'}\congruent{0}{2}$

So $\curl{\indi{A'}:A'\in\F'}$ is linearly independent over $\bb{F}_{2}$ and we conclude

as desired.$\square$