Mike asked in this question to find a model of vector space with additional operation $\wedge$ so that we have $a \wedge (a+b) = (a+b) \wedge b = a \wedge b$ and $\wedge$ is non-distributive.
I built set-theoretic vector space like this:
Let $(X,P(X))$ be some nonempty set $X$ and $P(X)$ the set of all subsets of $X$.
Define $+(A,B)$ to be the symmetric difference of the sets $A$ and $B$.
Let the field be a field with two elements: $0$ and $1$.
$+$ is commutative and associative.
Zero vector is the empty set.
Define $0 \cdot A$ to be the empty set and $1 \cdot A$ to be $A$.
And now the problem is how to define $\wedge$ so that $a \wedge (a+b) = (a+b) \wedge b = a \wedge b$ is true and is non-distributive.
But, I know of only few operations on arbitrary sets: of union, intersection, difference, symmetric difference, complementation, and that´s all.
My question would be:
Can we define on this set-theoretic vector space an operation $\wedge$ so that we have $a \wedge (a+b) = (a+b) \wedge b = a \wedge b$ and $\wedge$ is non-distributive?