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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.

$-A=A$

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?

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  • $\begingroup$ Doesn't the union operator work? $A\cup (A\Delta B)=A\cup B=(A\Delta B)\cup B$, but $A\cup B\neq (A\cup A)\Delta (A\cup B)=B\setminus A$ $\endgroup$ – Vsotvep Jun 18 at 9:11
  • $\begingroup$ @Vsotvep I am not sure, I have to put it on the paper. $\endgroup$ – Grešnik Jun 18 at 9:15

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