TITLE: Does this "reverse distributivity" ever occur: $a \circ (b\times c) = (a \times b) \circ (a \times c)$?
QUESTION [7 upvotes]: I was reading a book containing a typo to the effect that they defined the distributive property as:
$$
a \circ (b\times c) = (a \times b) \circ (a \times c) \tag{*}\label{*}
$$
which is wrong of course. I will call the property (*) "reverse distributivity" for now. It got me wondering:

Are there any examples of structures with this "reverse distributivity"? What can we say about such a structure? And are there names for these things?

Some findings so far:
If we assume the existence of neutral elements, then things quickly degenerate. Assume that $(M, \circ, 1_\circ, \times, 1_\times)$ is an algebraic structure with two binary operators satisfying (*), and where $1_\circ$ and $1_\times$ are neutral elements. Then we have:
$$
1_\circ  = (1_\circ \times 1_\times) \circ (1_\circ \times 1_\times)
\stackrel{\eqref{*}} = 1_\circ \circ (1_\times \times 1_\times)
= 1_\times
$$
so the identity elements are in fact equal. Let $1 := 1_\times = 1_\circ$. Then, for any $a,b\in M$:
$$
a \times b = 1 \circ (a \times b) \stackrel{\eqref{*}} =
(1 \times a ) \circ (1 \times b) = a \circ b
$$
so in fact the two compositions are the same. In this case (*) becomes
$$
a \circ (b\circ c) = (a \circ b) \circ (a \circ c)
$$
which seems to be known as self-distributivity and shows up in a number of places (e.g. group conjugation and logical implication).
But if we want two (different) compositions that satisfy (*), then this shows that they at least cannot both have neutral elements. (If we only assume the existence of $1_\times$, then we can show that $a \circ a = a \circ 1_\times$ for all $a$). I haven't gone much further than this.

REPLY [4 votes]: I will treat the case where $\times$ forms a monoid with unit element $1$. From monoids, one can go in two directions: towards group-like structures or towards lattice-like structures. The group-like cases are all trivial, but we can find many non-degenerate lattice-like examples.
So consider a monoid $(M,\times,1)$ and assume that we have an operation $\circ$ that satisfies the reverse distributivity law $a\circ(b \times c) = (a \times b)\circ(a \times c)$ for any $a,b,c \in M$. It follows then that have $b\circ c = (1 \times b)\circ(1\times c)=1\circ(b\times c)$ for all $b,c \in M$.
Define the function $f: M \rightarrow M$ as $f(x) = 1\circ x$. Then reverse distributivity becomes $f(a \times b \times c) = a\circ (b \times c) = (a \times b) \circ (a \times c) = f(a \times b \times a \times c)$. Vice versa, any function $f: M \rightarrow M$ satisfying $f(a \times b \times c) = f(a\times b \times a \times c)$ for all $a,b,c \in M$ gives rise to a reverse distributive operation via $a\circ b = f(a \times b)$. Therefore:

Any potential $f$ must satisfy $f(a) = f(a \times a)$ for all $a \in M$. This follows by setting $b,c = 1$ in the equation above.
If $M$ is a group, then all such functions $f$ are in fact constant. To see this, notice that $f(a\times b \times c) = f(a \times b \times a \times c)$, then set $c=1$ and $b=a^{-1}$ to conclude $f(1) = f(a \times a^{-1}) = f(a \times a^{-1} \times a) = f(a)$ for all $a \in M$.
If $M$ is an idempotent commutative monoid (semilattice), then the equality $f(a \times b \times c) = f(a \times a \times b \times c) = f(a \times b \times a \times c)$ holds for any function $f: M \rightarrow M$. This observation yields plenty of non-degenerate examples. For example, consider the Boolean algebra $\{\text{true},\text{false}\}$ of truth values, and set $\times$ to be disjunction $\vee$. Define $a \not\vee b = \neg(a \vee b)$. The pair $(\vee,\not\vee)$ satisfies reverse distributivity, i.e. $a \not\vee (b \vee c) = (a \vee b) \not\vee (a \vee c)$. Similarly for $\wedge$.

If the structure $(M,\times)$ is a general semigroup, then these observations fail. Interestingly, by analyzing all 14 non-constant binary operations, we can show that all suitable pairs of operations $(\times, \circ)$ over truth values have $a \circ b = f(a \times b)$ for some $f$. However, there is a three-element structure $(S,\times,\circ)$ with $(\times,\circ)$ satisfying reverse distributivity that does not have $a \circ b = f(a \times b)$ for any function $f$ and with $a\times b$ non-constant. E.g. Consider the two operations below.
x|012    o|012
-----    -----
0|121    0|011
1|211    1|011
2|111    2|011

Since $b \times c \neq 0$, we have $a \circ (b \times c) = 1 = 1 \circ 1 = (a \times b) \circ (a \times c)$, but not $a \circ b = f(a \times b)$, since that would give $0 = 0 \circ 0 = f(0 \times 0) = f(1) = f(1 \times 1) = 1 \circ 1 = 1$. 
edit: While I'm sure that someone, somewhere has already named these structures, I doubt that there's a widely-agreed-upon name for them, even among people who actively study these structures (especially given that we don't have such a name even for the self-distributive case, and even when equipped with additional axioms. Viz. racks, crystals, automorphic sets, etc.).