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.

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    $\begingroup$ You can always take $\times = $ anything and $x\circ y = C$ (any constant.) Both sides of the identity are just $C$. $\endgroup$ – Jair Taylor Jan 11 '20 at 21:20

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:

  1. 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.

  2. 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 $a \in M$.

  3. 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.).

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    $\begingroup$ This is nice. Let me just suggest some corrections (of some typos, certainly). In the end of the third paragraph, you mean $a\circ b = f(a\times b)$. And in the end of the BA example, you mean reverse distributivity (not commutativity); in the same example, I think the symbol $\not\vee$ doesn't read very well, since the slash is mostly concealed with the $\vee$ symbol. $\endgroup$ – amrsa Jan 12 '20 at 10:43
  • $\begingroup$ This is certainly the sort of answer I was hoping for. It shows that we get only more or less trivial situations if $(M,\times)$ has almost any sort of interesting structure. The semilattice case is of course less trivial, but $\circ$ still doesn't bring anything new to the table. This explains why we don't come across (or talk about) the property, which is actually what I was hoping to learn and should probably have written in the question. I think I will still study for completion the symmetric case of $(M,\times)$ just a monoid and $(M,\circ)$ having more structure, like in a ring. $\endgroup$ – Milten Jan 12 '20 at 12:02
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    $\begingroup$ One question: Can't we just shrink the Cayley tables in your examples to 2x2, i.e. $a\times b=1$, $a\circ b = b$ for $a,b\in\{0,1\}$. Then we still don't have $a\circ b=f(a\times b)$ for any $f$ by your own argument, right? So what you said about "all suitable pairs of operations $(\times,\circ)$ over truth values" doesn't seem to hold... $\endgroup$ – Milten Jan 12 '20 at 12:09
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    $\begingroup$ @Milten: Quite right! I got my notes mixed up. The intended statement was that " all suitable pairs of operations (×,∘) over truth values have a∘b=f(a×b) for some f or a×b constant", and I had a different (non-constant) intended example for the 3-value case. I'll edit the answer with the correct example, and fix the typos pointed out by amrsa. $\endgroup$ – Z. A. K. Jan 12 '20 at 12:16

For a yet trivial example in which $\circ$ and $\times$ denote different operations (and $\circ$ has a neutral element, and additionally, they're both associative and commutative), consider $M = \{0,1\}$, with $\circ$ being the usual product, and $\times$ a null product (that is, $x \times y = 0$, for all $x$ and $y$).

Then the equation $$x \circ (y \times x) = (x \times y) \circ (x \times z)$$ reduces to $$x \circ 0 = 0 \circ 0,$$ which is true.

It is not a very interesting example, but the answer to your question is yes, there exist such structures.
The example you deduced from the existence of identities for both operations (both operations being equal and self-distributive) is also perfectly legitimate.

There are many more others (some without any of the trivial aspects the above examples show), but concerning names, usually these structures are only named if they are found to be interesting in some context.
As a rule, the answer to the question "Are there such algebras satisfying these axioms" is yes, at least if the axioms are equalities. In these cases, there's always at least the trivial example, that is, the one-element algebra with whatever are the operations.

  • $\begingroup$ One might note that your example generalises to any case where $\circ$ has a right annihilating element $0$, i.e. $x\circ 0 = 0$, and then let $x\times y = 0$. This is a nice parallel to the other trivial solution $x\circ y =$ (a constant). $\endgroup$ – Milten Jan 12 '20 at 12:16
  • $\begingroup$ @Milten Yes, you have the example you yourself came up with, my example, the example given in a comment by Jair Taylor, the family of examples given in Z.A.K.'s answer, and if you want to, you can come up with a whole lot of other examples. For example, if $(A_i)$ is a family of examples, then so is $\prod_i A_i$, and also any subalgebra of that one and any homomorphic image. This enable us to construct infinitely many examples which are not trivial at all, but that doesn't mean they are interesting. $\endgroup$ – amrsa Jan 12 '20 at 12:21

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