The finite algebras $(A,\*,+)$ that satisfy the identity $(x\*y)+(y\*z)=(x+y)\*(y+z)$ are precisely the algebras such that the one-dimensional cellular automata produced by $\*$ and $+$ are commutative cellular automata. To be clear the operations $\*,+$ do not necessarily satisfy any associativity or any other well known laws. See the paper [Commuting Cellular Automata](http://www.santafe.edu/media/workingpapers/97-08-076.pdf) for details on this identity and its relation to cellular automata (or see [my answer here](https://mathoverflow.net/a/156514/22277) for less details). I want to know if the variety $V$ of algebras that satisfy $(x\*y)+(y\*z)=(x+y)\*(y+z)$ is generated by the collection of all its finite algebras. It suffices to show that the free algebras in $V$ are subdirect products of finite algebras in $V$. If the finite algebras $(A,\*,+)$ satisfy other kinds of identities not implies by $(x\*y)+(y\*z)=(x+y)\*(y+z)$, then what is an explicit example of such an identity? If the free algebras in $V$ are subdirect products of finite algebras in $V$, then what is a good explicit example of how to embed the free algebras in $V$ as subdirect products of finite algebras in $V$?


Part of the motivation for this question is that cellular automata are usually over a finite alphabet $A$, so I wonder if finite algebras that satisfy $(A,\*,+)$ that satisfy the identity $(x\*y)+(y\*z)=(x+y)\*(y+z)$ satisfy any other identities.



Is the variety of algebras $(A,*,+)$ that satisfy $(x*y)+(y*z)=(x+y)*(y+z)$ generated by its finite algebras?
**Theorem.** If $\mathcal V$ is a variety of finite signature and
$\mathcal V$ can be axiomatized by identities of the form $s\approx t$ where
$\textit{length}(s)=\textit{length}(t)$, then $\mathcal V$ is generated
by its finite members.


(For this theorem, say that $\textit{length}(s)=\textit{length}(t)$ if $s$ and $t$ have the same multiset of variables and the same number of operation symbols.)


*Proof.*
Under the length hypothesis of the theorem, there is a congruence
$\theta\_n$ on ${\mathbf F}\_{\mathcal V}(m)$ which
relates two elements iff they are equal or else they can be
each represented by terms of length greater than $n$.
The signature is finite, so there are finitely many elements of
${\mathbf F}\_{\mathcal V}(m)$ represented by terms of length at most $n$,
and therefore ${\mathbf F}\_{\mathcal V}(m)/\theta\_n$ is finite.
The intersection of the $\theta\_n$'s is trivial, so this yields
a representation
$$
{\mathbf F}\_{\mathcal V}(m)\hookrightarrow \prod\_n 
{\mathbf F}\_{\mathcal V}(m)/\theta\_n
$$
of ${\mathbf F}\_{\mathcal V}(m)$ as a subdirect product
of finite algebras. \\


Your variety has finite signature, and the defining identity $(x\*y)+(y\*z)=(x+y)\*(y+z)$ has the same multiset of variables on both sides, $\{x, y, y, z\}$, and the same number ($3$) of operation symbols on both sides.


**EDIT** (8/18/15): The above explains why the finitely generated free algebras in $\mathcal V$ are subdirect products of finite algebras. A question was raised about whether it is necessary to also show that the infinitely generated free algebras are subdirect products of finite algebras. It is not necessary to do this: the infinitely generated free algebras in $\mathcal V$ are subdirect products of finitely generated free algebras in $\mathcal V$.


