A conjecture in equational logic In an algebra with a single binary operation g, is there a single equational identity that generates the same set of identities as the set {g(x,y)=g(y,x) , g(g(x,y),z)=g(x,g(y,z))}? My conjecture is that there isn't.
 A: I use multiplicative notation for the operation, i.e., I write $xy$ instead of $g(x,y)$.
In the language of one binary operation, there is no single identity which is equivalent to the conjunction of the associative and commutative laws.
Case I. Each side of the supposed identity contains at least two multiplications, i.e., the identity is of the form $(t_1t_2)t_3=(t_4t_5)t_6$ or $(t_1t_2)t_3=t_4(t_5t_6)$ or $t_1(t_2t_3)=t_4(t_5t_6)$ for some terms $t_1,t_2,t_3,t_4,t_5,t_6$.
The commutative law is not a consequence of such an identity, or even the set of all such identities. To see this, consider the $4$-element structure $\{a,b,c,d\}$ with multiplication defined so that $ab=c$, while $xy=d$ whenever $x\ne a$ or $y\ne b$. In this structure $ab\ne ba$, while both sides of all the identities under consideration evaluate to $d$.
Case II. One side of the supposed identity contains at most one multiplication, i.e., the identity is of the form $xy=t$ or $xx=t$ or $x=t$ for some term $t$.
We may assume that the identity is a consequence of the associative and commutative laws, and so it holds in the structure $(\mathbb Z,+)$. Therefore, each variable occurs the same number of times on either side of the equality sign. This leaves us with the identities $xy=yx$,$\ xy=xy$,$\ xx=xx$, and $x=x$, none of which implies the associative law. Namely, the $2$-element structure $\{a,b\}$, where $aa=b$ and $ab=ba=bb=a$, is commutative but not associative; in fact, for any $x\in\{a,b\}$, we have
$$(ax)b=b(xa)=a,$$
$$a(xb)=(bx)a=b.$$
This is the unique (up to isomorphism) binary operation on a $2$-element set which is commutative but not associative; it can be interpreted as either of the truth-functions NOR or NAND.
