A characterization of the identities which are equivalent to the trivial identity

Question

This is a follow-up to my previous question, here: Is only the commutative identity equivalent to the commutative identity?. As in that question, let our signature be that of a single binary operation symbol $+$, which is just a symbol for an arbitrary binary operation on an arbitrary set. The trivial identity is $x=y$. However, there are other identities which are equivalent to it, such as $x+y=z$. I have a conjecture that any identity equivalent to the trivial identity is of the form $t=v$ (or its reverse $v=t$), for some term $t$ and a variable $v$ which does not appear in the term $t$. Is that conjecture correct? Or is there a single identity equivalent to the trivial identity which is not of that form?

Answer 1

Yes, this is correct.

• First, suppose $\mathcal{A}$ has more than two elements and $E$ is an identity of the form $v=t(x_1,...,x_n)$ (or $t=v$) with $v\not\in\{x_1,...,x_n\}$. Fix distinct $c,d\in\mathcal{A}$. If $c=t(c,...,c)$ in $\mathcal{A}$ then $E$ fails in $\mathcal{A}$ via the assignment $x_i\mapsto c, v\mapsto d$; if $c\not=t(c,...,c)$ in $\mathcal{A}$ then $E$ fails in $\mathcal{A}$ via the assignment $x_i\mapsto c, v\mapsto c$.

• Now suppose $s(x_1,...,x_n)=t(y_1,...,y_m)$ is an equation where neither $s$ or $t$ is a variable. Then $s=t$ is satisfied in any structure where all primitive functions are constant with the same value. Since nontrivial such structures exist, we get that $s=t$ does not force triviality. This means that WLOG $s$ is just the variable $x$; we now need to show that $x\not\in\{y_1,...,y_m\}$. But again, if $x$ were one of the $y_i$s then we could build a nontrivial model of $s=t$ by interpreting the basic functions as appropriate projections (e.g. if $t(y_1,...,y_m)$ is $f(y_1,...,y_m)$ for some primitive function $f$ and $x=y_i$, then we want $f$ to be projection onto the $i$th coordinate and so-on).