Follow up to a previous universal algebra question on decidability of consistency

Question

This is a follow up to my previous question on universal algebra and decidability, here: Is it decidable if a finite set of equations have only trivial models?. In that question, the answerer said that it is undecidable in the general case if a set of equations is consistent i.e. has only one-element models. However, I now want to ask about a special case of that question. Suppose we are given a single unary operation symbol $f$ and no other operation symbols. Is it decidable if a given finite set $S$ of equations in $f$ is consistent? What about if we have two unary operations $f$ and $g$? What about more than two?

Answer 1

Suppose we are given a single unary operation symbol $f$ and no other operation symbols. Is it decidable if a given finite set $$ of equations in $f$ is consistent?

Yes.

Any consistent set of identities in the language of one unary operation has a $2$-element model $\langle \{0,1\}; f\rangle$ where either (i) $f$ is the identity function or (ii) $f$ is a constant function. Therefore, to show that $S$ is consistent, it suffices to test whether either of these types of algebras is a model of $S$.

$S$ holds in $\langle \{0,1\},f\rangle$ when $f$ is the identity function unless some identity in $S$ has the form $f^m(x_i)\approx f^n(x_j)$, with $m,n\geq 0$ and $x_i\neq x_j$. (That is, $S$ holds in this model unless some identity in $S$ involves two distinct variables, in which case $S$ does not hold in this model.)

$S$ holds in $\langle \{0,1\},f\rangle$ when $f$ is the constant function unless some identity in $S$ has the form $f^m(x_i)\approx f^n(x_j)$ or $f^n(x_j)\approx f^m(x_i)$, with $m=0<n$ and with $x_i=x_j$ allowed. (That is, $S$ holds in this model unless some identity in $S$ has a variable on one side and a higher-weight term on the other side.)

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Appendix.
Here is a sketch of how to show that any consistent set $S$ of identities in the language of one unary operation has a $2$-element model.

Let $\mathcal V$ be the class of models of $S$, and assume that $\mathcal V$ is nontrivial. The statement that $\mathcal V$ is nontrivial is equivalent to the statement that the $2$-generated free algebra in $\mathcal V$ has more than one element.

If $\mathcal V$ contains a nontrivial model $A$ where $f$ is not surjective, then $A$ has a congruence with two classes, $\textrm{image}(f), A-\textrm{image}(f)$, the quotient modulo this congruence has two elements, and $f$ is constant on this quotient. This quotient is a $2$-element model in $\mathcal V$ with $f$ constant.

If $f$ is surjective on every model of $\mathcal V$, then argue that $f$ must be a permutation on some nontrivial model $B$ (e.g. the $2$-generated free algebra). Define a congruence on $B^2$ which has the diagonal of $B^2$ as one class and the off-diagonal of $B^2$ as the other class.) The quotient of $B^2$ modulo this congruence is a $2$-element model in $\mathcal V$ on which $f$ is the identity function.