The equational theory of regular languages has no finite set of axioms for general alphabets
For each $k &gt; 0$, the equation
$$
 A^{\ast} = (A^k)^{\ast}(1 + A + A^2 + \ldots + A^{k-1})
$$
holds for regular languages over every alphabet, hence if for some alphabet we have a finite axiomatization, this equation is provable from it. Now as shown in A. Ginsburg, Algebraic Theory of Automata, the following infinite system of axioms is complete for unary regular languages.
\begin{align*}
 A + B &amp; = B + A \\
 A + (B + C) &amp; = (A + B) + C \\ 
 A + A &amp; = A \\
 A + \emptyset &amp; = A \\
 AB &amp;= BA \\
 (AB)C &amp; = A(BC) \\
 A1 &amp; = A \\
 A \emptyset &amp; = \emptyset \\ 
 (A+B)C &amp; = AC + BC \\
 1^{\ast} &amp; = 1, \emptyset^{\ast} = 1 \\
 (AB^{\ast})^{\ast} &amp; = 1 + AA^{\ast}B^{\ast} \\
 (A + B)^{\ast} &amp; = A^{\ast} B^{\ast} \\
  A^{\ast} &amp; = (A^k)^{\ast}(1 + A + A^2 + \ldots + A^{k-1}), \quad k &gt; 0
\end{align*}
So we get a finite axiomatisation for unary languages, if we take but the last family of axioms, and all axioms from the finite axiom system for some alphabet that prove 
$$
 A^{\ast} = (A^k)^{\ast}(1 + A + A^2 + \ldots + A^{k-1})
$$
for each $k &gt; 0$. Then all of the above equations are derivable by this finite axiom system, and as the above one is complete, this finite axiom system would be complete, which is not possible. Hence there is no finite system of axioms for any alphabet.