How to prove that there is no 5-ary operation in a clone satisfying certain conditions Question:
How to prove that there is no 5-ary operation $f \in \mathcal{C}$ satisfying
$f(2, 1, 3, 4, 3) = 1$ and
$f(2, 1, 1, 4, 3) = 2$?
The $\mathcal{C} = Clo(\textbf{A})$ is a clone of $\textbf{A}$, where $\textbf{A} = ({1, 2, 3, 4}, \ast)$ with
\begin{array}{ |c|c|c|c|c| }
\hline
∗& 1& 2& 3& 4 \\
\hline 
1 &2 &3& 2& 1\\
\hline 
2 &1& 4& 3& 4\\
\hline 
3& 2& 1& 2& 1\\
\hline 
4& 3 &4& 3& 2\\
\hline
\end{array}
My thoughts:
I want to use invariant relations somehow, but not sure, how to proceed with that.
According to the Theorem (Geiger; Bodnarcuk, Kaluznin, Kotov, Romov), $Clo(\textbf{A})=Pol(Inv(\textbf{A}))$ for finite set $\textbf{A}$. (Where $Pol({A})$ = the clone of polynomial operations on $\textbf{A}$, $Inv({A})$ = all relations invariant under every $f$ in $\textbf{A}$).
So maybe I should prove that if $f$ is in $Pol(Inv(\textbf{A}))$ then it cannot be 5-ary?
Any advice is appreciated.
 A: The algebra $A$ in question has universe $\{1,2,3,4\}$ and a single binary operation given by the table
$$
\begin{array}{|c||c|c|c|c|}
\hline
∗& 1& 2& 3& 4 \\
\hline 
\hline
1 &2 &3& 2& 1\\
\hline 
2 &1& 4& 3& 4\\
\hline 
3& 2& 1& 2& 1\\
\hline 
4& 3 &4& 3& 2\\
\hline
\end{array}
$$
The question is whether the clone of $A$ contains an operation $f$ satisfying both $f(2, 1, 3, 4, 3) = 1$ and
$f(2, 1, 1, 4, 3) = 2$. If the clone of $A$ contained such an operation,
then we could apply it to the pairs
$(2,2), (1,1), (3,1), (4,4), (3,3)\in A\times A$ to produce $(1,2)$.
That is, the two equations involving $f$ acting on individual elements of $A$ can be combined into a statement about $f$ acting on pairs:
$$
f((2,2), (1,1), (3,1), (4,4), (3,3))=(1,2).
$$
If there were such an $f$, then it could be used to prove that the pair $(1,2)$ belongs to the subalgebra of $A^2$ (or `binary invariant relation' of $A$) that is generated by $(2,2), (1,1), (3,1), (4,4), (3,3)$. The subalgebra of $A^2$ that is generated by $(2,2), (1,1), (3,1), (4,4), (3,3)$
is exactly the congruence $\theta$ of $A$ that is generated by $(1,3)$. So, if $f$ existed, then $(1,2)$ would have to be a member of $\theta$.
But you can compute $\theta$ easily and see that it is the equivalence relation corresponding to the partition $13/2/4$, so $(1,2)\notin \theta$, so $f$ is not in the clone of $A$.
