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.

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 $A$.