First let me recall some (abbreviated, and possibly simplified to suit my situation) definitions:
Let $A$ be a finite set and $\mathbb{A}$ some set of relations on $A$. Let $m, n$ be positive integers.
We say that a relation $R$ on $A$ is pp-definable (positively primitively definable) from $\mathbb{A}$ if it can be defined using only
- relations from $\mathbb{A}$,
- conjunctions,
- existential quantifiers (for elements of $A$),
- equality (for elements of $A$).
We define $\mathrm{Pol}_m(\mathbb{A})$ to be the set of all maps $f:A^m\rightarrow A$ such that they are invariant under every relation in $\mathbb{A}$, i.e. whenever $R \in \mathbb{A}, R \subseteq A^n$ and $a_{ij} \in A, \; i=1,2, \dots, n, j=1,2, \dots, m,$ such that for every $j$, $(a_{1j}, \dots a_{nj}) \in R$, then $$(f(a_{11},a_{12}, \dots, a_{1m}), f(a_{21},a_{22}, \dots, a_{2m}), \dots, f(a_{n1},a_{n2}, \dots, a_{nm}) ) \in R$$.
Now to add some coding: Since the set $A$ is finite, we can consider every map $f:A^m \rightarrow A$ to be just an ordered $|A|^m$-tuple of elements of $A$: Simply enumerate elements of $A^n,$ i.e. suppose $A^m=\{v_1, \dots, v_{|A|^m}\}$, then one can identify $f$ with $(f(v_1), \dots, f(v_{|A|^m}))\in A^{|A|^m}$. This way one can view $\mathrm{Pol}_m(\mathbb{A})$ as $|A|^m$-ary relation on $A$.
Now I finally get to the point:
In my universal algebra course, it was stated as "observation" that $\mathrm{Pol}_m(\mathbb{A})$ is pp-definable from $\mathbb{A}$. How can it be proven?
I can see how this works if $\mathbb{A}$ is finite: Say $\mathbb{A}=\{R\}, \; R \subseteq A^n$. Then one can define $(a_1, \dots a_{|A|^m})$ to be in $\mathrm{Pol}_m(\mathbb{A})$ provided that $$\bigwedge_{(i_1, \dots, i_n)\in F}R(a_{i_1}, \dots, a_{i_n}),$$
where $F$ is some finite set of $n$-tuples of numbers from $\{1, \dots, |A|^m\}$ (determined by $R$ - basically one takes all the "testing conditions" for $f$ to be invariant under $R$ into account and demands that they are satisfied). If $\mathbb{A}$ is finite, one just uses the conjunction of such formulas for all the relations in $\mathbb{A}$. I fail to see, however, how does this work when $\mathbb{A}$ is infinite. Does the statement even hold in such case? Also, I do not see how (and if) the possibility of using existential quantifiers would be of any help.
Thanks in advance for any help.
EDIT: Adding some details to the argumentation in the $\mathbb{A}=\{R\}$ case:
Take $$S=\{(a_{ij})_{ij} \in A^{n\times m} \; | \; \forall j: (a_{1j}, \dots a_{nj})\in R \}$$ and then $$F=\{(i_1, \dots, i_n) \in \{1, \dots, |A|^m\}^n \; | \; v_{i_1}, \dots, v_{i_{n}} \text{are (in this order) rows of some matrix in }S\}.$$
Then "$(i_1, \dots, i_n) \in F$" means "there is some $m$-tuple of $n$-tuples in the relation $R$ which forces every polymorphism of $R$ to have the $n$-tuple of arguments for vectors $v_{i_1}, \dots, v_{i_n}$ again in the relation R", which is (after considering the coding of functions) precisely what the atomic formula $R(a_{i_1}, \dots, a_{i_n})$ (i.e. $(a_{i_1}, \dots, a_{i_n}) \in R$ ) ensures.
A classmate of mine figured it out.
For the infinite case, the key observation is that $\mathrm{Pol}_m(\mathbb{A})$ is a subset of $A^{A^m}$, which is a finite set. Say $A^{A^m}\setminus \mathrm{Pol}_m(\mathbb{A})=\{f_1, \dots, f_k\}$. Then for every $i$, there is a relation $R_i \in \mathbb{A}$ such that $f_i$ is not invariant under $R_i$. Then clearly $\mathrm{Pol}_m(\mathbb{A})=\mathrm{Pol}_m(\{R_1, \dots, R_k\})$ and thus, the pp-definition for the finite case can be used.