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example of polyadic algebra
Recall that the canonical example of a monadic algebra is that of a functional monadic algebra, which is a pair $(B,\exists)$ such that $B$ is the set of all functions from a nonempty set $X$ to a Boolean algebra $A$ such that, for each $f\in B$, the supremum and the infimum of $f(X)$ exist, and $\exists$ is a function on $B$ that maps each element $f$ to $f^{{\exists}}$, a constant element whose range is a singleton consisting of the supremum of $f(X)$.
The canonical example of a polyadic algebra is an extension (generalization) of a functional monadic algebra, known as the functional polyadic algebra. Instead of looking at functions from $X$ to $A$, we look at functions from $X^{I}$ (where $I$ is some set), the $I$fold cartesian power of $X$, to $A$. In this entry, an element $x\in X^{I}$ is written as a sequence of elements of $A$: $(x_{i})_{{i\in I}}$ where $x_{i}\in A$, or $(x_{i})$ for short.
Before constructing the functional polyadic algebra based on the sets $X,I$ and the Boolean algebra $A$, we first introduce the following notations:

for any $J\subseteq I$ and $x\in X^{I}$, define the subset (of $X^{I}$)
$[x]_{J}:=\{y\in X^{I}\mid x_{i}=y_{i}\mbox{ for every }i\notin J\},$ 
for any function $\tau:I\to I$ and any $f:X^{I}\to A$, define the function $f_{{\tau}}$ from $X^{I}$ to $A$, given by
$f_{{\tau}}(x_{i}):=f(x_{{\tau(i)}}).$
Now, let $B$ be the set of all functions from $X^{I}$ to $A$ such that
1. for every $f\in B$, every $J\subseteq I$ and every $x\in X^{I}$, the arbitrary join
$\bigvee f\left([x]_{J}\right)$ exists.
Before stating the next condition, we introduce, for each $f\in B$, a function $f^{{\exists J}}:X^{I}\to A$ as follows:
$f^{{\exists J}}(x):=\bigvee f\left([x]_{J}\right).$ Now, we are ready for the next condition:
2. if $f\in B$, then $f^{{\exists J}}\in B$,
3. if $f\in B$, then $f_{{\tau}}\in B$ for $\tau:I\to I$.
Note that if $A$ were a complete Boolean algebra, we can take $B$ to be $A^{{X^{I}}}$, the set of all functions from $X^{I}$ to $A$.
Next, define $\exists:P(I)\to B^{B}$ by $\exists(J)(f)=f^{{\exists J}}$, and let $S$ be the semigroup of functions on $I$ (with functional compositions as multiplications), then we call the quadruple $(B,I,\exists,S)$ the functional polyadic algebra for the triple $(A,X,I)$.
Remarks. Let $(B,I,\exists,S)$ be the functional polyadic algebra for $(A,X,I)$.

$(B,I,\exists,S)$ is a polyadic algebra. The proof of this is not difficult, but involved, and can be found in the reference below.

If $I$ is a singleton, then $(B,I,\exists,S)$ can be identified with the functional monadic algebra $(B,\exists)$ for $(A,X)$, for $S$ is just $I$, and $X^{I}$ is just $X$.

If $I$ is $\varnothing$, then $(B,I,\exists,S)$ can be identified with the Boolean algebra $A$, for $S=\varnothing$ and $X^{I}$ is a singleton, and hence the set of functions from $X^{I}$ to $A$ is identified with $A$.
References
 1 P. Halmos, Algebraic Logic, Chelsea Publishing Co. New York (1962).
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