example of polyadic algebra with equality

Recall that given a triple $(A,I,X)$ where $A$ is a Boolean algebra, $I$ and $X\ne \mathrm{\varnothing }$ are sets. we can construct a polyadic algebra $(B,I,\exists ,S)$ called the functional polyadic algebra for $(A,I,X)$. In this entry, we will construct an example of a polyadic algebra with equality called the functional polyadic algebra with equality from $(B,I,\exists ,S)$.

We start with a simpler structure. Let $B$ be an arbitrary Boolean algebra, $I$ and $X\ne \mathrm{\varnothing }$ are sets. Let $Y=X^I$, the set of all $I$-indexed $X$-valued sequences, and $Z=B^Y$, the set of all functions from $Y$ to $B$. Call the function $e:I\times I\to Z$ the functional equality associated with $(B,I,X)$, if for each $i,j\in I$, $e(i,j)$ is the function defined by

The quadruple $(B,I,X,e)$ is called a functional equality algebra.

Now, $B$ will have the additional structure of being a polyadic algebra. Start with a Boolean algebra $A$, and let $I$ and $X$ be defined as in the last paragraph. Then, as stated above in the first paragraph, and illustrated in here (http://planetmath.org/ExampleOfPolyadicAlgebra), $(B,I,\exists ,S)$ is a polyadic algebra (called the functional polyadic algebra for $(A,I,X)$). Using the $B$ just constructed, the quadruple $(B,I,X,e)$ is a functional equality algebra, and is called the functional polyadic algebra with equality for $(A,I,X)$.

It is not hard to show that $e$ is an equality predicate on $C=(B,I,\exists ,S)$, and as a result $(C,e)$ is a polyadic algebra with equality.