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# example of polyadic algebra with equality

Recall that given a triple $(A,I,X)$ where $A$ is a Boolean algebra, $I$ and $X\neq\varnothing$ 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 $X\neq\varnothing$ 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

$e(i,j)(x):=\left\{\begin{array}[]{ll}1&\textrm{if }x_{i}=x_{j},\\ 0&\textrm{otherwise.}\end{array}\right.$ |

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, $(B,I,\exists,S)$ 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.

# References

- 1
P. Halmos,
*Algebraic Logic*, Chelsea Publishing Co. New York (1962).

## Mathematics Subject Classification

03G15*no label found*

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