# Equality of Relations

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## Contents

## Theorem

Let $\mathcal R_1$ and $\mathcal R_2$ be relations on $S_1 \times T_1$ and $S_2 \times T_2$ respectively.

Then $\mathcal R_1$ and $\mathcal R_2$ are **equal** if and only if:

- $S_1 = S_2$
- $T_1 = T_2$
- $\tuple {s, t} \in \mathcal R_1 \iff \tuple {s, t} \in \mathcal R_2$

It is worth labouring the point that for two relations to be **equal**, not only must their domains be equal, but so must their codomains.

## Proof

This follows from set equality and Equality of Ordered Pairs.

A **relation** on $S \times T$ is an ordered triple, thus:

\(\displaystyle \mathcal R_1\) | \(=\) | \(\displaystyle \tuple {S_1, T_1, R_1}\) | where $R_1 \subseteq S_1 \times T_1$ | ||||||||||

\(\displaystyle \mathcal R_2\) | \(=\) | \(\displaystyle \tuple {S_2, T_2, R_2}\) | where $R_2 \subseteq S_2 \times T_2$ |

From Equality of Ordered Tuples, $\mathcal R_1 = \mathcal R_1$ if and only if:

\(\displaystyle S_1\) | \(=\) | \(\displaystyle S_2\) | |||||||||||

\(\displaystyle T_1\) | \(=\) | \(\displaystyle T_2\) | |||||||||||

\(\displaystyle R_1\) | \(=\) | \(\displaystyle R_2\) |

Then we have that $R_1$ and $R_2$ are sets of ordered pairs.

Thus by definition of set equality:

- $R_1 = R_2 \iff \paren {\forall \tuple {x, y} \in \S_1 \times T_2: \tuple {x, y} \in R_1 \iff \tuple {x, y} \in R_2}$

Hence the result.

$\blacksquare$

## Also known as

Some sources refer to this concept between two relations as being **equivalence**, rather than **equality**.

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 4$. Relations; functional relations; mappings