# Equality of Relations

## Theorem

Let $\RR_1$ and $\RR_2$ be relations on $S_1 \times T_1$ and $S_2 \times T_2$ respectively.

Then $\RR_1$ and $\RR_2$ are equal if and only if:

$S_1 = S_2$
$T_1 = T_2$
$\tuple {s, t} \in \RR_1 \iff \tuple {s, t} \in \RR_2$

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:

 $\displaystyle \RR_1$ $=$ $\displaystyle \tuple {S_1, T_1, R_1}$ where $R_1 \subseteq S_1 \times T_1$ $\displaystyle \RR_2$ $=$ $\displaystyle \tuple {S_2, T_2, R_2}$ where $R_2 \subseteq S_2 \times T_2$

From Equality of Ordered Tuples, $\RR_1 = \RR_1$ 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.