# Equality of Ordered Pairs/Lemma

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## Theorem

Let $\set {a, b}$ and $\set {a, d}$ be doubletons such that $\set {a, b} = \set {a, d}$.

Then:

- $b = d$

## Proof

We have that:

- $b \in \set {a, b}$

and so by definition of set equality:

- $b \in \set {a, d}$

So:

- $(1): \quad$ either $b = a$ or $b = d$.

First suppose that $b = a$.

Then:

- $\set {a, b} = \set {a, a} = \set a$

We have that:

- $d \in \set {a, d}$

and so by definition of set equality:

- $d \in \set {a, b}$

and so as $\set {a, b} = \set a$ it follows that:

- $d = a$

We have $b = a$ and $d = a$ and so:

- $b = d$

and so $b = d$.

$\Box$

Next suppose that $b \ne a$.

We have from $(1)$ that either $b = a$ or $b = d$.

As $b \ne a$ it must be the case that $b = d$.

$\Box$

So in either case we see that:

- $b = d$

$\blacksquare$

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $1.1: \ 12$ - 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Lemma $4.3$.