Equality of Ordered Pairs/Lemma

From ProofWiki
Jump to navigation Jump to search

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