Equality of Ordered Pairs/Necessary Condition/Proof from Empty Set Formalization

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Theorem

Let $\tuple {a, b}$ and $\tuple {c, d}$ be ordered pairs such that $\tuple {a, b} = \tuple {c, d}$.

Then $a = c$ and $b = d$.


Proof

First a lemma:

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

Then:

$b = d$

$\Box$


Let $\tuple {a, b} = \tuple {c, d}$.

From the empty set formalization:

$\set {\set {\O, a}, \set {\set \O, b} } = \set {\set {\O, c}, \set {\set \O, d} }$


First we note the special case where $a = \set \O$ and $b = \O$.

Then we have:

\(\ds \set {\set {\O, a}, \set {\set \O, b} }\) \(=\) \(\ds \set {\set {\O, \set \O}, \set {\set \O, \O} }\)
\(\ds \) \(=\) \(\ds \set {\set {\O, \set \O}, \set {\O, \set \O} }\) Definition of Set Equality: $\set {\O, \set \O } = \set {\set \O, \O}$
\(\ds \) \(=\) \(\ds \set {\set {\O, \set \O} }\) Doubleton Class of Equal Sets is Singleton Class
\(\ds \) \(=\) \(\ds \set {\set {\O, c}, \set {\set \O, d} }\) by definition

Thus we have that:

$c = d = \set {\O, \set \O}$

and so:

$c = \set \O$ and $d = \O$

leading to the result that $a = c$ and $b = d$.

$\Box$


Suppose otherwise, that either $a \ne \set \O$ or $b \ne \O$, or both.


Let $x \in \set {\set {\O, a}, \set {\set \O, b} }$.

Then either:

$x = \set {\O, a}$

or:

$x = \set {\set \O, b}$


First suppose $x = \set {\O, a}$.

Then:

$x = \set {\O, c}$ or $x = \set {\set \O, d}$

Aiming for a contradiction, suppose $x = \set {\set \O, d}$.

That is:

$\set {\O, a} = \set {\set \O, d}$

which means that $a = \set \O$ and $d = \O$.

But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\set \O, d}$.

Thus $x = \set {\O, c}$.

We have that:

$x = \set {\O, a}$ and $x = \set {\O, c}$

and so:

$\set {\O, a} = \set {\O, c}$

It follows from the lemma that:

$a = c$


Now suppose $x = \set {\set \O, b}$.

Then:

$x = \set {\O, c} \lor \set {\set \O, d}$

Aiming for a contradiction, suppose $x = \set {\O, c}$.

That is:

$\set {\set \O, b} = \set {\O, c}$

which means that $b = \O$ and $c = \set \O$.

But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\O, c}$.

Thus $x = \set {\set \O, d}$.

We have that:

$x = \set {\set \O, b}$ and $x = \set {\set \O, d}$

and so:

$\set {\set \O, b} = \set {\set \O, d}$

It follows from the lemma that:

$b = d$

$\blacksquare$



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