User:Dfeuer/Equality of Ordered Pairs implies Equality of Elements

Theorem
Let $a$, $b$, $c$, and $d$ be sets.

Let $(a,b) = (c,d)$, where $(a,b)$ and $(c,d)$ are ordered pairs.

Proof
Let $(a,b) = (c,d)$.

Then by the definition of ordered pair:


 * $\{\{a\}, \{a, b\}\} = \{\{c\}, \{c, d\}\}$.

By the definition of User:Dfeuer/Definition:Unordered Pair:


 * $\{a\} \in \{\{a\}, \{a, b\}\}$ and $\{a, b\} \in \{\{a\}, \{a, b\}\}$

Thus:


 * $\{a\} \in \{\{c\}, \{c, d\}\}$ and $\{a, b\} \in \{\{c\}, \{c, d\}\}$

By the definition of unordered pair:


 * $\{a\} = \{c\}$ or $\{a\} = \{c, d\}$

First suppose that $\{a\} = \{c, d\}$.

By User:Dfeuer/Equality of Ordered Pairs implies Equality of Elements/Lemma 1:


 * $a = c = d$

Thus $\{\{c\}, \{c, d\}\} = \{\{a\}, \{a, a\}\}$.

Applying User:Dfeuer/Equality of Ordered Pairs implies Equality of Elements/Lemma 1 twice:


 * $\{\{a\}, \{a, a\}\} = \{\{a\}, \{a\}\} = \{\{a\}\}$

So $\{\{c\}, \{c, d\}\} = \{\{a\}\}$.

As $\{a, b\} \in \{\{c\}, \{c, d\}\}$, $\{a, b\} \in \{\{a\}\}$.

By the definition of singleton, $\{a, b\} = \{a\}$.

By User:Dfeuer/Equality of Ordered Pairs implies Equality of Elements/Lemma 1, $b = a$.

Thus we have $a = b = c = d$, so $a = c$ and $b = d$.

Suppose instead that $\{a\} = \{c\}$.

Then $a = c$ by User:Dfeuer/Singletons are Equal iff Elements are Equal.

Thus we have:


 * $\{\{a\}, \{a, b\}\} = \{\{a\}, \{a, d\}\}$.

Thus User:Dfeuer/Equality of Ordered Pairs implies Equality of Elements/Lemma 2 shows that:


 * $\{a, b\} = \{a, d\}$

Applying User:Dfeuer/Equality of Ordered Pairs implies Equality of Elements/Lemma 2 again shows that $b = d$.

Thus we have $a = c$ and $b = d$ in either case.