Difference between revisions of "Ordered pair"

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$$\impliedby$$

Clearly if

$$a=a'$$ and $$b=b'$$ then $$(a,b)=\{\{a\},\{a,b\}\}=\{\{a'\},\{a',b'\}\}=(a',b')$$ and we're done.

$$\implies$$

Assume

$$(a,b)=\{\{a\},\{a,b\}\}=\{\{a'\},\{a',b'\}\}=(a',b')$$ .

If

$$a\ne b$$ then we must have $$\{a\}=\{a'\}$$ and $$\{a,b\}=\{a',b'\}$$ (as clearly $$\{a\}=\{a',b'\}$$ is false, there are either 2 or 1 elements not contained in $$\{a\}$$ that are in $$\{a',b'\}$$ - namely $$a'$$ and $$b'$$ )

Clearly

$$a=a'$$ , then $$\{a,b\}=\{a',b'\}\implies b=b'$$ .

If

$$a=b$$ then $$(a,a)=\{\{a\},\{a,a\}\}=\{\{a\}\}$$ , we know $$\{\{a\}\}=\{\{a'\},\{a',b'\}\}$$ so again using the Set theory axioms (namely Extensionality) we see $$a=a'=b'$$ so $$a=a'$$ and $$b=b'$$ holds here too. This completes the proof.