# tuple identity proof..

I have in one book little proof to create.

We know that tuple can be expressed as $$(a,b) = \{\{a\},\{a,b\}\}$$

and I have too prove that:

$$a = c \ \land b = d \implies (a,b) = (c,d)$$

I know this proof is very trivial but I'm a little too unexperienced in proofs so i want to ask if this proof I created is valid enought?

$$a=c\ \land b=d \implies a=c=x\ \land b=d=y \implies \{\{x\},\{x,y\}\} = \{\{a\}\{a,b\}\} = \{\{c\},\{c,d\}\} \implies (a,b) = (c,d)$$

Or is there any more basic proof?

• As it stands, your proof doesn't make any sense. To show $(a,b) = (c,d)$ you must show $\{\{a\},\{a,b\}\} = \{\{c\},\{c,d\}\}$ as set equality, i.e. all the elements in each are elements in the other. – Good Morning Captain Dec 28 '19 at 10:07
• but if $a = x$ then $\{a\} =\{x\}$ or can't I use this? – Patrik Bašo Dec 28 '19 at 10:10
• Sure, but more care is needed than that. See en.wikipedia.org/wiki/Ordered_pair#Kuratowski_definition for a proof of equality using this definition. – Good Morning Captain Dec 28 '19 at 10:14
• Note that the direction you want to prove is trivial, but it's the other direction that makes this definition of tuple useful: if $(a,b)=(c,d)$ then $a=c$ and $b=d$. – Berci Dec 28 '19 at 10:27
• IMV a proof is not needed. $a$ and $c$ are labels/notations of the same set and $b$ and $d$ are labels/notations for the same set. Then automatically $(a,b)$ and $(c,d)$ are labels/notations for the same set. What you are doing can be compared with e.g. "proving" the implication $x=y\implies x+1=y+1$. – drhab Dec 28 '19 at 10:59