# Re: [isabelle] Pairs/tuples

```Hi,

there are only pairs in Isabelle and the pair constructor * is
right-associative. Hence when you write

nat * nat * nat

it really is

nat * (nat * nat)

and a corresponding value would be (1, (2, 3)) (again you can save
parentheses by writing (1, 2, 3) instead). Thus getting the third
element is done by nested calls to snd, e.g.,

lemma "snd (snd (1, (2, 3))) = 3" by simp

You see that I do not need a sledgehammer to prove the fact ;). Mere
rewriting is enough. The same is true for your initial lemma

lemma "snd (snd ([(1, 2, 3)] ! 0)) = 3" by simp

hope this helps

chris

```
```On 03/31/2011 04:42 PM, Steve W wrote:
```
```Greetings,

I'm currently experimenting with pairs/tuples, like the following:

axiomatization
lst :: "((nat*nat)*nat) list"
where ax : "lst = [((1,2),3)]"

lemma "snd (lst ! 0) = (3::nat)"
using ax
sledgehammer

Does anyone know how I can prove that lemma since sledgehammer can't
seem to
find a proof?

Also, if lst was of type "(nat * nat * nat) list", then how do I read the
3rd element from a tuple?

TIA

Steve

```
```
```
```

```

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