Re: [isabelle] Pairs/tuples


Thanks for that. But if I have

lst :: "nat*nat*nat"
where ax : "lst = (1,2,3)"

lemma "snd (snd lst) = 3"
using ax
apply simp

doesn't seem to prove it. I guess I need some form of a substitution so that (1,2,3) is substituted into lst?


On Mar 31, 2011 4:20pm, Christian Sternagel <c.sternagel at> wrote:

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, eg,

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


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


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


lst :: "((nat*nat)*nat) list"

where ax : "lst = [((1,2),3)]"

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

using ax


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?



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