Re: [isabelle] Adding lemmas to HOL?
You can define the longest common prefix with the existing functions in List:
lcp xs ys = map fst (takeWhile (%(x, y). x = y) (zip xs ys))
In AFP/Coinductive, I have defined an operation llcp to return the longest common prefix
of two possibly infinite lists. There are also a few lemmas about llcp.
On 27/05/16 13:36, Joachim Breitner wrote:
so much â quantifiers. Given that for any two lists, the maximal common
prefix is unique, wouldnât it make sense to give this a name, define it
straight-forwardly using fun, and then possibly add lemmas?
Am Freitag, den 27.05.2016, 12:05 +0200 schrieb Christoph Dittmann:
I am looking for lemmas that show that from every pair of lists you
split off a maximal common prefix or suffix. I am looking for this:
shows "âps xs' ys'. xs = ps @ xs' â ys = ps @ ys'
â (xs' =  â ys' =  â hd xs' â hd ys')"
by (induct xs ys rule: list_induct2', blast, blast, blast)
(metis (no_types, hide_lams) append_Cons append_Nil list.sel(1))
shows "âss xs' ys'. xs = xs' @ ss â ys = ys' @ ss
â (xs' = Nil â ys' = Nil â last xs' â last ys')"
using maximal_common_prefix[of "rev xs" "rev ys"]
unfolding rev_swap rev_append by (metis last_rev rev_is_Nil_conv)
The proofs work, but I would have expected that these statements
exist somewhere in src/HOL/List.thy. Unfortunately, I couldn't find
these or something similar. I would be glad if someone corrected me.
These two lemmas look to me like they could be useful for many proofs
because they give a general way to decompose two lists.
So my question is, what is the process to propose new lemmas for HOL?
Do these two lemmas look reasonable enough (assuming they don't
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