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:

lemma maximal_common_prefix:
   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))

corollary maximal_common_suffix:
   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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