Re: [isabelle] newbie question

In general it is much better to use the "induct" method insead of the induct_tac. In your case you need to generalize over ys, since in the recursive call ys is modified by revH.

The following works for me:

lemma rev_revH: "revH xs ys = rev xs @ ys"
  by (induct xs arbitrary: ys, auto)

If you insist on induct_tac:

lemma rev_revH: "ALL ys. revH xs ys = rev xs @ ys"
  apply (induct_tac xs)
  apply simp_all


Tim Newsham wrote:
[possible duplicate, I jumped the gun on sending the first before I was properly subscribed].

I'm trying to prove a simple proof related to the "reverse" proof
in the tutorial (I have the rest of the tutorial theory here
as well including lemmas app_Nil2, app_assoc, rev_app and rev_rev):

--- snip ---
         revH :: "'a list => 'a list => 'a list"

  "revH [] ys = ys"
  "revH (x # xs) ys = revH xs (x # ys)"

  lemma rev_revH: "revH xs ys = rev xs @ ys"
  apply(induct_tac xs)

  lemma rev_rev2: "rev xs = revH xs []"
  apply(induct_tac xs)
--- snip ---

when I evaluate the first lemma it is able to automatically reduce the problem to the goal:

   forall a list.
       revH list ys = rev list @ ys ==>
       revH list (a # ys) = rev list @ a # ys

To me this seems to imply that this is solved, but I guess Isabelle doesn't see it that way. I tried to strengthen the proof by saying
"!! ys ." but that didn't seem to have any effect.  What do I need to
do here to complete this proof?

Tim Newsham

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