*To*: cl-isabelle-users at lists.cam.ac.uk*Subject*: [isabelle] proof solution [ newbie question ]*From*: "Gregory Kulczycki" <gregwk at vt.edu>*Date*: Sat, 7 Jul 2007 16:11:04 -0400*Sender*: gkulczycki at gmail.com

I've been experimenting with the Isabelle prover and have reached a point in a proof where I have the following subgoal: /\ xs ys zs. [| hd xs = hd ys; tl xs @ zs = tl ys |] ==> xs @ zs = ys This looks obvious to me (since xs @ zs = ys is equivalent to hd xs # tl xs @ zs = hd ys # tl ys for non-empty xs) , but I don't know how to proceed. Any suggestions? Thanks, Greg

**Follow-Ups**:**Re: [isabelle] proof solution [ newbie question ]***From:*Tobias Nipkow

**Re: [isabelle] proof solution [ newbie question ]***From:*Amine Chaieb

- Previous by Date: [isabelle] Starting Isabelle from within PolyML
- Next by Date: Re: [isabelle] proof solution [ newbie question ]
- Previous by Thread: Re: [isabelle] Starting Isabelle from within PolyML
- Next by Thread: Re: [isabelle] proof solution [ newbie question ]
- Cl-isabelle-users July 2007 archives indexes sorted by: [ thread ] [ subject ] [ author ] [ date ]
- Cl-isabelle-users list archive Table of Contents
- More information about the Cl-isabelle-users mailing list