*To*: Gregory Kulczycki <gregwk at vt.edu>*Subject*: Re: [isabelle] proof solution [ newbie question ]*From*: Tobias Nipkow <nipkow at in.tum.de>*Date*: Sun, 08 Jul 2007 10:43:46 +0200*Cc*: cl-isabelle-users at lists.cam.ac.uk*In-reply-to*: <15677a8b0707071311t75dcb4e3k69931327f984a1ef@mail.gmail.com>*References*: <15677a8b0707071311t75dcb4e3k69931327f984a1ef@mail.gmail.com>*User-agent*: Thunderbird 2.0.0.0 (Macintosh/20070326)

Tobias Gregory Kulczycki schrieb:

I've been experimenting with the Isabelle prover and have reached apoint ina 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

**References**:**[isabelle] proof solution [ newbie question ]***From:*Gregory Kulczycki

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