# Re: [isabelle] substitution

Hi,
> Hello,
> I had this problem when proving a result. I was trying to prove a result of
> the form
> [i+1< list.length l; (i+k+2) < list.length l) ===> ((take i l)@drop (i+2)
> l)!(k+i) = l!(i+2+k)"
>
> when I try to prove the same result in the next step subtituting n for i+k,
> as in the following, it does not seem to work. Using [of ...] does not seem
> to help either. Any suggestions?
Try this:
lemma assumes il: "i+1< length l" and ikl: "(i+k+2) < length l"
shows "((take i l)@drop (i+2) l)!(k+i) = l!(i+2+k)"
proof -
have lt: "length (take i l) = i" using il by auto
have "((take i l)@drop (i+2) l)!(k+i) = drop (i+2) l ! k"
unfolding nth_append using il lt by simp
also have "... = l!(i + 2 + k)"
by (rule nth_drop, insert ikl, simp)
finally show ?thesis by simp
qed
for finding the useful lemmas I usually recommend find_theorems which will give
you pointers to the essential lemmas nth_append and nth_drop.
find_theorems "(_ @ _) ! _ = _"
find_theorems "drop _ _ ! _ = _"
Cheers,
René
--
René Thiemann mailto:rene.thiemann at uibk.ac.at
Computational Logic Group http://cl-informatik.uibk.ac.at/~thiemann/
Institute of Computer Science phone: +43 512 507-6434
University of Innsbruck

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