# Re: [isabelle] Instantiating to a lambda expression

My guess is that this instantiation is possible, but it's asking too much to expect the automation to invent this lambda-expression.
Larry Paulson
On 15 Jan 2011, at 15:52, John Munroe wrote:
> Hi,
>
> If i have
>
> consts
> foo :: "real => real"
> bar :: "real => real"
>
> and an axiom
>
> same_ax: "ALL (g1::real=>real) g2 x y. (y > x & g1 y - g1 x = g2 y -
> g2 x) --> g1 = g2"
>
> I can get a proof quite fine with:
>
> lemma lem1: "foo = bar"
> proof -
> have "ALL x y. y > x --> foo y - foo x = bar y - bar x"
> sorry
> then obtain r1 r2 where #: "r2 > r1" and ##: "foo r2 - foo r1 = bar
> r2 - bar r1"
> by auto
> then show ?thesis
> using same_ax
> by auto
> qed
>
> However, if I change the type of bar to:
>
> bar :: "real => real => real", the following won't go through:
>
> lemma lem1: "ALL x. foo x = bar x 3"
> proof -
> have "ALL x y. y > x --> foo y - foo x = bar y 3 - bar x 3"
> sorry
> then obtain r1 r2 where #: "r2 > r1" and ##: "foo r2 - foo r1 = bar
> r2 3 - bar r1 3"
> by auto
> then show ?thesis
> using same_ax
> by auto
>
> How come g1 or g2 can't be instantiated to %x. bar x 3, which is of
> type real=>real?
>
> Thanks
>
> John
>

*This archive was generated by a fusion of
Pipermail (Mailman edition) and
MHonArc.*