|[isabelle] Domain proof||Roger H.||1/17/14 8:10 AM|
how can i prove
lemma "dom (SOME b. dom b = A) = A"
|Re: [isabelle] Domain proof||Peter Lammich||1/17/14 8:40 AM|
On Fr, 2014-01-17 at 18:03 +0200, Roger H. wrote:You have to show that there is such a beast b, ie,
obtain b where "dom b = A" ...
sledgehammer (*Should find a proof now, using the rules for SOME,
> Thank you!
|Re: [isabelle] Domain proof||Manuel Eberl||1/17/14 9:23 AM|
On my system, the nicest proof sledgehammer finds (after I provide a
suitable witness, as Peter suggested) is something like "by (metis
(lifting, full_types) dom_const dom_restrict inf_top.left_neutral
someI_ex)", which takes over a second to run.
These proofs often get a bit tricky because Higher Order Unification
does strange things with the someI/someI_ex rules and therefore, one
often has to instantiate them manually. I would prove your lemmas like this:
by (rule someI_ex[where P = "λb. dom b = A"],
rule exI[where x = "(λ_. Some undefined) |` A"], simp)
Or, alternatively, if you prefer a more readable Isar proof:
let ?f = "(λ_. Some undefined) |` A"
have "dom ?f = A" by simp
thus ?thesis by (rule someI[where P = "λb. dom b = A"])
In case you're wondering what "(λ_. Some undefined) |` A" is: "(λ_. Some
undefined)" is simply a partial function that is defined everywhere and
always returns "undefined", which is some fixed value of your codomain
type about which you know nothing – except that it exists. |` A then
restricts this function to A, i.e. returns "None" everywhere except for
values in A. You could also write "(λx. if x ∈ A then Some undefined
Note that without the explicit "where" instantiations in someI and
someI_ex, it does not work because unification does not produce the
unifier I want. In fact, I'm curious as to why this happens myself. Can
anybody explain this?