Re: [isabelle] Evaluation of record expressions
Hello Thomas, hello all,
you're right, I really failed...Thanks for your counter example. Well the reason behind this is that i work with a custom package of theorems and used the wrong lemma which resulted in an unsolveable goal. So I'm really sorry.
Anyway, I managed to correct this and now Isabelle outputs this subgoal:
0 < (a::real) ⟹
a < 1 ⟹
0 < b ⟹
b < 1 ⟹
0 < a + b - a * b ⟹
∀s. t s = 0 ∨ t s = 1 ⟹
∀s. c s = 0 ∨ c s = 1 ⟹
(THE ba. isLub UNIV
(range (λs. (if t s = 0 ∧ c s = 0 then 1 else 0) + (if t s = 0 ∧ c s = 1 then 1 else 0) * a / (a + b - a * b) +
(if t s = 1 ∧ c s = 1 then 1 else 0) * (1 - b) * a / (a + b - a * b)))
≤ (THE ba. isLub UNIV
(range (λs. (if c s ≠ 1 then 1 else 0) *
((if t s = 0 ∧ c s = 0 then 1 else 0) + (if t s = 0 ∧ c s = 1 then 1 else 0) * a / (a + b - a * b) +
(if t s = 1 ∧ c s = 1 then 1 else 0) * (1 - b) * a / (a + b - a * b))))
The remaining functions are all defined in Set.thy/Lubs.thy and now I am stuck on how to prove this. This subgoal just shows that the upper bound of
the first lambda function is less or equal than the upper bound of the second function. Any help would be appreciated. I hope I didn't forget anything this time. Thanks!
Am 15-10-13, schrieb Thomas Sewell <thomas.sewell at nicta.com.au>:
> Hey all.
> I had a look at this on the assumption that this was somehow related to the record package.
> Looking at this test lemma, it would seem that
> a) it is false in the case where "t s = 1 & c s = 1 & a * 2 = 1 & b * 2 = 1"
> b) it seems to have nothing to do with the record package
> c) nothing needs to be evaluated, instead, hypotheses need to be proven, which is hard when they're false.
> You can see that the lemma is false via:
> apply (rule allI)
> apply (case_tac "t s = 1 & c s = 1 & a = 0.5 & b = 0.5")
> apply (elim conjE, simp only: )
> apply (simp add: field_simps)
> (the fact that all these tools are conservative means the original goal was false also)
> I'm not exactly sure what you're looking for.
> Good luck,
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