Isabelle Failed to refine any pending goal during instantiation - isabelle

datatype 'a list = Cons 'a "'a list" | Nil
instantiation list :: (order) order
begin
fun less_eq_list :: "'a list ⇒ 'a list ⇒ bool" where
"less_eq_list Nil Nil = True" |
"less_eq_list (Cons _ _) Nil = True" |
"less_eq_list Nil (Cons _ _) = False" |
"less_eq_list (Cons _ a) (Cons _ b) = less_eq_list a b"
instance
proof
fix x y:: "'a list"
show "x ≤ x"
apply(induct_tac x)
apply(auto)
done
(* at this point the state is
show x ≤ x
Successful attempt to solve goal by exported rule:
?x2 ≤ ?x2
proof (state)
this:
x ≤ x
goal (3 subgoals):
1. ⋀x y. (x < y) = (x ≤ y ∧ ¬ y ≤ x)
2. ⋀x y z. x ≤ y ⟹ y ≤ z ⟹ x ≤ z
3. ⋀x y. x ≤ y ⟹ y ≤ x ⟹ x = y
*)
show "(x < y) = (x ≤ y ∧ ¬ y ≤ x)"
(* I get an error here
Failed to refine any pending goal
Local statement fails to refine any pending goal
Failed attempt to solve goal by exported rule:
(?x2 < ?y2) = (?x2 ≤ ?y2 ∧ ¬ ?y2 ≤ ?x2)
*)
qed
end
What is wrong with this? The proof of "x ≤ x" worked like a charm. Somehow "(x < y) = (x ≤ y ∧ ¬ y ≤ x)" doesn't match any subgoal.

Class order is a subclass of preorder, which in turn is a subclass of ord. Class ord requires you to define both less_eq (≤) and less (<). In your code, you have correctly defined less_eq_list but forgot to define less_list, and that's why you got an error when trying to prove (x < y) = (x ≤ y ∧ ¬ y ≤ x).

Related

Why do I get this exception on an induction rule for a lemma?

I am trying to prove the following lemma (which is the meaning formula for the addition of two Binary numerals).
It goes like this :
lemma (in th2) addMeaningF_2: "∀m. m ≤ n ⟹ (m = (len x + len y) ⟹ (evalBinNum_1 (addBinNum x y) = plus (evalBinNum_1 x) (evalBinNum_1 y)))"
I am trying to perform strong induction. When I apply(induction n rule: less_induct) on the lemma, it throws an error.
exception THM 0 raised (line 755 of "drule.ML"):
infer_instantiate_types: type ?'a of variable ?a
cannot be unified with type 'b of term n
(⋀x. (⋀y. y < x ⟹ ?P y) ⟹ ?P x) ⟹ ?P ?a
Can anyone explain this?
Edit:
For more context
locale th2 = th1 +
fixes
plus :: "'a ⇒ 'a ⇒ 'a"
assumes
arith_1: "plus n zero = n"
and plus_suc: "plus n (suc m) = suc ( plus n m)"
len and evalBinNum_1 are both recursive functions
len gives us the length of a given binary numeral, while evalBinNum_1 evaluates binary numerals.
fun (in th2) evalBinNum_1 :: "BinNum ⇒ 'a"
where
"evalBinNum_1 Zero = zero"|
"evalBinNum_1 One = suc(zero)"|
"evalBinNum_1 (JoinZero x) = plus (evalBinNum_1 x) (evalBinNum_1 x)"|
"evalBinNum_1 (JoinOne x) = plus (plus (evalBinNum_1 x) (evalBinNum_1 x)) (suc zero)"
The problem is that Isabelle cannot infer the type of n (or the bound occurrence of m) when trying to use the induction rule less_induct. You might want to add a type annotation such as (n::nat) in your lemma. For the sake of generality, you might want to state that the type of n is an instance of the class wellorder, that is, (n::'a::wellorder). On another subject, I think there is a logical issue with your lemma statement: I guess you actually mean ∀m. m ≤ (n::nat) ⟶ ... ⟶ ... or, equivalently, ⋀m. m ≤ (n::nat) ⟹ ... ⟹ .... Finally, it would be good to know the context of your problem (e.g., there seems to be a locale th2 involved) for a more precise answer.

Isabelle structure proof

There is a set of some structures. I'm trying to prove that the cardinality of the set equals some number. Full theory is too long to post here. So here is a simplified one just to show the idea.
Let the objects (which I need to count) are sets containing natural numbers from 1 to n. The idea of the proof is as follows. I define a function which transforms sets to lists of 0 and 1. Here is the function and its inverse:
fun set_to_bitmap :: "nat set ⇒ nat ⇒ nat ⇒ nat list" where
"set_to_bitmap xs x 0 = []"
| "set_to_bitmap xs x (Suc n) =
(if x ∈ xs then Suc 0 else 0) # set_to_bitmap xs (Suc x) n"
fun bitmap_to_set :: "nat list ⇒ nat ⇒ nat set" where
"bitmap_to_set [] n = {}"
| "bitmap_to_set (x#xs) n =
(if x = Suc 0 then {n} else {}) ∪ bitmap_to_set xs (Suc n)"
value "set_to_bitmap {1,3,7,8} 1 8"
value "bitmap_to_set (set_to_bitmap {1,3,7,8} 1 8) 1"
Then I plan to prove that 1) a number of 0/1 lists with length n equals 2^^n,
2) the functions are bijections,
3) so the cardinality of the original set is 2^^n too.
Here are some auxiliary definitions and lemmas, which seems useful:
definition "valid_set xs n ≡ (∀a. a ∈ xs ⟶ 0 < a ∧ a ≤ n)"
definition "valid_bitmap ps n ≡ length ps = n ∧ set ps ⊆ {0, Suc 0}"
lemma length_set_to_bitmap:
"valid_set xs n ⟹
x = Suc 0 ⟹
length (set_to_bitmap xs x n) = n"
apply (induct xs x n rule: set_to_bitmap.induct)
apply simp
sorry
lemma bitmap_members:
"valid_set xs n ⟹
x = Suc 0 ⟹
set_to_bitmap xs x n = ps ⟹
set ps ⊆ {0, Suc 0}"
apply (induct xs x n arbitrary: ps rule: set_to_bitmap.induct)
apply simp
sorry
lemma valid_set_to_valid_bitmap:
"valid_set xs n ⟹
x = Suc 0 ⟹
set_to_bitmap xs x n = ps ⟹
valid_bitmap ps n"
unfolding valid_bitmap_def
using bitmap_members length_set_to_bitmap by auto
lemma valid_bitmap_to_valid_set:
"valid_bitmap ps n ⟹
x = Suc 0 ⟹
bitmap_to_set ps x = xs ⟹
valid_set xs n"
sorry
lemma set_to_bitmap_inj:
"valid_set xs n ⟹
valid_set xy n ⟹
x = Suc 0 ⟹
set_to_bitmap xs x n = ps ⟹
set_to_bitmap ys x n = qs ⟹
ps = qs ⟹
xs = ys"
sorry
lemma set_to_bitmap_surj:
"valid_bitmap ps n ⟹
x = Suc 0 ⟹
∃xs. set_to_bitmap xs x n = ps"
sorry
lemma bitmap_to_set_to_bitmap_id:
"valid_set xs n ⟹
x = Suc 0 ⟹
bitmap_to_set (set_to_bitmap xs x n) x = xs"
sorry
lemma set_to_bitmap_to_set_id:
"valid_bitmap ps n ⟹
x = Suc 0 ⟹
set_to_bitmap (bitmap_to_set ps x) x n = ps"
sorry
Here is a final lemma:
lemma valid_set_size:
"card {xs. valid_set xs n} = 2 ^^ n"
Does this approach seem valid? Are there any examples of such a proof? Could you suggest an idea on how to prove the lemmas? I'm stuck because the induction with set_to_bitmap.induct seems to be not applicable here.
In principle, that kind of approach does work: if you have a function f from a set A to a set B and an inverse function to it, you can prove bij_betw f A B (read: f is a bijection from A to B), and that then implies card A = card B.
However, there are a few comments that I have:
You should use bool lists instead of nat lists if you can only have 0 or 1 in them anyway.
It is usually better to use existing library functions than to define new ones yourself. Your two functions could be defined using library functions like this:
set_to_bitmap :: nat ⇒ nat ⇒ nat set ⇒ bool list
set_to_bitmap x n A = map (λi. i ∈ A) [x..<x+n]
bitmap_to_set :: nat ⇒ bool list ⇒ nat set
bitmap_to_set n xs = (λi. i + n) ` {i. i < length xs ∧ xs ! i}```
Side note: I would use upper-case letters for sets, not something like xs (which is usually used for lists).
Perhaps this is because you simplified your problem, but in its present form, valid_set A n is simply the same as A ⊆ {1..n} and the {A. valid_set A n} is simply Pow {1..n}. The cardinality of that is easy to show with results from the library:
lemma "card (Pow {1..(n::nat)}) = 2 ^ n"
by (simp add: card_Pow)`
As for your original questions: Your first few lemmas are provable, but for the induction to go through, you have to get rid of some of the unneeded assumptions first. The x = Suc 0 is the worst one – there is no way you can use induction if you have that as an assumption, because as soon as you do one induction step, you increase x by 1 and so you won't be able to apply your induction hypothesis. The following versions of your first three lemmas go through easily:
lemma length_set_to_bitmap:
"length (set_to_bitmap xs x n) = n"
by (induct xs x n rule: set_to_bitmap.induct) auto
lemma bitmap_members:
"set (set_to_bitmap xs x n) ⊆ {0, Suc 0}"
by (induct xs x n rule: set_to_bitmap.induct) auto
lemma valid_set_to_valid_bitmap: "valid_bitmap (set_to_bitmap xs x n) n"
unfolding valid_bitmap_def
using bitmap_members length_set_to_bitmap by auto
I also recommend not adding "abbreviations" like ps = set_to_bitmap xs x n as an assumption. It doesn't break anything, but it tends to complicate things needlessly.
The next lemma is a bit trickier. Due to your recursive definitions, you have to generalise the lemma first (valid_bitmap requires the set to be in the range from 1 to n, but once you make one induction step it has to be from 2 to n). The following works:
lemma valid_bitmap_to_valid_set_aux:
"bitmap_to_set ps x ⊆ {x..<x + length ps}"
by (induction ps x rule: bitmap_to_set.induct)
(auto simp: valid_bitmap_def valid_set_def)
lemma valid_bitmap_to_valid_set:
"valid_bitmap ps n ⟹ valid_set (bitmap_to_set ps 1) n"
using valid_bitmap_to_valid_set_aux unfolding valid_bitmap_def valid_set_def
by force
Injectivity and surjectivity (which is your ultimate goal) should follow from the fact that the two are inverse functions. Proving that will probably be doable with induction, but will require a few generalisations and auxiliary lemmas. It should be easier if you stick to the non-recursive definition using library functions that I sketched above.

How to define a termination order for the function with fmmap_keys?

I'm trying to define a supremum operation for a datatype based on fmap:
datatype t = A | B | C "(nat, t) fmap"
abbreviation
"supc f xs ys ≡
fmmap_keys
(λk x. f x (the (fmlookup ys k)))
(fmfilter (λk. k |∈| fmdom ys) xs)"
fun sup_t (infixl "⊔" 65) where
"A ⊔ _ = A"
| "B ⊔ B = B"
| "B ⊔ _ = A"
| "C xs ⊔ C ys = C (supc (⊔) xs ys)"
| "C xs ⊔ _ = A"
And get the error:
Unfinished subgoals:
(a, 1, <):
1. ⋀ys x. size (the (fmlookup ys x)) < Suc (∑x∈fset (fset_of_fmap ys). Suc (case x of (a, x) ⇒ size x))
(a, 1, <=):
1. ⋀ys x. size (the (fmlookup ys x)) ≤ Suc (∑x∈fset (fset_of_fmap ys). Suc (case x of (a, x) ⇒ size x))
(a, 2, <):
1. ⋀xs xa. size xa < Suc (∑x∈fset (fset_of_fmap xs). Suc (case x of (a, x) ⇒ size x))
(a, 2, <=):
1. ⋀xs xa. size xa ≤ Suc (∑x∈fset (fset_of_fmap xs). Suc (case x of (a, x) ⇒ size x))
(a, 3, <):
1. False
Calls:
a) (C xs, C ys) ~> (xa, the (fmlookup ys x))
Measures:
1) λp. size (snd p)
2) λp. size (fst p)
3) size
Result matrix:
1 2 3
a: ? ? <=
Could not find lexicographic termination order.
If I simplify the function passed as the first argument to fmmap_keys, then the error disappears:
abbreviation
"supc f xs ys ≡
fmmap_keys
(λk x. x)
(fmfilter (λk. k |∈| fmdom ys) xs)"
So I guess, that the error is caused by a complex recursive call of sup_t. The only possible source of non-termination is structures of the form C («[x ↦ C (...)]»). But an external C is removed on each recursive call so the function should terminate.
Could you suggest how to fix this error or redefine supc?
UPDATE
Here is an alternative definition:
abbreviation
"supc f xs ys ≡
fmap_of_list (map
(λ(k, x). (k, f x (the (fmlookup ys k))))
(sorted_list_of_fmap (fmfilter (λk. k |∈| fmdom ys) xs)))"
function sup_t (infixl "⊔" 65) where
"A ⊔ _ = A"
| "B ⊔ x = (if x = B then B else A)"
| "C xs ⊔ x = (case x of C ys ⇒ C (supc sup_t xs ys) | _ ⇒ A)"
by pat_completeness auto
termination
apply auto
I have to prove the following subgoal:
⋀a b. sup_t_dom (a, b)
How to unfold sup_t_dom?
Please find a potentially viable solution in the code listing below.
Background
The issue that you have encountered is described partially in the document "Defining Recursive Functions in Isabelle/HOL" written by Alexander Krauss (also known as "Tutorial on Function Definitions" in Isabelle documentation) and more comprehensively in the PhD thesis "Automating Recursive Definitions and Termination Proofs in Higher-Order Logic" that was also written by Alexander Krauss. In particular, see Chapter 4 in the tutorial and section 3.3 in the thesis.
Size of t
From the aforementioned references, it is possible to infer that one way to prove the termination of sup_t is to provide a suitable measure function. In this case, it is apparent that a measure function that is associated with the size of the datatype might be suitable for the application. Unfortunately, t is a nested type and (in this particular case) the default function size does not seem to capture the recursive nature of the datatype - this is not always the case (see section 3.3.2 in the thesis).
As the first step, I provided a new size function for t. The definition is based on the total number of Cs contained within x::t (the definition should be easy to modify to suit your needs for other applications).
Measure and Termination
I found that the measure function (λ(xs, ys). size ys) is suitable to prove the termination of sup_t. Also, this measure function is used in Isabelle to prove the termination of sup_t if it is declared with the command fun. However, in this case, it was not possible to prove that the arguments of recursive calls indeed decrease with respect to the relation that was established by the measure automatically. However, it would be sufficient to show that "size (the (fmlookup x k)) < size (C x)".
Unfortunately, the function supc (as stated in your question) is not self-certifying with respect to the property that the first argument that is passed to (λk x. f x (the (fmlookup ys k))) is in the domain of ys. Therefore, the (fmlookup ys k) can take the value the None. Given that this issue seems to be nearly orthogonal to the main topic of the question, I decided not to investigate it further and made an amendment to the function supc to ensure that it is guaranteed to return a concrete term of t (you may want to prove explicitly that the function specified below is identical in its behaviour to the one that you provided in your question or, otherwise, provide a better alternative that would be self-certifying):
abbreviation
"supc f xs ys ≡
fmmap_keys
(λk x. if (k |∈| fmdom ys) then (f x (the (fmlookup ys k))) else A)
(fmfilter (λk. k |∈| fmdom ys) xs)"
After this modification, the previous goal "size (the (fmlookup x k)) < size (C x)" was changed to "(k |∈| fmdom ys) ⟹ size (the (fmlookup x k)) < size (C x)", which could be easily proven (see lemma measure_cond). If this lemma is declared as an introduction rule then the termination of sup_t can be proven automatically if it is declared with the command fun.
Remarks
The main reason why I decided to investigate this issue and provide an answer is that I knew very little about some of the main topics of the question and wanted to learn them. As a result, my answer may be suboptimal due to the lack of experience/knowledge in these areas. Of course, if you also have doubts about whether the solution that I proposed here is optimal for the application, it may be worth trying to ask the question on the mailing list.
theory termination_problem
imports
Complex_Main
"HOL-Library.Finite_Map"
begin
datatype (plugins del: "size") t = A | B | C "(nat, t) fmap"
abbreviation "tcf ≡ (λ v::(nat × nat). (λ r::nat. snd v + r))"
interpretation tcf: comp_fun_commute tcf
proof
fix x y
show "tcf y ∘ tcf x = tcf x ∘ tcf y"
proof -
fix z
have "(tcf y ∘ tcf x) z = snd y + snd x + z" by auto
also have "(tcf x ∘ tcf y) z = snd y + snd x + z" by auto
ultimately have "(tcf y ∘ tcf x) z = (tcf x ∘ tcf y) z" by auto
then show "(tcf y ∘ tcf x) = (tcf x ∘ tcf y)" by auto
qed
qed
instantiation t :: size
begin
primrec t_size :: "t ⇒ nat" where
AR: "t_size A = 0" |
BR: "t_size B = 0" |
CR: "t_size (C x) =
(Suc 0) + ffold tcf 0 (fset_of_fmap (fmmap t_size x))"
definition size_t where
size_t_def: "size_t = t_size"
instance ..
end
lemma ffold_rec_exp:
assumes "k |∈| fmdom x"
and "ky = (k, the (fmlookup (fmmap t_size x) k))"
shows "ffold tcf 0 (fset_of_fmap (fmmap t_size x)) =
tcf ky (ffold tcf 0 ((fset_of_fmap (fmmap t_size x)) |-| {|ky|}))"
using assms tcf.ffold_rec by auto
lemma elem_le_ffold:
assumes "k |∈| fmdom x"
shows "t_size (the (fmlookup x k)) <
(Suc 0) + ffold tcf 0 (fset_of_fmap (fmmap t_size x))"
using ffold_rec_exp assms by auto
lemma measure_cond [intro]:
assumes "k |∈| fmdom x"
shows "size (the (fmlookup x k)) < size (C x)"
using assms elem_le_ffold size_t_def by auto
abbreviation
"supc f xs ys ≡
fmmap_keys
(λk x. if (k |∈| fmdom ys) then (f x (the (fmlookup ys k))) else A)
(fmfilter (λk. k |∈| fmdom ys) xs)"
fun sup_t (infixl "⊔" 65) where
"A ⊔ _ = A"
| "B ⊔ x = (if x = B then B else A)"
| "C xs ⊔ x = (case x of C ys ⇒ C (supc sup_t xs ys) | _ ⇒ A)"
(*Examples*)
abbreviation "list_1 ≡ fmap_of_list [(1::nat, B)]"
abbreviation "list_2 ≡ fmap_of_list [(1::nat, A), (2::nat, A)]"
value "(C list_1) ⊔ (C list_2)"
abbreviation "list_3 ≡ fmap_of_list [(1::nat, B), (3::nat, A)]"
abbreviation "list_4 ≡ fmap_of_list [(2::nat, A), (4::nat, B)]"
value "(C list_3) ⊔ (C list_4)"
abbreviation "list_5 ≡ fmap_of_list [(1::nat, B), (2::nat, B)]"
abbreviation "list_6 ≡ fmap_of_list [(2::nat, B), (4::nat, B)]"
value "(C list_5) ⊔ (C list_6)"
abbreviation "list_7 ≡
fmap_of_list [(1::nat, B), (2::nat, C list_5), (3::nat, A)]"
abbreviation "list_8 ≡ fmap_of_list [(2::nat, B), (4::nat, B)]"
value "(C list_7) ⊔ (C list_8)"
abbreviation "list_9 ≡
fmap_of_list [(1::nat, B), (2::nat, C list_5), (3::nat, A)]"
abbreviation "list_10 ≡ fmap_of_list [(2::nat, C list_6), (3::nat, B)]"
value "(C list_9) ⊔ (C list_10)"
end

How to define a linear ordering on a type?

I'm trying to define a conjunction function for 4-valued logic (false, true, null, and error). In my case the conjunction is equivavlent to min function on linear order false < error < null < true.
datatype bool4 = JF | JT | BN | BE
instantiation bool4 :: linear_order
begin
fun leq_bool4 :: "bool4 ⇒ bool4 ⇒ bool" where
"leq_bool4 JF b = True"
| "leq_bool4 BE b = (b = BE ∨ b = BN ∨ b = JT)"
| "leq_bool4 BN b = (b = BN ∨ b = JT)"
| "leq_bool4 JT b = (b = JT)"
instance proof
fix x y z :: bool4
show "x ⊑ x"
by (induct x) simp_all
show "x ⊑ y ⟹ y ⊑ z ⟹ x ⊑ z"
by (induct x; induct y) simp_all
show "x ⊑ y ⟹ y ⊑ x ⟹ x = y"
by (induct x; induct y) simp_all
show "x ⊑ y ∨ y ⊑ x"
by (induct x; induct y) simp_all
qed
end
definition and4 :: "bool4 ⇒ bool4 ⇒ bool4" where
"and4 a b ≡ minimum a b"
I guess there must be an easier way to define a linear order in Isabelle HOL. Could you suggest a simplification of the theory?
You can use the "Datatype_Order_Generator" AFP entry.
Then it's as simple as importing "$AFP/Datatype_Order_Generator/Order_Generator" and declaring derive linorder "bool4". Note that the constructors must be declared in the order you want them when defining your datatype.
Details on how to download and use the AFP locally can be found here.

What happens during function proofs

I am trying to proof a property of the icmp6 checksum function (sum 16bit integers, add carry, invert 16bit integer).
I defined the functions in isabelle. (I know my proofs are terrible)
But for some reason, isabelle can't proof something about the icmp_csum function, it wants to have.
When I replace the oops in the paste with done it produces thousands of lines that just says:
"linarith_split_limit exceeded (current value is 9)"
theory Scratch
imports Main Int List
begin
fun norm_helper :: "nat ⇒ nat" where
"norm_helper x = (let y = divide x 65536 in (y + x - y * 65536))"
lemma "x ≥ 65536 ⟹ norm_helper x < x" by simp
lemma h: "norm_helper x ≤ x" by simp
fun normalize :: "nat ⇒ nat" where
"normalize x = (if x ≥ 65536
then normalize (norm_helper x)
else x)"
inductive norm_to :: "nat ⇒ nat ⇒ bool" where
"(x < 65536) ⟹ norm_to x x"
| "norm_to y z ⟹ y = norm_helper x ⟹ norm_to x z"
lemma ne: "norm_to x y ⟹ y = normalize x"
apply (induct x y rule: norm_to.induct) by simp+
lemma i: "norm_to x y ⟹ x ≥ y"
apply (induct x y rule: norm_to.induct) by simp+
lemma l: "norm_to x y ⟹ y < 65536"
apply (induct x y rule: norm_to.induct) by simp+
lemma en: "y = normalize x ⟹ norm_to x y"
apply (induct x rule: normalize.induct)
proof -
fix x :: nat
assume 1: "(x ≥ 65536 ⟹ y = Scratch.normalize (norm_helper x) ⟹ norm_to (norm_helper x) y)"
assume 2: "y = Scratch.normalize x"
show "norm_to x y"
proof (cases "x ≥ 65536")
show "¬ 65536 ≤ x ⟹ norm_to x y"
using norm_to.intros(1)[of x] 2 by simp
{
assume s: "65536 ≤ x"
have d: "y = normalize (norm_helper x)" using 2 s by simp
show "65536 ≤ x ⟹ norm_to x y"
using 1 d norm_to.intros(2)[of "norm_helper x" y x]
by blast
}
qed
qed
lemma "normalize x ≤ x" using en i by simp
lemma n[simp]: "normalize x < 65536" using en l by blast
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
| "sum (x#xs) = x + sum xs"
fun csum :: "nat list ⇒ nat" where
"csum xs = normalize (sum xs)"
fun invert :: "nat ⇒ nat" where
"invert x = 65535 - x"
lemma c: "csum xs ≤ 65535" using n[of "sum xs"] by simp
lemma ic: "invert (csum xs) ≥ 0" using c[of xs] by blast
lemma asdf:
assumes "xs = ys"
shows "invert (csum xs) = invert (csum ys)"
using HOL.arg_cong[of "csum xs" "csum ys" invert,
OF HOL.arg_cong[of xs ys csum]] assms(1)
by blast
function icmp_csum :: "nat list ⇒ nat" where
"icmp_csum xs = invert (csum xs)"
apply simp
apply (rule asdf)
apply simp
oops
end
I have no idea why there is tracing output from linarith there, but given that your definition is neither recursive nor performs pattern matching, you can write it as a definition:
definition icmp_csum :: "nat list ⇒ nat" where
"icmp_csum xs = invert (csum xs)"
Another possibility is to change invert to a definition instead of a fun. (In general, if it's neither recursive nor performs pattern matching, definition is preferable because it has much less overhead than fun.)
NB, just import Main, not Main Int List.
Edit: An explanation from Tobias Nipkow on the mailing list:
This is a known issue. In the outdated LNCS 2283 you can find a discussion what to do about it in Section 3.5.3 Simplification and Recursive Functions. The gist: don't use "if", use pattern matching or "case". Or disable if_split.

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