Can't prove trivial lemma about function with non-standard recursion - coq

I'm having a great difficulty trying to prove even very simple lemmas about a function I defined. This is my definition:
Require Import List.
Require Export Omega.
Require Export FunInd.
Require Export Recdef.
Notation "A :: B" := (cons A B).
Notation "[]" := nil.
Notation "[[ A ]]" := (A :: nil).
Inductive tm :=
| E: nat -> tm
| L: list tm -> tm.
Definition T := list tm.
Fixpoint add_list (l: list nat) : nat :=
match l with
| [] => 0
| n :: l' => n + (add_list l')
end.
Fixpoint depth (t: tm) : nat :=
match t with
| E _ => 1
| L l => 1 + (add_list (map depth l))
end.
Definition sum_depth (l: T) := add_list (map depth l).
Function sum_total (l: T) {measure sum_depth l} : nat :=
match l with
| [] => 0
| [[E n]] => n
| [[L li]] => sum_total li
| E n :: l' => n + (sum_total l')
| L li :: l' => (sum_total li) + (sum_total l')
end.
Proof.
- auto.
- intros; unfold sum_depth; subst. simpl; omega.
- intros; subst; unfold sum_depth; simpl; omega.
- intros; subst; unfold sum_depth; simpl; omega.
Defined.
The inductive type can't be changed.
I can prove simple propositions like Lemma test : forall n, sum_total [[E n]] = n. using the compute tactic, but another trivial lemma like Lemma test2 : forall l, sum_total [[L l]] = sum_total l. hangs.

First, it seems OK that the compute tactic "hangs" on the goal you mention (because when using the Function … Proof. … Defined. definition methodology, your function sum_total incorporates some proof terms, which are not intended to be computed − all the more on an arbitrary argument l; maybe a tactic such as simpl or cbn would be more suitable in this context).
Independently of my comment on list notations, I had a closer look on your formalization and it seems the Function command is unneeded in your case, because sum_total is essentially structural, so you could use a mere Fixpoint, provided the inductive type you are looking at is slightly rephrased to be defined in one go as a mutually-defined inductive type (see the corresponding doc of the Inductive command in Coq's refman which gives a similar, typical example of "tree / forest").
To elaborate on your example, you may want to adapt your definition (if it is possible for your use case) like this:
Inductive tm :=
| E: nat -> tm
| L: T -> tm
with T :=
Nil : T
| Cons : forall (e : tm) (l : T), T.
Notation "[[ A ]]" := (Cons A Nil).
Fixpoint sum_total (l: T) {struct l} : nat :=
match l with
| Nil => 0
| [[E n]] => n
| [[L li]] => sum_total li
| Cons (E n) l' => n + (sum_total l')
| Cons (L li) l' => (sum_total li) + (sum_total l')
end.
(* and the lemma you were talking about is immediate *)
Lemma test2 : forall l, sum_total [[L l]] = sum_total l.
reflexivity.
Qed.
Otherwise (if you cannot rephrase your tm inductive like this), another solution would be to use another strategy than Function to define your sum_total function, e.g. Program Fixpoint, or the Equations plugin (which are much more flexible and robust than Function when dealing with non-structural recursion / dependently-typed pattern matching).
Edit: as the OP mentions the inductive type itself can't be changed, there is a direct solution, even when using the mere Function machinery: relying on the "equation lemma" that is automatically generated by the definition.
To be more precise, if you take your script as is, then you get the following lemma "for free":
Search sum_total "equation".
(*
sum_total_equation:
forall l : T,
sum_total l =
match l with
| [] => 0
| [[E n]] => n
| E n :: (_ :: _) as l' => n + sum_total l'
| [[L li]] => sum_total li
| L li :: (_ :: _) as l' => sum_total li + sum_total l'
end
*)
So you could easily state and prove the lemma you are interested in by doing:
Lemma test2 : forall l, sum_total [[L l]] = sum_total l.
intros l.
rewrite sum_total_equation.
reflexivity.
Qed.

Here is an answer that doesn't require changing the inductive type.
There is a simple definition of sum_total that is both comparatively easy to understand and gives (almost) the lemma you are looking for by compute.
Fixpoint sum_tm (t : tm) : nat :=
match t with
| E n => n
| L li => list_sum (map sum_tm li)
end.
Definition sum_total (l : T) : nat := list_sum (map sum_tm l).
Lemma test2 : forall l, sum_total [[L l]] = sum_total l + 0.
reflexivity.
Qed.
(list_sum comes from the List module.)
Notice how the definition of sum_tm and sum_total exactly follows the structure of the definition of term and T, with list_sum (composed with map) corresponding to the use of list. This pattern is in general effective for these problems with nested inductives.
If you want to get rid of the + 0, you can define a different version of list_sum that includes a case for the singleton list (and you can fuse this with map if you want, though it is not necessary).
That would look like replacing list_sum with list_sum_alt defined as
Fixpoint list_sum_alt (l : list nat) : nat :=
match l with
| [] => 0
| [[n]] => n
| n :: li => n + list_sum_alt li
end.
With this definition, test2 holds by compute.

Related

Conversion from not equal to equal in nat

I have function,whose output is some natural number.I have proved a lemma,that output of this function cannot be zero. It means output is equal to some natural number S m.I want to convert the above lemma.
Theorem greater:forall (m :nat)(l:list nat),
m=?0=false ->
0=? (f1 + m)=false->
(f1 + m)= S m.
The statement you entered does not type check. Regardless, I don't see how it could hold -- for instance, if by l you mean f1 : nat, then the statement would imply that 3 = 2.
Require Import Coq.Arith.Arith.
Theorem greater:forall (m :nat)(f1:nat),
m=?0=false ->
0=? (f1 + m)=false->
(f1 + m)= S m.
Admitted.
Lemma contra : False.
Proof.
pose proof (greater 1 2 eq_refl eq_refl).
easy.
Qed.
Proving that something that is not zero is a successor can be done as follows:
Require Import Coq.Arith.Arith.
Lemma not_zero_succ :
forall n, n <> 0 ->
exists m, n = S m.
Proof. destruct n as [|n]; eauto; easy. Qed.
Edit The complete statement you wrote below is also contradictory:
Require Import Coq.Arith.Arith.
Require Import Coq.Lists.List.
Import ListNotations.
Fixpoint lt_numb (n: nat) (l: list nat) : nat :=
match l with
| nil => 0
| h::tl =>
if h <? n then S (lt_numb n tl) else lt_numb n tl
end.
Fixpoint greatest (large: nat) (l: list nat) : nat :=
match large with
| O => 0
| S m' => (lt_numb large l) + (greatest m' l)
end.
Definition change (n: nat) (l: list nat) : list nat :=
match l with
| nil => l
| h::tl => if n <? h then l else n::tl
end.
Fixpoint g_value (elements: nat) (l: list nat) : nat :=
match l with
| nil => 0
| [n] => n
| h :: l =>
match elements with
| O => h
| S elements' => g_value elements' (change h l)
end
end.
Theorem no_elements : forall (m n z :nat)(l:list nat),
m=?0=false -> greatest(g_value (length (n :: l)) (n :: l) + m) (n :: l) = (S z).
Proof. Admitted.
Goal False.
pose proof (no_elements 1 0 1 [] eq_refl).
simpl in H.
discriminate.
Qed.

Defining a function that returns one element satisfying the condition

I want to declare a function that yeilds the element (b, n) that the b is equal to true.
Require Export List.
Import Coq.Lists.List.ListNotations.
Definition lstest := list (bool * nat).
Fixpoint existbool (l : lstest) : option (bool * nat) :=
match l with
| [] => None
| (b, n) :: l' => if b then Some (b, n) else existbool l'
end.
The function always get the first element satisfyting b = true. I want to express that there exists an element satisfyting b = true and returns the element. How can I define such a function?
In the following function the type of existbool_ex tells you that we output a pair contained in the list with its first element true (assuming we output a Some).
(* These are all from the standard library *)
Locate "{ _ : _ | _ }".
Print sig.
Print In.
Print fst.
(* Defining Property here to shorten code for exist *)
Definition P l (x : bool * nat) := fst x = true /\ In x l.
Fixpoint existbool_ex (l : list (bool * nat)) :
option {x : bool * nat | fst x = true /\ In x l} :=
match l return option {x : bool * nat | P l x} with
| [] => None
| x' :: l' =>
match x' with
| (true,n) as ans =>
Some (exist (P (ans :: l')) ans (conj eq_refl (or_introl eq_refl)))
| (false,n) =>
match existbool_ex l' with
| None => None
| Some (exist _ x a) =>
match a with
| conj Heq Hin =>
Some (exist (P ((false, n) :: l')) x (conj Heq (or_intror Hin)))
end
end
end
end.
(* Note the as pattern got desugared into a let binding. *)
Print existbool_ex.
(* However we have a somewhat sane extraction, (tail recursive) *)
Require Extraction.
Extraction existbool_ex.
You could write a function get_number that requires a proof that the list has a true value somewhere.
Definition has_true (l : lstest):= exists n, In (true, n) l.
get_number is defined with the help of refine which lets us leave 'holes' (written _) in the proof term to fill in later. Here we have two holes; one for the absurd case when the list is [], and one where we construct the proof term for the recursive call.
Fixpoint get_number (l:lstest) (H: has_true l) : nat.
refine (
match l as l' return l' = _ -> nat with
| (true, n)::_ => fun L => n
| (false, _)::l' => fun L => get_number l' _
| [] => fun L => _
end eq_refl).
now exfalso; subst l; inversion H.
now subst l; inversion H; inversion H0;
[congruence | eexists; eauto].
Defined.
The function uses the convoy pattern so that the match statement does not forget the shape of l in the different branches.
If you want to, you can prove rewriting lemmas to make it easier to use.
Lemma get_number_false l m H: exists H', get_number ((false, m)::l) H = get_number l H'.
Proof. eexists; reflexivity. Qed.
Lemma get_number_true l m H: get_number ((true, m)::l) H = m.
Proof. reflexivity. Qed.
Lemma get_number_nil H m: get_number [] H <> m.
Proof. now inversion H. Qed.
Lemma get_number_proof_irrel l H1 H2: get_number l H1 = get_number l H2.
Proof. induction l as [ | [[|] ?] l']; eauto; now inversion H1. Qed.

Teach coq to check termination

Coq, unlike many others, accepts an optional explicit parameter,which can be used to indicate the decreasing structure of a fixpoint definition.
From Gallina specification, 1.3.4,
Fixpoint ident params {struct ident0 } : type0 := term0
defines the syntax. but from it, we've known that it must be an identifier, instead of a general measure.
However, in general, there are recursive functions, that the termination is not quite obvious,or it in fact is, but just difficult for the termination checker to find a decreasing structure. For example, following program interleaves two lists,
Fixpoint interleave (A : Set) (l1 l2 : list A) : list A :=
match l1 with
| [] => []
| h :: t => h :: interleave l2 t
end
This function clearly terminates, while Coq just couldn't figure it out. The reason is neither l1 nor l2 are decreasing every cycle. But what if we consider a measure, defined to be length l1 + length l2? Then this measure clearly decreases every recursion.
So my question is, in the case of sophisticated situation, where code is not straightforward to be organized in a termination checkable way, how do you educate coq and convince it to accept the fixpoint definition?
You have multiple options and all of them boil down to structural recursion in the end.
Preamble
From Coq Require Import List.
Import ListNotations.
Set Implicit Arguments.
Structural recursion
Sometimes you can reformulate your algorithm in a structurally recursive way:
Fixpoint interleave1 {A} (l1 l2 : list A) {struct l1} : list A :=
match l1, l2 with
| [], _ => l2
| _, [] => l1
| h1 :: t1, h2 :: t2 => h1 :: h2 :: interleave1 t1 t2
end.
Incidentally, in some cases you can use a trick with nested fixes -- see this definition of Ackermann function (it wouldn't work with just Fixpoint).
Program Fixpoint
You can use Program Fixpoint mechanism which lets you write your program naturally and later prove that it always terminates.
From Coq Require Import Program Arith.
Program Fixpoint interleave2 {A} (l1 l2 : list A)
{measure (length l1 + length l2)} : list A :=
match l1 with
| [] => l2
| h :: t => h :: interleave2 l2 t
end.
Next Obligation. simpl; rewrite Nat.add_comm; trivial with arith. Qed.
Function
Another option is to use the Function command which can be somewhat limited compared to Program Fixpoint. You can find out more about their differences here.
From Coq Require Recdef.
Definition sum_len {A} (ls : (list A * list A)) : nat :=
length (fst ls) + length (snd ls).
Function interleave3 {A} (ls : (list A * list A))
{measure sum_len ls} : list A :=
match ls with
| ([], _) => []
| (h :: t, l2) => h :: interleave3 (l2, t)
end.
Proof.
intros A ls l1 l2 h t -> ->; unfold sum_len; simpl; rewrite Nat.add_comm; trivial with arith.
Defined.
Equations plugin
This is an external plugin which addresses many issues with defining functions in Coq, including dependent types and termination.
From Equations Require Import Equations.
Equations interleave4 {A} (l1 l2 : list A) : list A :=
interleave4 l1 l2 by rec (length l1 + length l2) lt :=
interleave4 nil l2 := l2;
interleave4 (cons h t) l2 := cons h (interleave4 l2 t).
Next Obligation. rewrite Nat.add_comm; trivial with arith. Qed.
The code above works if you apply this fix.
Fix / Fix_F_2 combinators
You can learn more about this (manual) approach if you follow the links from this question about mergeSort function. By the way, the mergeSort function can be defined without using Fix if you apply the nested fix trick I mentioned earlier. Here is a solution which uses Fix_F_2 combinator since we have two arguments and not one like mergeSort:
Definition ordering {A} (l1 l2 : list A * list A) : Prop :=
length (fst l1) + length (snd l1) < length (fst l2) + length (snd l2).
Lemma ordering_wf' {A} : forall (m : nat) (p : list A * list A),
length (fst p) + length (snd p) <= m -> Acc (#ordering A) p.
Proof.
unfold ordering; induction m; intros p H; constructor; intros p'.
- apply Nat.le_0_r, Nat.eq_add_0 in H as [-> ->].
intros contra%Nat.nlt_0_r; contradiction.
- intros H'; eapply IHm, Nat.lt_succ_r, Nat.lt_le_trans; eauto.
Defined.
Lemma ordering_wf {A} : well_founded (#ordering A).
Proof. now red; intro ; eapply ordering_wf'. Defined.
(* it's in the stdlib but unfortunately opaque -- this blocks evaluation *)
Lemma destruct_list {A} (l : list A) :
{ x:A & {tl:list A | l = x::tl} } + { l = [] }.
Proof.
induction l as [|h tl]; [right | left]; trivial.
exists h, tl; reflexivity.
Defined.
Definition interleave5 {A} (xs ys : list A) : list A.
refine (Fix_F_2 (fun _ _ => list A)
(fun (l1 l2 : list A)
(interleave : (forall l1' l2', ordering (l1', l2') (l1, l2) -> list A)) =>
match destruct_list l1 with
| inright _ => l2
| inleft pf => let '(existT _ h (exist _ tl eq)) := pf in
h :: interleave l2 tl _
end) (ordering_wf (xs,ys))).
Proof. unfold ordering; rewrite eq, Nat.add_comm; auto.
Defined.
Evaluation tests
Check eq_refl : interleave1 [1;2;3] [4;5;6] = [1;4;2;5;3;6].
Check eq_refl : interleave2 [1;2;3] [4;5;6] = [1;4;2;5;3;6].
Check eq_refl : interleave3 ([1;2;3], [4;5;6]) = [1;4;2;5;3;6].
Fail Check eq_refl : interleave4 [1;2;3] [4;5;6] = [1;4;2;5;3;6]. (* Equations plugin *)
Check eq_refl : interleave5 [1;2;3] [4;5;6] = [1;4;2;5;3;6].
Exercise: what happens with this last check if you comment out destruct_list lemma?
You can use something called a measure instead of a structural argument for termination. For this, I believe you have to use the Program Fixpoint mechanism, which is a little involved and will make your proofs look uglier (because it generates a structural recursion out of the proof that you provide, so that the function you will actually use is not quite the function you wrote).
Details here:
https://coq.inria.fr/refman/program.html
It also seems like something called Equations can deal with measures?
cf. http://mattam82.github.io/Coq-Equations/examples/RoseTree.html
https://www.irif.fr/~sozeau/research/coq/equations.en.html

How can I split a list in half in coq?

It looks definitely simple task until I actually try to work on it. My method is to use twin pointers to avoid asking the length of the list ahead of time, but the difficulties come from the implication that I know for sure one list is "no emptier" than another. Specifically, in pseudo-coq:
Definition twin_ptr (heads, tail, rem : list nat) :=
match tail, rem with
| _, [] => (rev heads, tail)
| _, [_] => (rev heads, tail)
| t :: tl, _ :: _ :: rm => twin_ptr (t :: heads) tl rm
end.
Definition split (l : list nat) := twin_ptr [] l l
But definitely it's not going to compile because the match cases are incomplete. However, the missing case by construction doesn't exist.
What's your way of implementing it?
I you are not afraid of dependent types, you can add a proof that rem is shorter than tail as an argument of twin_ptr. Using Program to help manage these dependent types, this could give the following.
Require Import List. Import ListNotations.
Require Import Program.
Require Import Arith.
Require Import Omega.
Program Fixpoint twin_ptr
(heads tail rem : list nat)
(H:List.length rem <= List.length tail) :=
match tail, rem with
| a1, [] => (rev heads, tail)
| a2, [a3] => (rev heads, tail)
| t :: tl, _ :: _ :: rm => twin_ptr (t :: heads) tl rm _
| [], _::_::_ => !
end.
Next Obligation.
simpl in H. omega.
Qed.
Next Obligation.
simpl in H. omega.
Qed.
Definition split (l : list nat) := twin_ptr [] l l (le_n _).
The exclamation mark means that a branch is unreachable.
You can then prove lemmas about twin_ptr and deduce the properties of split from them. For example,
Lemma twin_ptr_correct : forall head tail rem H h t,
twin_ptr head tail rem H = (h, t) ->
h ++ t = rev head ++ tail.
Proof.
Admitted.
Lemma split_correct : forall l h t,
split l = (h, t) ->
h ++ t = l.
Proof.
intros. apply twin_ptr_correct in H. assumption.
Qed.
Personally, I dislike to use dependent types in functions, as resulting objects are more difficult to manipulate. Instead, I prefer defining total functions and give them the right hypotheses in the lemmas.
You do not need to maintain the invariant that the second list is bigger than the third. Here is a possible solution:
Require Import Coq.Arith.PeanoNat.
Require Import Coq.Arith.Div2.
Require Import Coq.Lists.List.
Import ListNotations.
Section Split.
Variable A : Type.
Fixpoint split_aux (hs ts l : list A) {struct l} : list A * list A :=
match l with
| [] => (rev hs, ts)
| [_] => (rev hs, ts)
| _ :: _ :: l' =>
match ts with
| [] => (rev hs, [])
| h :: ts => split_aux (h :: hs) ts l'
end
end.
Lemma split_aux_spec hs ts l n :
n = div2 (length l) ->
split_aux hs ts l = (rev (rev (firstn n ts) ++ hs), skipn n ts).
Proof.
revert hs ts l.
induction n as [|n IH].
- intros hs ts [|x [|y l]]; easy.
- intros hs ts [|x [|y l]]; simpl; try easy.
intros Hn.
destruct ts as [|h ts]; try easy.
rewrite IH; try congruence.
now simpl; rewrite <- app_assoc.
Qed.
Definition split l := split_aux [] l l.
Lemma split_spec l :
split l = (firstn (div2 (length l)) l, skipn (div2 (length l)) l).
Proof.
unfold split.
rewrite (split_aux_spec [] l l (div2 (length l))); trivial.
now rewrite app_nil_r, rev_involutive.
Qed.
End Split.
May I suggest going via a more precise type? The main idea is to define a function splitting a Vector.t whose nat index has the shape m + n into a Vector.t of size m and one of size n.
Require Import Vector.
Definition split_vector : forall a m n,
Vector.t a (m + n) -> (Vector.t a m * Vector.t a n).
Proof.
intros a m n; induction m; intro v.
- firstorder; constructor.
- destruct (IHm (tl v)) as [xs ys].
firstorder; constructor; [exact (hd v)|assumption].
Defined.
Once you have this, you've reduced your problem to defining the floor and ceil of n / 2 and proving that they sum to n.
Fixpoint div2_floor_ceil (n : nat) : (nat * nat) := match n with
| O => (O , O)
| S O => (O , S O)
| S (S n') => let (p , q) := div2_floor_ceil n'
in (S p, S q)
end.
Definition div2_floor (n : nat) := fst (div2_floor_ceil n).
Definition div2_ceil (n : nat) := snd (div2_floor_ceil n).
Lemma plus_div2_floor_ceil : forall n, div2_floor n + div2_ceil n = n.
Proof.
refine
(fix ih n := match n with
| O => _
| S O => _
| S (S n') => _
end); try reflexivity.
unfold div2_floor, div2_ceil in *; simpl.
destruct (div2_floor_ceil n') as [p q] eqn: eq.
simpl.
replace p with (div2_floor n') by (unfold div2_floor ; rewrite eq ; auto).
replace q with (div2_ceil n') by (unfold div2_ceil ; rewrite eq ; auto).
rewrite <- plus_n_Sm; do 2 f_equal.
apply ih.
Qed.
Indeed, you can then convert length xs into ceil (length xs / 2) + floor (length xs / 2) and use split_vector to get each part.
Definition split_list a (xs : list a) : (list a * list a).
Proof.
refine
(let v := of_list xs in
let (p , q) := split_vector a (div2_floor _) (div2_ceil _) _ in
(to_list p, to_list q)).
rewrite plus_div2_floor_ceil; exact v.
Defined.

How can I automate counting within proofs in Coq?

I have a function count that counts how many times a given predicate is provable when applied to elements of a list. It is defined as follows:
Parameter T : Type.
Parameter dec: forall (p: T -> Prop) (w: T), {p w} + {~ (p w)}.
Fixpoint count (p: T -> Prop) (l: list T) := match l with
| nil => 0
| (cons head tail) => if (dec p head) then (1 + (count p tail)) else (count p tail)
end.
I then use this function to state lemmas like the following:
Parameter a b c: T.
Parameter q: T -> Prop.
Axiom Aa: (q a).
Axiom Ab: (q b).
Axiom Ac: ~ (q c).
Lemma example: (count q (cons a (cons b (cons c nil)))) = 2.
My proofs of such lemmas tend to be quite tedious:
Lemma example: (count q (cons a (cons b (cons c nil)))) = 2.
Proof.
unfold count.
assert (q a); [apply Aa| auto].
assert (q b); [apply Ab| auto].
assert (~ (q c)); [apply Ac| auto].
destruct (dec q a); [auto | contradiction].
destruct (dec q b); [auto | contradiction].
destruct (dec q c); [contradiction | auto].
Qed.
What can I do to automate such tedious proofs that involve computation with my count function?
This is typically the kind of cases where you are better off proving things by reflection. See how things go smoothly (of course I modified a bit your example to avoid all these axioms):
Require Import List.
Import ListNotations.
Fixpoint count {T : Type} (p : T -> bool) (l : list T) :=
match l with
| [] => 0
| h :: t => if p h then S (count p t) else (count p t)
end.
Inductive T := a | b | c.
Definition q x :=
match x with
| a => true
| b => true
| c => false
end.
Lemma example: (count q [a; b; c]) = 2.
Proof.
reflexivity.
Qed.
I realize that your definition of count was taking a propositional predicate on type T (but with the assumption that all predicates on type T are decidable) and instead I propose to define count to take a boolean predicate. But you may realize that having a decidable propositional predicate or having a boolean predicate is actually equivalent.
E.g. from your axioms, I can define a function which transform any propositional predicate into a boolean one:
Parameter T : Type.
Parameter dec: forall (p: T -> Prop) (w: T), {p w} + {~ (p w)}.
Definition prop_to_bool_predicate (p : T -> Prop) (x : T) : bool :=
if dec p x then true else false.
Of course, because there are axioms involved in your example, it won't actually be possible to compute with the boolean predicate. But I'm assuming that you put all these axioms for the purpose of the example and that your actual application doesn't have them.
Answer to your comment
As I told you, as soon as you have defined some function in terms of an axiom (or of a Parameter since this is the same thing), there is no way you can compute with it anymore.
However, here is a solution where the decidability of propositional predicate p is a lemma instead. I ended the proof of the lemma with Defined instead of Qed to allow computing with it (otherwise, it wouldn't be any better than an axiom). As you can see I also redefined the count function to take a predicate and a proof of its decidability. The proof by reflection still works in that case. There is no bool but it is strictly equivalent.
Require Import List.
Import ListNotations.
Fixpoint count {T : Type}
(p : T -> Prop) (dec : forall (w: T), {p w} + {~ (p w)}) (l : list T) :=
match l with
| [] => 0
| h :: t => if dec h then S (count p dec t) else (count p dec t)
end.
Inductive T := a | b | c.
Definition p x := match x with | a => True | b => True | c => False end.
Lemma dec_p: forall (w: T), {p w} + {~ (p w)}.
Proof.
intros []; simpl; auto.
Defined.
Lemma example2: (count p dec_p [a; b; c]) = 2. Proof. reflexivity. Qed.
Let's create our custom hint database and add your axioms there:
Hint Resolve Aa : axiom_db.
Hint Resolve Ab : axiom_db.
Hint Resolve Ac : axiom_db.
Now, the firstorder tactic can make use of the hint database:
Lemma example: count q (cons a (cons b (cons c nil))) = 2.
Proof.
unfold count.
destruct (dec q a), (dec q b), (dec q c); firstorder with axiom_db.
Qed.
We can automate our solution using the following piece of Ltac:
Ltac solve_the_probem :=
match goal with
|- context [if dec ?q ?x then _ else _] =>
destruct (dec q x);
firstorder with axioms_db;
solve_the_probem
end.
Then, unfold count; solve_the_probem. will be able to prove the lemma.