Coq: unary to binary convertion - coq

Task: write a function to convert natural numbers to binary numbers.
Inductive bin : Type :=
| Z
| A (n : bin)
| B (n : bin).
(* Division by 2. Returns (quotient, remainder) *)
Fixpoint div2_aux (n accum : nat) : (nat * nat) :=
match n with
| O => (accum, O)
| S O => (accum, S O)
| S (S n') => div2_aux n' (S accum)
end.
Fixpoint nat_to_bin (n: nat) : bin :=
let (q, r) := (div2_aux n 0) in
match q, r with
| O, O => Z
| O, 1 => B Z
| _, O => A (nat_to_bin q)
| _, _ => B (nat_to_bin q)
end.
The 2-nd function gives an error, because it is not structurally recursive:
Recursive call to nat_to_bin has principal argument equal to
"q" instead of a subterm of "n".
What should I do to prove that it always terminates because q is always less then n.

Prove that q is (almost always) less than n:
(* This condition is sufficient, but a "better" one is n <> 0
That makes the actual function slightly more complicated, though *)
Theorem div2_aux_lt {n} (prf : fst (div2_aux n 0) <> 0) : fst (div2_aux n 0) < n.
(* The proof is somewhat involved...
I did it by proving
forall n k, n <> 0 ->
fst (div2_aux n k) < n + k /\ fst (div2_aux (S n) k) < S n + k
by induction on n first *)
Then proceed by well-founded induction on lt:
Require Import Arith.Wf_nat.
Definition nat_to_bin (n : nat) : bin :=
lt_wf_rec (* Recurse down a chain of lts instead of structurally *)
n (fun _ => bin) (* Starting from n and building a bin *)
(fun n rec => (* At each step, we have (n : nat) and (rec : forall m, m < n -> bin) *)
match div2_aux n 0 as qr return (fst qr <> 0 -> fst qr < n) -> _ with (* Take div2_aux_lt as an argument; within the match the (div2_aux_lt n 0) in its type is rewritten in terms of the matched variables *)
| (O, r) => fun _ => if r then Z else B Z (* Commoning up cases for brevity *)
| (S _ as q, r) => (* note: O is "true" and S _ is "false" *)
fun prf => (if r then A else B) (rec q (prf ltac:(discriminate)))
end div2_aux_lt).
I might suggest making div2_aux return nat * bool.
Alternatively, Program Fixpoint supports these kinds of induction, too:
Require Import Program.
(* I don't like the automatic introing in program_simpl and
now/easy can solve some of our obligations. *)
#[local] Obligation Tactic := (now program_simpl) + cbv zeta.
(* {measure n} is short for {measure n lt}, which can replace the
core language {struct arg} when in a Program Fixpoint
(n can be any expression and lt can be any well-founded relation
on the type of that expression) *)
#[program] Fixpoint nat_to_bin (n : nat) {measure n} : bin :=
match div2_aux n 0 with
| (O, O) => Z
| (O, _) => B Z
| (q, O) => A (nat_to_bin q)
| (q, _) => B (nat_to_bin q)
end.
Next Obligation.
intros n _ q [_ mem] prf%(f_equal fst).
simpl in *.
subst.
apply div2_aux_lt.
auto.
Defined.
Next Obligation.
intros n _ q r [mem _] prf%(f_equal fst).
specialize (mem r).
simpl in *.
subst.
apply div2_aux_lt.
auto.
Defined.

Related

Reversing a vector in Coq

I am trying to reverse a vector in Coq. My implementation is as follows:
Fixpoint vappend {T : Type} {n m} (v1 : vect T n) (v2 : vect T m)
: vect T (plus n m) :=
match v1 in vect _ n return vect T (plus n m) with
| vnil => v2
| x ::: v1' => x ::: (vappend v1' v2)
end.
Theorem plus_n_S : forall n m, plus n (S m) = S (plus n m).
Proof.
intros. induction n; auto.
- simpl. rewrite <- IHn. auto.
Qed.
Theorem plus_n_O : forall n, plus n O = n.
Proof.
induction n.
- reflexivity.
- simpl. rewrite IHn. reflexivity.
Qed.
Definition vreverse {T : Type} {n} (v : vect T n) : vect T n.
induction v.
- apply [[]].
- rewrite <- plus_n_O. simpl. rewrite <- plus_n_S.
apply (vappend IHv (t ::: [[]])).
Show Proof.
Defined.
The problem is, when I try to compute the function, it produces something like:
match plus_n_O (S (S O)) in (_ = y) return (vect nat y) with
...
and couldn't get further. What's the problem here? How can I fix this?
The problem is that your functions use opaque proofs, plus_n_S and plus_n_O. To compute vreverse, you need to compute these proofs, and if they are opaque, the computation will be blocked.
You can fix this issue by defining the functions transparently. Personally, I prefer not to use proof mode when doing this, since it is easier to see what is going on. (I have used the standard library definition of vectors here.)
Require Import Coq.Vectors.Vector.
Import VectorNotations.
Fixpoint vappend {T : Type} {n m} (v1 : t T n) (v2 : t T m)
: t T (plus n m) :=
match v1 in t _ n return t T (plus n m) with
| [] => v2
| x :: v1' => x :: vappend v1' v2
end.
Fixpoint plus_n_S n m : n + S m = S (n + m) :=
match n with
| 0 => eq_refl
| S n => f_equal S (plus_n_S n m)
end.
Fixpoint plus_n_O n : n + 0 = n :=
match n with
| 0 => eq_refl
| S n => f_equal S (plus_n_O n)
end.
Fixpoint vreverse {T : Type} {n} (v : t T n) : t T n :=
match v in t _ n return t T n with
| [] => []
| x :: v =>
eq_rect _ (t T)
(eq_rect _ (t T) (vappend (vreverse v) [x]) _ (plus_n_S _ 0))
_ (f_equal S ( plus_n_O _))
end.
Compute vreverse (1 :: 2 :: 3 :: []).

Lemma related to the counting of numbers in list

I want to solve a lemma which relate two lists after removing a number from the list with the help of following functions. Here is code
Theorem remove_decr_count: forall (l : list nat),
leb (count 0 (remove_one 0 s)) (count 0 s) = true.
Used functions are
Fixpoint remove_one (v:nat) (l:list nat) : list nat:=
match l with
| [] => []
| h :: t => if beq_nat v h then t else h :: remove_one v t
end.
Fixpoint leb (n m:nat) : bool :=
match n, m with
| O, _ => true
| S _, O => false
| S n', S m' => leb n' m'
end.
Fixpoint count (v:nat) (l:list nat) : nat :=
match l with
| [] => 0
| h :: t => (if beq_nat h v then 1 else 0) + (count v t)
end.
One way to proceed is by induction on the list l (warning: you used s in the theorem's definition, though), and then by case, on whether the head of the list is 0 or not. Rewrites are used to guide the proof.
Using the SSReflect tactics language, the proof could proceed like this (I replaced beq_nat by ==, and added the leb1 lemma, which is also proved by induction, here on n).
From Coq Require Import Init.Prelude Unicode.Utf8.
From mathcomp Require Import all_ssreflect.
Fixpoint remove_one (v:nat) (l:list nat) : list nat:=
match l with
| nil => nil
| cons h t => if v == h then t else cons h (remove_one v t)
end.
Fixpoint count (v:nat) (l:list nat) : nat :=
match l with
| nil => 0
| cons h t => (if h == v then 1 else 0) + (count v t)
end.
Fixpoint leb (n m:nat) : bool :=
match n, m with
| O, _ => true
| S _, O => false
| S n', S m' => leb n' m'
end.
Lemma leb1 (n : nat) : leb n (S n).
Proof. by elim: n. Qed.
Theorem remove_decr_count: forall (l : list nat),
leb (count 0 (remove_one 0 l)) (count 0 l).
Proof.
elim=> [|h t IH] //=.
- have [] := boolP (h == 0) => eqh0.
by rewrite eq_sym eqh0 leb1.
- by rewrite eq_sym ifN //= ifN.
Qed.

Transform the lemma leaving previous state on Coq

I want to partially derive functions whose input is a dependent list.
I tried to define deriveP with proving.
Derive is a function in Coquelicot.Derive.
Definition deriveP {P A B}(k:nat)(I:Euc (S P) -> Euc A -> Euc B)
(input:Euc A)(train:Euc B)(p :Euc (S P))
:(lt k (S P)) -> (lt ((S P)-(k+1)) (S P)) -> R.
intros.
pose fk := firstk k (S P) p H.
pose lk := lastk ((S P)-(k+1)) (S P) p H0.
pose pk := EucNth k p.
apply arith_basic in H.
exact ( Derive (fun PK => EucSum (QuadraticError (I (fk +++ (PK ::: lk)) input) train )) pk ).
I can not apply arith_basic poposed by Tiago because H is used in fk.
I can apply arith_basic to H before I make fk, but then I can not make fk because There is not k < P.+1.
I want to apply arith_basic to H while leaving k < P.+1.
Please help me.
(***********************************************************)
This is dependent list of R.
Require Import Coq.Reals.Reals.
Require Import Coquelicot.Coquelicot.
Inductive Euc:nat -> Type:=
|RO : Euc 0
|Rn : forall {n:nat}, R -> Euc n -> Euc (S n).
Notation "[ ]" := RO.
Notation "[ r1 , .. , r2 ]" := (Rn r1 .. ( Rn r2 RO ) .. ).
Infix ":::" := Rn (at level 60, right associativity).
Basic list operation.
Definition head {n} (v : Euc (S n)) : R :=
match v with
| x ::: _ => x
end.
Definition tail {n} (v : Euc (S n)) : Euc n :=
match v with
| _ ::: v => v
end.
(* extract the last element *)
Fixpoint last {n} : Euc (S n) -> R :=
match n with
| 0%nat => fun v => head v
| S n => fun v => last (tail v)
end.
(* eliminate last element from list *)
Fixpoint but_last {n} : Euc (S n) -> Euc n :=
match n with
| 0%nat => fun _ => []
| S n => fun v => head v ::: but_last (tail v)
end.
(* do the opposite of cons *)
Fixpoint snoc {n} (v : Euc n) (x : R) : Euc (S n) :=
match v with
| [] => [x]
| y ::: v => y ::: snoc v x
end.
(* extract last k elements *)
Fixpoint lastk k n : Euc n -> (lt k n) -> Euc k :=
match n with
|0%nat => fun _ (H : lt k 0) => False_rect _ (Lt.lt_n_O _ H)
|S n => match k with
|S m => fun v H => snoc (lastk m n (but_last v) (le_S_n _ _ H)) (last v)
|0%nat => fun _ H => []
end
end.
(* extract first k elements *)
Fixpoint firstk k n : Euc n -> (lt k n) -> Euc k :=
match n with
|0%nat => fun _ (H :lt k 0) => False_rect _ (Lt.lt_n_O _ H)
|S n => match k with
|S m => fun v H => (head v) ::: firstk m n (tail v) (le_S_n _ _ H)
|0%nat => fun _ _ => []
end
end.
(* extract nth element *)
(* 0 origine *)
Fixpoint EucNth (k:nat) :forall {n}, Euc (S n) -> R:=
match k with
| 0%nat => fun _ e => head e
| S k' => fun n =>
match n return Euc (S n) -> R with
| 0%nat => fun e => head e
| S n' => fun v => EucNth k' (tail v)
end
end.
Fixpoint EucAppend {n m} (e:Euc n) (f:Euc m) :Euc (n+m):=
match e with
|[] => f
|e' ::: es => e' ::: (EucAppend es f)
end.
Infix "+++" := EucAppend (at level 60, right associativity).
Fixpoint QuadraticError {n : nat} (b : Euc n) : Euc n -> Euc n.
refine (match b in Euc n return Euc n -> Euc n with
|#Rn m x xs => _
|#RO => fun H => []
end).
remember (S m).
intro H; destruct H as [| k y ys].
inversion Heqn0.
inversion Heqn0.
subst; exact ((x - y)^2 ::: QuadraticError _ xs ys).
Defined.
Fixpoint EucSum {A}(e:Euc A) :R:=
match e with
| [] => 0%R
| e' ::: es => e' + (EucSum es)
end.
Your lemma k + S (P - (k + 1)) = P can be solved just with basic algebraic operations.
Particularly you just need two lemmas to make this easier:
Theorem minus_assoc : forall y z, z < y -> z + (y - z) = y.
intro y.
induction y.
intros;inversion H.
intros.
destruct z.
trivial.
rewrite PeanoNat.Nat.sub_succ.
rewrite <- (IHy _ (le_S_n _ _ H)) at 2; trivial.
Qed.
Theorem minus_S : forall x y, y < x -> S (x - (S y)) = x - y.
intro.
induction x.
intros.
inversion H.
intros.
destruct y.
simpl.
rewrite PeanoNat.Nat.sub_0_r; trivial.
rewrite PeanoNat.Nat.sub_succ.
apply IHx.
exact (le_S_n _ _ H).
Qed.
Now you just have to rewrite your goal to a trivial preposition :
Theorem arith_basic : forall k P, k < P -> k + S (P - (k + 1)) = P.
intros.
rewrite PeanoNat.Nat.add_1_r.
rewrite minus_S.
auto.
rewrite minus_assoc.
assumption.
trivial.
Qed.
Most of these kinds of goals can solve by lia tactic which automatically solves arithmetics goals of Z, nat, positive, and N.
Theorem arith_basic : forall k P, k < P -> k + S (P - (k + 1)) = P.
intros;lia.
Qed
Even though I recommend automation, proving by hands can help understand your main goal which may be not able to be solved by only automation.
I have solved on my own.
We can duplicate lemma in the sub-goal with generalize tactic.
Definition deriveP {P A B}(k:nat)(I:Euc (S P) -> Euc A -> Euc B)
(input:Euc A)(train:Euc B)(p :Euc (S P))
:(lt k (S P)) -> (lt ((S P)-(k+1)) (S P)) -> R.
intros.
generalize H.
intro H1.
apply arith_basic in H1.
pose lk := lastk ((S P)-(k+1)) (S P) p H0.
pose fk := firstk k (S P) p H.
pose pk := EucNth k p.
rewrite (_: (P.+1)%nat = (k + (P.+1 - (k + 1)%coq_nat)%coq_nat.+1)%coq_nat) in I.
exact ( Derive (fun PK => EucSum (QuadraticError (I (fk +++ (PK ::: lk)) input) train )) pk ).
apply H1.
Defined.

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.

Coq rewriting using lambda arguments

We have a function that inserts an element into a specific index of a list.
Fixpoint inject_into {A} (x : A) (l : list A) (n : nat) : option (list A) :=
match n, l with
| 0, _ => Some (x :: l)
| S k, [] => None
| S k, h :: t => let kwa := inject_into x t k
in match kwa with
| None => None
| Some l' => Some (h :: l')
end
end.
The following property of the aforementioned function is of relevance to the problem (proof omitted, straightforward induction on l with n not being fixed):
Theorem inject_correct_index : forall A x (l : list A) n,
n <= length l -> exists l', inject_into x l n = Some l'.
And we have a computational definition of permutations, with iota k being a list of nats [0...k]:
Fixpoint permute {A} (l : list A) : list (list A) :=
match l with
| [] => [[]]
| h :: t => flat_map (
fun x => map (
fun y => match inject_into h x y with
| None => []
| Some permutations => permutations
end
) (iota (length t))) (permute t)
end.
The theorem we're trying to prove:
Theorem num_permutations : forall A (l : list A) k,
length l = k -> length (permute l) = factorial k.
By induction on l we can (eventually) get to following goal: length (permute (a :: l)) = S (length l) * length (permute l). If we now simply cbn, the resulting goal is stated as follows:
length
(flat_map
(fun x : list A =>
map
(fun y : nat =>
match inject_into a x y with
| Some permutations => permutations
| None => []
end) (iota (length l))) (permute l)) =
length (permute l) + length l * length (permute l)
Here I would like to proceed by destruct (inject_into a x y), which is impossible considering x and y are lambda arguments. Please note that we will never get the None branch as a result of the lemma inject_correct_index.
How does one proceed from this proof state? (Please do note that I am not trying to simply complete the proof of the theorem, that's completely irrelevant.)
There is a way to rewrite under binders: the setoid_rewrite tactic (see ยง27.3.1 of the Coq Reference manual).
However, direct rewriting under lambdas is not possible without assuming an axiom as powerful as the axiom of functional extensionality (functional_extensionality).
Otherwise, we could have proved:
(* classical example *)
Goal (fun n => n + 0) = (fun n => n).
Fail setoid_rewrite <- plus_n_O.
Abort.
See here for more detail.
Nevertheless, if you are willing to accept such axiom, then you can use the approach described by Matthieu Sozeau in this Coq Club post to rewrite under lambdas like so:
Require Import Coq.Logic.FunctionalExtensionality.
Require Import Coq.Setoids.Setoid.
Require Import Coq.Classes.Morphisms.
Generalizable All Variables.
Instance pointwise_eq_ext {A B : Type} `(sb : subrelation B RB eq)
: subrelation (pointwise_relation A RB) eq.
Proof. intros f g Hfg. apply functional_extensionality. intro x; apply sb, (Hfg x). Qed.
Goal (fun n => n + 0) = (fun n => n).
setoid_rewrite <- plus_n_O.
reflexivity.
Qed.