Question

我想部分派生输入为从属列表的函数。

我试图通过证明来定义deriveP。

Derive是Coquelicot.Derive中的函数。

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 ).

我无法应用Tiago构成的arith_basic，因为在fk中使用了H。在制作fk之前，我可以将arith_basic应用于H，但是由于没有k < P.+1，因此无法制作fk。

我想在离开k < P.+1的同时将arith_basic应用于H。

请帮助我。

（************************************************** *************）

这是R的从属列表。

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).

基本列表操作。

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.

Answer 1

您的引理k + S（P-（k + 1））= P可以用基本的代数运算来求解。尤其是，您只需要两个引理即可简化此操作：

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.

现在，您只需要将目标重写为一个简单的介词即可：

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.

这些目标中的大多数都可以通过对策来解决，这种对策可以自动解决Z，nat，positive和N的算术目标。

Theorem arith_basic : forall k P, k < P -> k + S (P - (k + 1)) = P.
  intros;lia.
Qed

尽管我建议自动化，但是通过动手进行证明可以帮助您了解主要目标，而这可能仅靠自动化无法解决。

Answer 2

我自己解决了。

我们可以使用generalize策略在子目标中复制引理。

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.

转换引理，将先前状态保留在Coq

2 个答案: