我正在編寫一個關於使用Coq系統的快速排列算法的程序驗證的論文。我已經在Coq中定義了一個快速排序,但是我的主管和我自己用舒適的方法寫出實際的證明並不是很舒服。有沒有人可以幫助公正證明的這一部分?以下是我們已經想出迄今:使用Coq快速排除
Inductive nat : Type :=
| O : nat
| S : nat -> nat.
Check (S (S (S (S O)))).
Definition isZero (n:nat) : bool :=
match n with
O => true
|S p => false
end.
Inductive List: Set :=
| nil: List
| cons: nat -> List -> List.
Fixpoint Concat (L R: List) : List :=
match L with
| nil => R
| cons l ls => cons l (Concat ls R)
end.
Fixpoint Less (n m:nat) :=
match m with
O => false
|S q => match n with
O => true
|S p => Less p q
end
end.
Fixpoint Lesseq (n m:nat) :=
match n with
O => true
|S p => match m with
O => false
|S q => Lesseq p q
end
end.
Fixpoint Greatereq (n m:nat) :=
match n with
O => true
|S p => match m with
O => true
|S q => Greatereq p q
end
end.
Fixpoint Allless (l:List) (n:nat) : List :=
match l with
nil => nil
|cons m ls => match Less n m with
false => Allless ls n
|true => cons m (Allless ls n)
end
end.
Fixpoint Allgeq (l:List) (n:nat) : List :=
match l with
nil => nil
|cons m ls => match Greatereq n m with
false => Allgeq ls n
|true => cons m (Allgeq ls n)
end
end.
Fixpoint qaux (n:nat) (l:List) : List :=
match n with
O => nil
|S p => match l with
nil => nil
|cons m ls => let low := Allless ls m in
(let high := Allgeq ls m in
Concat (qaux p low) (cons m (qaux p high)))
end
end.
Fixpoint length (l:List) : nat :=
match l with
nil => O
|cons m ls => S (length ls)
end.
Fixpoint Quicksort (l:List) : List := qaux (length l) l.
我知道一個證明的工作,我們需要一個引理或者定理但當時我不知道在哪裏後啓動的。感謝您的幫助:)