功能和定义:间的平等
问题说明
我对 a列表包含的命题 P （或重复l ）有一个归纳定义重复元素，以及对其取反的函数定义 Q （或 no_repeats l ）。
I have an inductive definition of the proposition P (or repeats l) that a lists contains repeating elements, and a functional definition of it's negation Q (or no_repeats l).
我想证明 P<-> 〜Q 和〜P<->问。我已经能够显示这四个含义中的三个，但是〜Q-> P 似乎有所不同，因为我无法从〜Q 中提取数据。
I want to show that P <-> ~ Q and ~ P <-> Q. I have been able to show three of the four implications, but ~ Q -> P seems to be different, because I'm unable to extract data from ~Q.
Require Import List.
Variable A : Type.
Inductive repeats : list A -> Prop := (* repeats *)
repeats_hd l x : In x l -> repeats (x::l)
| repeats_tl l x : repeats l -> repeats (x::l).
Fixpoint no_repeats (l: list A): Prop :=
match l with nil => True | a::l' => ~ In a l' /\ no_repeats l' end.
Lemma not_no_repeats_repeats: forall l, (~ no_repeats l) -> repeats l.
induction l; simpl. tauto. intros.
对 l 进行归纳后，第二个情况是
After doing induction on l, the second case is
IHl : ~ no_repeats l -> repeats l
H : ~ (~ In a l /\ no_repeats l)
============================
repeats (a :: l)
是否有可能推论从这里开始\ /〜no_repeats l （足够了吗？）
Is it possible to deduce In a l \/ ~ no_repeats l (which is sufficient) from this?
正确答案
您的陈述暗示 A 上的相等支持双重否定消除：
Your statement implies that equality on A supports double negation elimination:
Require Import List.
Import ListNotations.
Variable A : Type.
Inductive repeats : list A -> Prop := (* repeats *)
repeats_hd l x : In x l -> repeats (x::l)
| repeats_tl l x : repeats l -> repeats (x::l).
Fixpoint no_repeats (l: list A): Prop :=
match l with nil => True | a::l' => ~ In a l' /\ no_repeats l' end.
Hypothesis not_no_repeats_repeats: forall l, (~ no_repeats l) -> repeats l.
Lemma eq_nn_elim (a b : A) : ~ a <> b -> a = b.
Proof.
intros H.
assert (H' : ~ no_repeats [a; b]).
{ simpl. intuition. }
apply not_no_repeats_repeats in H'.
inversion H'; subst.
{ subst. simpl in *. intuition; tauto. }
inversion H1; simpl in *; subst; intuition.
inversion H2.
Qed.
并非每种类型都支持 eq_nn_elim 只能通过在 A 上放置其他假设来证明 not_no_repeats_repeats 。足以假设 A 具有可判定的相等性；即：
Not every type supports eq_nn_elim, which means that you can only prove not_no_repeats_repeats by placing additional hypotheses on A. It should suffice to assume that A has decidable equality; that is:
Hypothesis eq_dec a b : a = b \/ a <> b.
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