在coq中,如何做;诱导n方程:Hn“;以一种不';不要搞乱归纳假设?
在使用归纳法时,我希望使用假设在coq中,如何做;诱导n方程:Hn“;以一种不';不要搞乱归纳假设?,coq,Coq,在使用归纳法时,我希望使用假设n=0和n=sn'来区分案例 Section x. Variable P : nat -> Prop. Axiom P0: P 0. Axiom PSn : forall n, P n -> P (S n). Theorem Pn: forall n:nat, P n. Proof. intros n. induction n. - (* = 0 *) apply P0. - (* = S n *)
n=0
和n=sn'
来区分案例
Section x.
Variable P : nat -> Prop.
Axiom P0: P 0.
Axiom PSn : forall n, P n -> P (S n).
Theorem Pn: forall n:nat, P n.
Proof. intros n. induction n.
- (* = 0 *)
apply P0.
- (* = S n *)
apply PSn. assumption.
Qed.
理论上,我可以用归纳法n eqn:Hn来实现这一点,但这似乎打乱了归纳假设:
Theorem Pn2: forall n:nat, P n.
Proof. intros n. induction n eqn: Hn.
- (* Hn : n = 0 *)
apply P0.
- (* Hn : n = S n0 *)
(*** 1 subgoals
P : nat -> Prop
n : nat
n0 : nat
Hn : n = S n0
IHn0 : n = n0 -> P n0
______________________________________(1/1)
P (S n0)
****)
Abort.
End x.
有没有一种简单的方法可以得到我想要的东西?我不确定这是否比你在第二次尝试中所做的更容易,但你可以先“记住”
n
Theorem Pn: forall n:nat, P n.
Proof. intro n. remember n. induction n.
- (*P : nat -> Prop
n0 : nat
Heqn0 : n0 = 0
============================
P n0
*)
subst. apply P0.
- (* P : nat -> Prop
n : nat
n0 : nat
Heqn0 : n0 = S n
IHn : n0 = n -> P n0
============================
P n0
*)
哦,我想我知道了 应用归纳假设将你的目标从(pn)改为(p(构造器n'),因此我认为一般来说,你可以根据目标来创建方程n=构造器n' 我认为有一种策略可以做到这一点:
(* like set (a:=b) except introduces a name and hypothesis *)
Tactic Notation
"provide_name" ident(n) "=" constr(v)
"as" simple_intropattern(H) :=
assert (exists n, n = v) as [n H] by (exists v; reflexivity).
Tactic Notation
"induction_eqn" ident(n) "as" simple_intropattern(HNS)
"eqn:" ident(Hn) :=
let PROP := fresh in (
pattern n;
match goal with [ |- ?FP _ ] => set ( PROP := FP ) end;
induction n as HNS;
match goal with [ |- PROP ?nnn ] => provide_name n = nnn as Hn end;
unfold PROP in *; clear PROP
).
它适用于我的示例:
Theorem Pn_3: forall n:nat, P n.
Proof.
intros n.
induction_eqn n as [|n'] eqn: Hn.
- (* n: nat, Hn: n = 0; Goal: P 0 *)
apply P0.
- (* n': nat, IHn': P n';
n: nat, Hn: n = S n'
Goal: P (S n') *)
apply PSn. exact IHn'.
Qed.
马特几乎是对的,你只是忘记了通过还原记忆中的
n
来概括一下你的目标:
Theorem Pn2: forall n:nat, P n.
Proof. intros n. remember n. revert n0 Heqn0.
induction n as [ | p hi]; intros m heq.
- (* heq : n = 0 *) subst. apply P0.
- (* heq : n = S n0 *)
(*
1 subgoal
P : nat -> Prop
p : nat
hi : forall n0 : nat, n0 = p -> P n0
m : nat
heq : m = S p
______________________________________(1/1)
P m
*) subst; apply (PSn p). apply hi. reflexivity.
谷歌搜索“coq感应eqn”的结果表明,这一点已被打破,而且“eqn:”一词似乎最近已从参考手册中删除。链接:,在这种情况下,Heqn0仍然被归纳假设捕获,使其无法使用(因为n0=sn/\n0=n是一个矛盾)。