什么是正确的证明术语,以便 ssreflect 教程与 exact: hAiB 示例一起工作?
What is the right proof term so that the ssreflect tutorial work with the exact: hAiB example?
我正在阅读 ssreflect 的教程 https://hal.inria.fr/inria-00407778/document,他们有证据:
Variables A B C : Prop.
Hypotheses (hAiBiC : A -> B -> C) (hAiB : A -> B) (hA : A).
Lemma HilbertS2 :
C.
Proof.
apply: hAiBiC; first by apply: hA.
exact: hAiB.
Qed.
但它实际上不起作用,因为目标是
B
这让我感到困惑...这是因为 coq 版本更改而无法正常工作的原因是什么?或者别的什么?无论如何,确切的论点应该是什么?
我想我确实理解 exact
论点的作用。它通过确保给定的证明项(程序)具有当前目标的类型来完成当前子目标。例如
Theorem add_easy_induct_1_exact:
forall n:nat,
n + 0 = n.
Proof.
exact (fun n : nat =>
nat_ind (fun n0 : nat => n0 + 0 = n0) eq_refl
(fun (n' : nat) (IH : n' + 0 = n') =>
eq_ind_r (fun n0 : nat => S n0 = S n') eq_refl IH) n).
Qed.
用于加法交换律的证明。
Module ssreflect1.
(* Require Import ssreflect ssrbool eqtype ssrnat. *)
From Coq Require Import ssreflect ssrfun ssrbool.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Theorem three_is_three:
3 = 3.
Proof. by []. Qed.
(*
*)
Lemma HilbertS :
forall A B C : Prop,
(A -> B -> C) -> (A -> B) -> A -> C.
(* A ->(B -> C)*)
Proof.
move=> A B C. (* since props A B C are the 1st things in the assumption stack, this pops them and puts them in the local context, note using the same name as the proposition name.*)
move=> hAiBiC hAiB hA. (* pops the first 3 premises from the hypothesis stack with those names into the local context *)
move: hAiBiC. (* put hAiBiC tactic back *)
apply.
by [].
(* move: hAiB.
apply. *)
by apply: hAiB.
(* apply: hAiB.
by [].dd *)
Qed.
Variables A B C : Prop.
Hypotheses (hAiBiC : A -> B -> C) (hAiB : A -> B) (hA : A).
Lemma HilbertS2 :
C.
Proof.
apply: hAiBiC; first by apply: hA.
exact: hAiB.
Qed.
Lemma HilbertS2 :
C.
Proof.
(* apply: hAiBiC; first by apply: hA. *)
apply: hAiBiC. (* usually we think of : as pushing to the goal stack, so match c with conclusion in
selected hypothesis hAiBiC and push the replacement, so put A & B in local context. *)
by apply: hA. (* discharges A *)
exact: hAiB.
End ssreflect1.
我正在使用的完整脚本。为什么不将假设置于当地环境中?
你的例子失败的原因可能是你没有打开一个部分。然后,您声明的各种假设将被视为“公理”,而不是在目标的上下文中。
另一方面,如果您在发布的文本片段之前开始一个部分,则一切正常,因为 exact: hAiB.
策略之前的目标也包含假设 hA
,即exact:
成功所必需的。
这是完整的脚本(在 coq 8.15.0 上测试过)
From mathcomp Require Import all_ssreflect.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Section sandbox.
Variables A B C : Prop.
Hypotheses (hAiBiC : A -> B -> C) (hAiB : A -> B) (hA : A).
Lemma HilbertS2 :
C.
Proof.
apply: hAiBiC; first by apply: hA.
exact: hAiB.
Qed.
End sandbox.
我正在阅读 ssreflect 的教程 https://hal.inria.fr/inria-00407778/document,他们有证据:
Variables A B C : Prop.
Hypotheses (hAiBiC : A -> B -> C) (hAiB : A -> B) (hA : A).
Lemma HilbertS2 :
C.
Proof.
apply: hAiBiC; first by apply: hA.
exact: hAiB.
Qed.
但它实际上不起作用,因为目标是
B
这让我感到困惑...这是因为 coq 版本更改而无法正常工作的原因是什么?或者别的什么?无论如何,确切的论点应该是什么?
我想我确实理解 exact
论点的作用。它通过确保给定的证明项(程序)具有当前目标的类型来完成当前子目标。例如
Theorem add_easy_induct_1_exact:
forall n:nat,
n + 0 = n.
Proof.
exact (fun n : nat =>
nat_ind (fun n0 : nat => n0 + 0 = n0) eq_refl
(fun (n' : nat) (IH : n' + 0 = n') =>
eq_ind_r (fun n0 : nat => S n0 = S n') eq_refl IH) n).
Qed.
用于加法交换律的证明。
Module ssreflect1.
(* Require Import ssreflect ssrbool eqtype ssrnat. *)
From Coq Require Import ssreflect ssrfun ssrbool.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Theorem three_is_three:
3 = 3.
Proof. by []. Qed.
(*
*)
Lemma HilbertS :
forall A B C : Prop,
(A -> B -> C) -> (A -> B) -> A -> C.
(* A ->(B -> C)*)
Proof.
move=> A B C. (* since props A B C are the 1st things in the assumption stack, this pops them and puts them in the local context, note using the same name as the proposition name.*)
move=> hAiBiC hAiB hA. (* pops the first 3 premises from the hypothesis stack with those names into the local context *)
move: hAiBiC. (* put hAiBiC tactic back *)
apply.
by [].
(* move: hAiB.
apply. *)
by apply: hAiB.
(* apply: hAiB.
by [].dd *)
Qed.
Variables A B C : Prop.
Hypotheses (hAiBiC : A -> B -> C) (hAiB : A -> B) (hA : A).
Lemma HilbertS2 :
C.
Proof.
apply: hAiBiC; first by apply: hA.
exact: hAiB.
Qed.
Lemma HilbertS2 :
C.
Proof.
(* apply: hAiBiC; first by apply: hA. *)
apply: hAiBiC. (* usually we think of : as pushing to the goal stack, so match c with conclusion in
selected hypothesis hAiBiC and push the replacement, so put A & B in local context. *)
by apply: hA. (* discharges A *)
exact: hAiB.
End ssreflect1.
我正在使用的完整脚本。为什么不将假设置于当地环境中?
你的例子失败的原因可能是你没有打开一个部分。然后,您声明的各种假设将被视为“公理”,而不是在目标的上下文中。
另一方面,如果您在发布的文本片段之前开始一个部分,则一切正常,因为 exact: hAiB.
策略之前的目标也包含假设 hA
,即exact:
成功所必需的。
这是完整的脚本(在 coq 8.15.0 上测试过)
From mathcomp Require Import all_ssreflect.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Section sandbox.
Variables A B C : Prop.
Hypotheses (hAiBiC : A -> B -> C) (hAiB : A -> B) (hA : A).
Lemma HilbertS2 :
C.
Proof.
apply: hAiBiC; first by apply: hA.
exact: hAiB.
Qed.
End sandbox.