Isabelle 如何将可传递关系从元素提升到列表?
我试图证明列表元素上的传递关系等价于列表上的传递关系(在某些条件下) 这是第一个引理:Isabelle 如何将可传递关系从元素提升到列表?,isabelle,theorem-proving,Isabelle,Theorem Proving,我试图证明列表元素上的传递关系等价于列表上的传递关系(在某些条件下) 这是第一个引理: lemma list_all2_rtrancl1: "(list_all2 P)⇧*⇧* xs ys ⟹ list_all2 P⇧*⇧* xs ys" apply (induct rule: rtranclp_induct) apply (simp add: list.rel_refl) by (smt list_all2_trans rtranclp.rtrancl_into_rtra
lemma list_all2_rtrancl1:
"(list_all2 P)⇧*⇧* xs ys ⟹
list_all2 P⇧*⇧* xs ys"
apply (induct rule: rtranclp_induct)
apply (simp add: list.rel_refl)
by (smt list_all2_trans rtranclp.rtrancl_into_rtrancl)
这是一个对称引理:
lemma list_all2_rtrancl2:
"(⋀x. P x x) ⟹
list_all2 P⇧*⇧* xs ys ⟹
(list_all2 P)⇧*⇧* xs ys"
apply (erule list_all2_induct)
apply simp
我想一个关系应该是反身的。但也许我应该用另一种假设。假设P是可传递的,但P不是可传递的,可以证明引理。我卡住了。你能建议选择什么样的假设以及如何证明这个引理吗
对于最后一个引理(xs=[0]
和ys=[2]
)的具体情况,Nitfick似乎给了我一个错误的反例:
我可以证明引理适用于这个例子:
lemma list_all2_rtrancl2_example_0_2:
"list_all2 (λx y. x = y ∨ Suc x = y)⇧*⇧* [0] [2] ⟹
(list_all2 (λx y. x = y ∨ Suc x = y))⇧*⇧* [0] [2]"
apply (rule_tac ?b="[1]" in converse_rtranclp_into_rtranclp; simp)
apply (rule_tac ?b="[2]" in converse_rtranclp_into_rtranclp; simp)
done
使用
listrel
而不是list\u all2
可能是可行的。实际上,如下所示,它们是等效的(请参见set\u listrel\u eq\u list\u all2
)。然而,在标准库中有几个关于listrel
的定理没有与list\u all2
等价的定理
还提供了
列表所有2
的直接证据(遗留):
list\u all2\u induct
应用于列表;基本情况并不重要。因此,它仍然显示(lp)*x#xs y#ys
如果(lp*)xs ys
,(lp)*xs ys
和P*xy
zs
(例如xs
),从而(lp)xs zs
和(lp)+zs ys
P*x y
和px
,通过基于P*
的传递性质的归纳,(lp)x#xs y#zs
。因此,同样,(lp)*x#xs y#zs
(lp)+zsys
和pyy
,通过归纳,(lp)+y#zsy#ys
。因此,同样,(lp)*y#zs y#ys
(lp)*x#xs y#ys
作为旁注,在我看来(我对Nittick知之甚少),Nittick不应该在没有任何警告的情况下提供无效的反例。我相信,通常,当
吹毛求疵怀疑某个反例可能无效时,它会通知用户该示例“可能是虚假的”。如果此问题未在其他地方记录,则提交错误报告可能很有用
伊莎贝尔版本:伊莎贝尔2020非常感谢您的帮助!我已经向邮件列表发送了关于挑剔的错误报告。谢谢!对我来说,校样里有很多新东西。我不懂感应装置和传输包。这是研究它们的一个很好的理由。@Denis我很惊讶你提到你不理解归纳集。从你的其他问题来看,你似乎理解归纳定义的谓词。在HOL:“px中,谓词和集合之间存在明显的对应关系⟷ x∈ {x.px}“
。当然,有许多书籍/课堂讲稿为归纳集提供了理论背景,例如,见此。@Denis在《高阶逻辑的证明助手》(PA)一书的第7.1节中有一个很好的例子,演示了HOL中归纳定义集和归纳定义谓词之间的对应关系以及《Isabelle/HOL的具体语义》(CS)一书中的第4.5.1节。PA给出了使用“归纳集”构造偶数集的示例,CS给出了确定自然数是否使用“归纳集”的谓词构造示例。
lemma list_all2_rtrancl2_example_0_2:
"list_all2 (λx y. x = y ∨ Suc x = y)⇧*⇧* [0] [2] ⟹
(list_all2 (λx y. x = y ∨ Suc x = y))⇧*⇧* [0] [2]"
apply (rule_tac ?b="[1]" in converse_rtranclp_into_rtranclp; simp)
apply (rule_tac ?b="[2]" in converse_rtranclp_into_rtranclp; simp)
done
lemma set_listrel_eq_list_all2:
"listrel {(x, y). r x y} = {(xs, ys). list_all2 r xs ys}"
using list_all2_conv_all_nth listrel_iff_nth by fastforce
lemma listrel_tclosure_1: "(listrel r)⇧* ⊆ listrel (r⇧*)"
by
(
simp add:
listrel_rtrancl_eq_rtrancl_listrel1
listrel_subset_rtrancl_listrel1
rtrancl_subset_rtrancl
)
lemma listrel_tclosure_2: "refl r ⟹ listrel (r⇧*) ⊆ (listrel r)⇧*"
by
(
simp add:
listrel1_subset_listrel
listrel_rtrancl_eq_rtrancl_listrel1
rtrancl_mono
)
context
includes lifting_syntax
begin
lemma listrel_list_all2_transfer[transfer_rule]:
"((=) ===> (=) ===> (=) ===> (=))
(λr xs ys. (xs, ys) ∈ listrel {(x, y). r x y}) list_all2"
unfolding rel_fun_def using set_listrel_eq_list_all2 listrel_iff_nth by blast
end
lemma list_all2_rtrancl_1:
"(list_all2 r)⇧*⇧* xs ys ⟹ list_all2 r⇧*⇧* xs ys"
proof transfer
fix r :: "'a ⇒ 'a ⇒ bool" and xs :: "'a list" and ys:: "'a list"
assume "(λxs ys. (xs, ys) ∈ listrel {(x, y). r x y})⇧*⇧* xs ys"
then have "(xs, ys) ∈ (listrel {(x, y). r x y})⇧*"
unfolding rtranclp_def rtrancl_def by auto
then have "(xs, ys) ∈ listrel ({(x, y). r x y}⇧*)"
using listrel_tclosure_1 by auto
then show "(xs, ys) ∈ listrel {(x, y). r⇧*⇧* x y}"
unfolding rtranclp_def rtrancl_def by auto
qed
lemma list_all2_rtrancl_2:
"reflp r ⟹ list_all2 r⇧*⇧* xs ys ⟹ (list_all2 r)⇧*⇧* xs ys"
proof transfer
fix r :: "'a ⇒ 'a ⇒ bool" and xs :: "'a list" and ys :: "'a list"
assume as_reflp: "reflp r" and p_in_lr: "(xs, ys) ∈ listrel {(x, y). r⇧*⇧* x y}"
from as_reflp have refl: "refl {(x, y). r x y}"
using reflp_refl_eq by fastforce
from p_in_lr have "(xs, ys) ∈ listrel ({(x, y). r x y}⇧*)"
unfolding rtranclp_def rtrancl_def by auto
with refl have "(xs, ys) ∈ (listrel {(x, y). r x y})⇧*"
using listrel_tclosure_2 by auto
then show "(λxs ys. (xs, ys) ∈ listrel {(x, y). r x y})⇧*⇧* xs ys"
unfolding rtranclp_def rtrancl_def by auto
qed
lemma list_all2_rtrancl2:
assumes as_r: "(⋀x. P x x)"
shows "(list_all2 P⇧*⇧*) xs ys ⟹ (list_all2 P)⇧*⇧* xs ys"
proof(induction rule: list_all2_induct)
case Nil then show ?case by simp
next
case (Cons x xs y ys) show ?case
proof -
from as_r have lp_xs_xs: "list_all2 P xs xs" by (rule list_all2_refl)
from Cons.hyps(1) have x_xs_y_zs: "(list_all2 P)⇧*⇧* (x#xs) (y#xs)"
proof(induction rule: rtranclp_induct)
case base then show ?case by simp
next
case (step y z) then show ?case
proof -
have rt_step_2: "(list_all2 P)⇧*⇧* (y#xs) (z#xs)"
by (rule r_into_rtranclp, rule list_all2_Cons[THEN iffD2])
(simp add: step.hyps(2) lp_xs_xs)
from step.IH rt_step_2 show ?thesis by (rule rtranclp_trans)
qed
qed
from Cons.IH have "(list_all2 P)⇧*⇧* (y#xs) (y#ys)"
proof(induction rule: rtranclp_induct)
case base then show ?case by simp
next
case (step ya za) show ?case
proof -
have rt_step_2: "(list_all2 P)⇧*⇧* (y#ya) (y#za)"
by (rule r_into_rtranclp, rule list_all2_Cons[THEN iffD2])
(simp add: step.hyps(2) as_r)
from step.IH rt_step_2 show ?thesis by (rule rtranclp_trans)
qed
qed
with x_xs_y_zs show ?thesis by simp
qed
qed