问题描述
假设我已经定义了一个归纳集,例如,归纳集“偶数”使得:
inductive_set Even :: "int set"
where ZERO : "0 ∈ Even"
| PLUS :"x ∈ Even ⟹x+2 ∈ Even"
| MIN :"x ∈ Even ⟹ x-2 ∈ Even"
lemma aux : "x= ((x::int)-2) + 2" by simp
通过用 2-2+2 代替 2 来证明 lemma : "2 ∈ Even"
相当容易
但我想知道如何证明 lemma : "1 ∉ Even"
?
编辑:
(*Javier Diaz's answer*)
lemma Even_divisible_by_2: "n ∈ Even ⟹ 2 dvd n"
by (induction rule: Even.induct) (simp,presburger+)
lemma "1 ∉ Even"
proof
assume "1 ∈ Even"
then have "2 dvd 1"
using Even_divisible_by_2 by fastforce
then show False
using odd_one by blast
qed
3 的等效方法是什么?
lemma "3 ∉ Even"
proof
assume "3 ∈ Even"
then have "2 dvd 3"
(*how to continue?*)
qed
提前致谢
解决方法
我将首先证明一个中间结果,即归纳集中的每个数字都可以被 2 整除:
lemma Even_divisible_by_2: "n ∈ Even ⟹ 2 dvd n"
by (induction rule: Even.induct) simp_all
然后用反证法证明你的结果:
lemma "1 ∉ Even"
proof
assume "1 ∈ Even"
then have "2 dvd 1"
using Even_divisible_by_2 by fastforce
then show False
using odd_one by blast
qed
我强烈建议您使用 Isabelle/Isar 而不是证明脚本。
注意:根据博文作者的要求,我将按照上述证明的风格添加 3 ∉ Even
的证明:
lemma "3 ∉ Even"
proof
assume "3 ∈ Even"
then have "2 dvd 3"
using Even_divisible_by_2 by fastforce
then show False
using odd_numeral by blast
qed
替代解决方案:@user9716869 基于 Even_divisible_by_2
的使用提供了以下更通用和更有效的解决方案:
lemma n2k1_not_Even: "odd n ⟹ n ∉ Even"
using Even_divisible_by_2 by auto
lemma "1 ∉ Even" and "3 ∉ Even" and "11 ∉ Even"
by (simp_all add: n2k1_not_Even)