嵌套析取的证明(规则disjE)

时间:2013-04-09 12:12:27

标签: isabelle isar

在Isar风格的Isabelle样张中,这很有效:

from `a ∨ b` have foo
proof
  assume a
  show foo sorry
next
  assume b
  show foo sorry
qed

此处proof调用的隐式规则是rule conjE。但是我应该把它放在那里使其不仅仅是一个分离:

from `a ∨ b ∨ c` have foo
proof(?)
  assume a
  show foo sorry
next
  assume b
  show foo sorry
next
  assume c
  show foo sorry
qed

3 个答案:

答案 0 :(得分:3)

在写这个问题时,我有了一个想法,结果证明是我想要的:

from `a ∨ b ∨ c` have foo
proof(elim disjE)
  assume a
  show foo sorry
next
  assume b
  show foo sorry
next
  assume c
  show foo sorry
qed

答案 1 :(得分:3)

进行此类案例分析的另一种规范方法如下:

{ assume a
  have foo sorry }
moreover
{ assume b
  have foo sorry }
moreover
{ assume c
  have foo sorry }
ultimately
have foo using `a ∨ b ∨ c` by blast

也就是说,让一个自动工具“弄清楚”最后的细节。在考虑算术案例时(by arith作为最后一步),这种方法尤其有效。

更新:使用新的consider语句,可按以下方式完成:

notepad
begin
  fix A B C assume "A ∨ B ∨ C"
  then consider A | B | C by blast
  then have "something"
  proof (cases)
    case 1
    show ?thesis sorry
  next
    case 2
    show ?thesis sorry
  next
    case 3
    show ?thesis sorry
  qed
end

答案 2 :(得分:1)

除了区分大小写之外,您似乎可以使用更通用的induct方法来进行出价。对于三种情况,这可以这样工作:证明引理disjCases3

lemma disjCases3[consumes 1, case_names 1 2 3]:
  assumes ABC: "A ∨ B ∨ C"
  and AP: "A ⟹ P"
  and BP: "B ⟹ P"
  and CP: "C ⟹ P"
  shows "P"
proof -
  from ABC AP BP CP show ?thesis by blast
qed

您可以按如下方式使用此引理:

from `a ∨ b ∨ c` have foo
proof(induct rule: disjCases3)
  case 1 thus ?case 
     sorry
next
  case 2 thus ?case 
     sorry
next
  case 3 thus ?case 
     sorry
qed

缺点是您需要一堆引理来涵盖任意数量的案例,disjCases2disjCases3disjCases4disjCases5等等,但其他情况似乎如此工作得很好。