Question

我在Isabelle / HOL中遇到了一个证明饮酒者原则的证据。我确实了解整个证明；但是，我对饮酒者原则证明中的下一个证明案例不太了解：

  assume "∀x. drunk x"
  then have "drunk a ⟶ (∀x. drunk x)" for a ..
  then show ?thesis ..

为什么证明还证明“ 醉⟶（∀x。drunk x）”？我认为∀x就足够了。醉x 以显示∃x。醉x 。

整个证明如下：

theorem Drinker's_Principle: "∃x. drunk x ⟶ (∀x. drunk x)"
proof cases
  assume "∀x. drunk x"
  then have "drunk a ⟶ (∀x. drunk x)" for a ..
  then show ?thesis ..
next
  assume "¬ (∀x. drunk x)"
  then have "∃x. ¬ drunk x" by (rule de_Morgan)
  then obtain a where "¬ drunk a" ..
  have "drunk a ⟶ (∀x. drunk x)"
  proof
    assume "drunk a"
    with ‹¬ drunk a› show "∀x. drunk x" by contradiction
  qed
  then show ?thesis ..
qed

Answer 1

我认为足够了。喝醉了x以显示∃x。醉x。

这就是我们实际上正在做的：语句末尾的for a

then have "drunk a ⟶ (∀x. drunk x)" for a ..

伊萨尔（Isar）所说的是，有一个证人“喝醉了⟶（∀x。drunk x）”。然后，证明的下一行使用它来证明具有存在前提的语句的版本。

饮酒者原则

1 个答案: