我试图在Isabelle中完成一个现在有效的证据:
lemma axiom1: " x = y ⟹ Δ x y z = 0"
proof -
assume 1[simp]: "x = y"
have 1: "Δ x y z = Δ y z x" by (rule axiom0_a)
also have "… = Δ y z y" by simp
also have "… = Δ z y y" by (rule axiom0_a)
moreover have "Δ y y z = - Δ z y y" by (rule axiom0_b)
moreover have "⋀r. ((r::real) = - r ⟹ r = 0)" by simp
ultimately show "Δ x y z = 0" by simp
qed
但是,我不得不手动将该假设添加到简化器中。我的问题是,简化器x=y
中的附加规则是否仍然是本证明的局部,或者是否会用于其他证明(这会导致一些问题)?另外,我认为这些假设会自动添加到简化器中:这个假设不是因为它会给循环带来一些危险吗?
答案 0 :(得分:1)
1
也有效的情况下使用proof
时,它不再是您当前子目标中的假设。如果您使用结构化证明,通过assumes
和shows
以结构化方式记下引理通常会更好:
lemma axiom1:
assumes 1[simp]: "x = y"
shows "Δ x y z = 0"
...