我如何表明,如果假设不暗示,那就与说命题等于false(COQ)一样?
我想证明两个字符不等。我的环境目前看起来像这样:
H: c1 = c2 -> not
_____________________
Goal: (c1 =? c2)%char = false
它不允许我应用H或重写H来解决目标,而H不可逆转。目标直接源于假设,那么我该如何证明这一点?我需要助手引理吗?
I want to prove two characters are not equal. My environment currently looks like this:
H: c1 = c2 -> not
_____________________
Goal: (c1 =? c2)%char = false
It won't allow me to apply H or rewrite H to solve the goal, and H is not invertible. The goal directly follows from the hypothesis, so how can I prove this? Do I need a helper lemma?
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(1)
看起来您的假设中有一个错别字
h
?您是指
不是(C1 = C2)
(或c1<> c2
,还是c1 = c2 - > false
)?如果是这样,您可以应用标准库的引理:
也可能有趣的是,查看
eqb_neq
的证明,以了解如何证明这种引理。Looks like there is a typo in your hypothesis
H
?Do you mean
not (c1 = c2)
(orc1 <> c2
, orc1 = c2 -> False
) ?If so, you may apply a lemma of Standard Library:
It may be interesting to look at the proof of
eqb_neq
too, in order to understand how to prove this kind of lemmas.