如何证明(forall x, P x /\ Q x) -> (对于所有 x,P x)
如何证明 (forall x, P x /\ Q x) -> (forall x, P x) 在 Coq 中? 已经尝试了几个小时,但无法弄清楚如何将先行词分解为 Coq 可以消化的内容。 (显然我是个新手:)
How does one prove (forall x, P x /\ Q x) -> (forall x, P x) in Coq? Been trying for hours and can't figure out how to break down the antecedent to something that Coq can digest. (I'm a newb, obviously :)
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只需应用 H 即可更快地完成此操作,但是此脚本
应该更清楚。
You can do it more quickly by just applying H, but this script
should be more clear.
尝试
Try
实际上,当我发现这个时,我就明白了这一点:
计算机科学家数学 2< /a>
在第 5 课中,他解决了完全相同的问题并使用“cut (P x /\ Q x)”,将目标从“P x”重写为“P x /\ Q x -> P x” 。 从那里你可以做一些操作,当目标只是“P x /\ Q x”时,你可以应用“forall x : P x /\ Q x”,其余的就很简单了。
Actually, I figured this one out when I found this:
Mathematics for Computer Scientists 2
In lesson 5 he solves the exact same problem and uses "cut (P x /\ Q x)" which re-writes the goal from "P x" to "P x /\ Q x -> P x". From there you can do some manipulations and when the goal is just "P x /\ Q x" you can apply "forall x : P x /\ Q x" and the rest is straightforward.
直观地,如果对于所有 x,P(x) AND Q(x) 成立,那么对于所有 x,P(x) 成立。
Intuitivly, if for all x, P(x) AND Q(x) hold, then for all x, P(x) holds.
答案如下:
您可以在此文件中找到其他已解决的练习:https: //cse.buffalo.edu/~knepley/classes/cse191/ClassNotes.pdf
here is the answer:
you can find other solved exercises in this file: https://cse.buffalo.edu/~knepley/classes/cse191/ClassNotes.pdf