[英]How to complete this proof of commutativity with `replace`?
On this documentation, it is mentioned how replace
could be used to complete the proof, but it ends up using rewrite
, which seems to be a syntax sugar that writes replace
for you. 在本文档中,提到了如何使用replace
来完成证明,但最终使用rewrite
,这似乎是为您编写replace
的语法糖。 I'm interested in understanding how to use it explicitly. 我有兴趣了解如何显式使用它。
If I understand correctly, it could be used to rewrite S k = S (plus k 0)
as S (plus k 0) = S (plus k 0)
, given a proof that k = plus k 0
, which would then be provable by reflexivity. 如果我理解正确,它可以用来将S k = S (plus k 0)
重写为S (plus k 0) = S (plus k 0)
,并证明k = plus k 0
,那么这是可证明的通过反射。 But if we instance it as replace {P = \\x => S x = S (plus k 0)} {x = k} {y = plus k 0} rec
, we'll now need a proof of S k = S (plus k 0)
, which is what we wanted to prove to begin with. 但是,如果我们将其replace {P = \\x => S x = S (plus k 0)} {x = k} {y = plus k 0} rec
为replace {P = \\x => S x = S (plus k 0)} {x = k} {y = plus k 0} rec
,我们现在需要S k = S (plus k 0)
的证明S k = S (plus k 0)
,这就是我们想证明的开始。 In short, I'm not sure what exactly P
should be. 简而言之,我不确定P
应该是什么。
Ah, it is fairly obvious in retrospect. 嗯,回想起来很明显。 If we let: 如果我们让:
P = \x => S x = S (plus k 0)
Then, we can prove it for x = (plus k 0)
(by reflexivity). 然后,我们可以证明x = (plus k 0)
(通过自反)。 Now, if we let y = k
, then, by using replace
, we gain a proof of S k = S (plus k 0)
, which is what we need. 现在,如果让y = k
,那么通过使用replace
,我们得到了S k = S (plus k 0)
的证明。 Or, in other words: 或者,换句话说:
plusCommZ : (m : Nat) -> m = plus m 0
plusCommZ Z = Refl
plusCommZ (S k) = replace
{P = \x => S x = S (plus k 0)}
{x = plus k 0}
{y = k}
(sym (plusCommZ k))
Refl
Completes the proof. 完成证明。 We could do it the other way around with P = \\x => S x = S k
. 我们可以用P = \\x => S x = S k
来进行另一种方式。
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