[英]Type Hierarchy in Agda
I am trying to figure out how type hierarchies work in Agda. 我试图找出类型层次结构如何在Agda中工作。
Assuming I define a set type X: 假设我定义了一个集合类型X:
X : Set
and then proceed to constructing an inductive type 然后继续构建归纳型
data Y : X -> Set where
What is the type of X -> Set
? 什么是X -> Set
? Is it Set or Type? 是设置还是类型?
Thank you! 谢谢!
Well, why not ask Agda itself? 好吧,为什么不问阿格达呢? I'm going to use excellent Agda mode for Emacs. 我将为Emacs使用出色的Agda模式。 We start with: 我们从:
module Hierarchy where
postulate
X : Set
data Y : X → Set where
-- empty
We have to load the file using Cc Cl
; 我们必须使用Cc Cl
加载文件; this typechecks the file, turns ?
这个typechecks文件,转?
s into holes, does syntax highlighting and so on. 进入洞,语法高亮等等。
Now, there is a command "Infer (deduce) type" available via Cc Cd
, so let's use that: 现在,有一个命令“推断(演绎)类型”可以通过Cc Cd
,所以让我们使用:
> C-c C-d
Expression:
> Y
X → Set
Right, that makes sense. 对,这是有道理的。 We defined Y : X → Set
, so it should come as no surprise. 我们定义了Y : X → Set
,所以它应该不足为奇。 Let's ask again: 我们再问一遍:
> C-c C-d
Expression:
> X → Set
Set₁
So, there you have it: Y : X → Set : Set₁
. 所以,你有它: Y : X → Set : Set₁
。
While the first part answers the question and shows you how to check this stuff yourself, doing that everytime would be dull, at least. 虽然第一部分回答了问题,并向您展示了如何自己检查这些东西,但每次这样做都会变得乏味,至少。 Here's how it works: 以下是它的工作原理:
To avoid paradoxes, we require 为避免悖论,我们要求
Set i : Set (i + 1)
which gives you the (infinite) hierarchy of Set
s. 它为您提供了Set
的(无限)层次结构。 If you had Set : Set
(which Agda allows via the --type-in-type
flag), you could derive contradiction such as this one . 如果你已经Set : Set
(这阿格达通过允许--type-in-type
标志),可以推导出矛盾,比如这一个 。
This also gives us a simple rule for functions: 这也为我们提供了一个简单的函数规则:
A : Set i
B : A → Set j
(a : A) → B a : Set (max i j)
Applying this to your example: 将此应用于您的示例:
X : Set
Set : Set₁
X → Set : Set (max 0 1)
X → Set : Set₁
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