Update to true \== false post
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Michael Zhang 2023-05-08 00:37:13 -05:00
parent a08f1206c2
commit 2f68aae3dd
Signed by: michael
GPG Key ID: BDA47A31A3C8EE6B
1 changed files with 37 additions and 14 deletions

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@ -125,23 +125,46 @@ Let's make sure this isn't broken by trying to apply this to something that's
actually true:
```
2-map : → Type
2-map 2 =
2-map 2 = ⊥
2-map else =
-- 2≢2 : 2 ≢ 2
-- 2≢2 p = transport (λ i → 2-map (p i)) tt
data NotBool : Type where
true1 : NotBool
true2 : NotBool
same : true1 ≡ true2
```
I commented the lines out because they don't compile, but if you tried to
compile it, it would fail with:
In this data type, we have a path over `true1` and `true2` that is a part of the
definition of the `NotBool` type. Since this is an intrinsic equality, we can't
map `true1` and `true2` to divergent types. Let's see what happens:
```
notbool-map : NotBool → Type
notbool-map true1 =
notbool-map true2 = ⊥
```
Ok, we've defined the same thing that we did before, but Agda gives us this
message:
```text
!=<
when checking that the expression transport (λ i → 2-map (p i)) tt
has type ⊥
Errors:
Incomplete pattern matching for notbool-map. Missing cases:
notbool-map (same i)
when checking the definition of notbool-map
```
That's because with identical terms, you can't simultaneously assign them to
different values, or else it would not be a proper function.
Agda helpfully notes that we still have another case in the inductive type to
pattern match on. Let's just go ahead and give it some value:
```text
notbool-map (same i) =
```
If you give it ``, it will complain that `⊥ != of type Type`, but if you give
it `⊥`, it will also complain! Because pattern matching on higher inductive
types requires a functor over the path, we must provide a function that
satisfies the equality `notbool-map true1 ≡ notbool-map true2`, which unless we
have provided the same type to both, will not be possible.
So in the end, this type `NotBool → Type` is only possible to write if the two
types we mapped `true1` and `true2` can be proven equivalent. But this also
means we can't use it to prove `true1 ≢ true2`, which is exactly the property we
wanted to begin with.