## Imports

These are some imports that are required for code on this page to work properly.

```
{-# OPTIONS --cubical #-}
open import Cubical.Foundations.Prelude
open import Data.Bool
open import Data.Unit
open import Data.Empty
¬_ : Set → Set
¬ A = A → ⊥
infix 4 _≢_
_≢_ : ∀ {A : Set} → A → A → Set
x ≢ y = ¬ (x ≡ y)
```

The other day, I was trying to prove `true ≢ false`

in Agda. I would write the
statement like this:

```
true≢false : true ≢ false
```

For many “obvious” statements, it suffices to just write `refl`

since the two
sides are trivially true via rewriting. For example:

```
open import Data.Nat
1+2≡3 : 1 + 2 ≡ 3
1+2≡3 = refl
```

This is saying that using the way addition is defined, we can just rewrite the left side so it becomes judgmentally equal to the right:

```
-- For convenience, here's the definition of addition:
-- _+_ : Nat → Nat → Nat
-- zero + m = m
-- suc n + m = suc (n + m)
```

- 1 + 2
- suc zero + suc (suc zero)
- suc (zero + suc (suc zero))
- suc (suc (suc zero))
- 3

However, in cubical Agda, naively using `refl`

with the inverse statement
doesn’t work. I’ve commented it out so the code on this page can continue to
compile.

```
-- true≢false = refl
```

It looks like it’s not obvious to the interpreter that this statement is actually true. Why is that

## Intuition

Well, in constructive logic / constructive type theory, proving something is
false is actually a bit different. You see, the definition of the `not`

operator, or $\neg$, was:

```
-- ¬_ : Set → Set
-- ¬ A = A → ⊥
```

The code is commented out because it was already defined at the top of the page in order for the code to compile.

This roughly translates to, “give me any proof of A, and I’ll produce a value of
the bottom type.” Since the bottom type $\bot$ is a type without values, being
able to produce a value represents logical falsehood. So we’re looking for a way
to ensure that any proof of `true ≢ false`

must lead to $\bot$.

The strategy here is we define some kind of “type-map”. Every time we see
`true`

, we’ll map it to some arbitrary inhabited type, and every time we see
`false`

, we’ll map it to empty.

```
bool-map : Bool → Type
bool-map true = ⊤
bool-map false = ⊥
```

This way, we can use the fact that transporting over a path (the path supposedly given to us as the witness that true ≢ false) will produce a function from the inhabited type we chose to the empty type!

```
true≢false p = transport (λ i → bool-map (p i)) tt
```

I used `true`

here, but I could equally have just used anything else:

```
bool-map2 : Bool → Type
bool-map2 true = 1 ≡ 1
bool-map2 false = ⊥
true≢false2 : true ≢ false
true≢false2 p = transport (λ i → bool-map2 (p i)) refl
```

## Note on proving divergence on equivalent values

Let’s make sure this isn’t broken by trying to apply this to something that’s actually true:

```
data NotBool : Type where
true1 : NotBool
true2 : NotBool
same : true1 ≡ true2
```

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:

```
Errors:
Incomplete pattern matching for notbool-map. Missing cases:
notbool-map (same i)
when checking the definition of notbool-map
```

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:

```
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.