# Quest 2 - Sigma Types

We are still trying to express and “prove” the statement

The statement

There exists a natural that is even.

We will achieve this by the end of this quest.

## Part 0 - Existence / Dependent Pair / Total Space of Bundles

Recall from Quest 1 - Dependent Types
that we defined `isEven`

.
What’s left is to be able write down “existence”.
In maths we might write

```
∃ x ∈ ℕ, isEven x
```

which in `agda`

notation is

```
Σ ℕ isEven
```

This is called a *sigma type*, which has three interpretations:

the proposition “there exists an even natural”

the construction “keep a recipe

`n`

of naturals together with a recipe of`isEven n`

”the total space of the bundle

`isEven`

over`ℕ`

, which is the space obtained by putting together all the fibers. Pictorially, it looks likewhich can also be viewed as the subset of even naturals, since the fibers are either empty or singleton. (It is a

*subsingleton bundle*).

## Part 1 - Making terms in Sigma Types

Making a term of this type has three interpretations:

(giving a proof that there existence of an even natural amounts to giving) a natural

`n : ℕ`

and a proof`hn : isEven n`

that`n`

is even.pairing a recipe

`n : ℕ`

with a recipe`hn : isEven n`

.(giving a point in the total space is giving) a point

`n : ℕ`

downstairs together with a point`hn : isEven n`

in its fiber.

Now you can prove that there exists an even natural:

Formulate the statement you need. Make sure you have it of the form

Name : Statement Name = {!!}

Load the file, go to the hole and refine the goal.

If you formulated the statement right it should split into

`{!!} , {!!}`

and you can check the types of terms the holes require.Fill the holes. There are many proofs you can do!

In general when `A : Type`

is a type and `B : A → Type`

is a
predicate/dependent construction/bundle over `A`

,
we can write the sigma type `Σ A B`

whose terms are pairs `a , b`

where `a : A`

and `b : B a`

.
In the special case when `B`

is not dependent on `a : A`

,
i.e. it looks like `λ a → C`

for some `C : Type`

then
`Σ A B`

is just

the proposition “

`A`

and`C`

” since giving a proof of this is the same as giving a proof of`A`

and a proof of`C`

a recipe

`a : A`

together with a recipe`c : C`

`B`

is now a*trivial bundle*since the fibers`B a`

are constant with respect to`a : A`

. In other words it is just a*product*`Σ A B ≅ A × C`

. For this reason, some refer to the sigma type as the*dependent product*, but we will avoid this terminology.

```
_×_ : Type → Type → Type
A × C = Σ A (λ a → C)
```

`agda`

supports the notation `_×_`

(without spaces)
which means from now on you can write `A × C`

(with spaces).

## Part 2 - Using Terms in Sigma Types

There are two ways of using a term in a sigma type.
We can extract the first part using `fst`

or the second part using `snd`

.
Given `x : Σ A B`

there are three interpretations of `fst`

and `snd`

:

Viewing

`x`

as a proof of existence`fst x`

provides the witness of existence and`snd`

provides the proof that the witness`fst x`

has the desired propertyViewing

`x`

as a recipe`fst`

extracts the first component and`snd`

extracts the second componentViewing

`x`

as a point in the total space of a bundle`fst x`

is the point that`x`

is over in the base space and`snd x`

is the point in the fiber that`x`

represents. In particular you can interpret`fst`

as projection from the total space to the base space, collapsing fibers.

For example to define a map that takes an even natural and divides it by two we can do

```
div2 : Σ ℕ isEven → ℕ
div2 x = {!!}
```

Load the file, go to the hole and case on

`x`

. You might want to rename`fst₁`

and`snd₁`

.div2 : Σ ℕ isEven → ℕ div2 (fst₁ , snd₁) = {!!}

Case on

`fst₁`

and tell`agda`

what to give for`0 , *`

, i.e. what “zero divided by two” ought to be.div2 : Σ ℕ isEven → ℕ div2 (zero , snd₁) = {!!} div2 (suc fst₁ , snd₁) = {!!}

Navigate to the second hole and case on

`fst₁`

again. Notice that`agda`

knows there is no term looking like`1 , *`

so it has skipped that case for us.div2 : Σ ℕ isEven → ℕ div2 (zero , snd₁) = 0 div2 (suc (suc fst₁) , snd₁) = {!!}

`(n + 2) / 2`

should just be`n/2 + 1`

so try writing in`suc`

and refining the goalHow do you write down

`n/2`

? Hint: we are in the “inductive step”.Try dividing some terms by

`2`

:Use

`C-c C-n`

and write`div2 (2 , tt)`

for example.Try dividing

`36`

by`2`

.

*Important observation* :
the two proofs `2 , tt`

and `36 , tt`

of the statement
“there exists an even natural” are not “the same” in any sense,
since if they were `div2 (2 , tt)`

would be “the same” `div2 (36/2 , tt)`

,
and hence `1`

would be “the same” as `18`

.

“The same”

Are they “the same”? What is “the same”?