Motivating Aya features: an incomplete story ​

Great. I expect you to have basic experience with interactive theorem proving. This is another Aya tutorial for interactive theorem prover users. If you find a bug, open an issue on GitHub!

This tutorial will use some basic Aya syntax. I hope those are sufficiently intuitive, or you can look up this tutorial.

Here's a little prelude, which you do not need to understand now.

prim I prim coe
prim intervalInv
inline def ~ ⇒ intervalInv
variable A B : Type
def infix = (a b : A) ⇒ ⟦ i ⟧ A { i := b | ~ i := a }
def refl {a : A} : a = a ⇒ fn i ⇒ a

open data Nat
| zero
| suc Nat

A journey begins ​

You went to a website and a random person has asked the following question:

open data Bool | true | false
def not Bool : Bool
| true ⇒ false
| false ⇒ true

def id (x : Bool) ⇒ x

def Goal ⇒ (fn x ⇒ not (not x)) = id

// And yes, below is the syntax for typed holes in Aya:
// def question : Goal => {??}

You realize that there is no way to prove it in your type theory. However, you are very smart and realized you can instead show the following:

def Goal' (x : Bool) ⇒ not (not x) = id x

This is pretty much the same theorem!

Now, suppose we need to show a propositional equality between two records. This means we have to show they're memberwise equal. One record has a member \ x ⇒ not (not x), and the other has id. This time, you cannot cheat by changing the goal type. You post the question on some mailing list and people are telling you that the alternative version of the theorem you have shown does not imply the original, unless "function extensionality" is a theorem in your type theory.

To have function extensionality as a theorem, you came across two distinct type theories: observational type theory and cubical type theory. Aya chose the latter (for now).

Cubical ​

Here's the proof of function extensionality in Aya:

def funExt (f g : A → B) (p : ∀ a → f a = g a) : f = g
   ⇒ fn i a ⇒ p a i

This is because Aya has a "cubical" equality type that is not inductively defined. We may also prove the action-on-path theorem, commonly known as cong, but renamed to pmap to avoid a potential future naming clash:

def pmap (f : A → B) {a b : A} (p : a = b) : f a = f b
   ⇒ fn i ⇒ f (p i)

We may also invert a path:

def sym {a b : A} (p : a = b) : b = a ⇒ fn i ⇒ p (~ i)

However, we cannot yet define transitivity of equality because we do not have the traditional elimination rule of the equality type -- the J rule. This will need some advanced proving techniques that are beyond the scope of this simple tutorial, so I'll skim them. First, we need type-safe coercion:

def cast (p : A ↑ = B) : A → B ⇒ coe 0 1 (fn i ⇒ p i)

Then, from q : b = c we construct the equivalence (a = b) = (a = c) and coerce along this equivalence:

def concat {a b c : A} (p : a = b) (q : b = c) : a = c ⇒
  cast (\i ⇒ a = q i) p

Note that at this point you can already do a bunch of familiar proofs about some simple types such as natural numbers or sized vectors. These are left as exercises, and you are encouraged to try yourself if you are not very sure about how it feels to prove things in Aya.

Jesper's master thesis ​

Here's a bonus feature. Remember the +-comm proof that you need two lemmas? It is standard to define + in the following way:

example def infix + Nat Nat : Nat
| 0, n ⇒ n
| suc m, n ⇒ suc (m + n)

And then you prove that a + 0 = a and a + suc b = suc (a + b). It is tempting to have | n, 0 => n as a computation rule as well, but this is incompatible with the usual semantics of pattern matching, which is compiled to elimination principles during type checking. However, you can do that in Aya. You may also add the other lemma as well.

overlap def infix + Nat Nat : Nat
| 0, n ⇒ n
| n, 0 ⇒ n
| suc m, n ⇒ suc (m + n)
| m, suc n ⇒ suc (m + n)
tighter =

This makes all of them definitional equality. So, +-comm can be simplified to just one pattern matching:

def +-comm (a b : Nat) : a + b = b + a
| 0, _ ⇒ refl
| suc _, _ ⇒ pmap suc (+-comm _ _)

Heterogeneous equality ​

When working with indexed families, you may want to have heterogeneous equality to avoid having mysterious coercions. For example, consider the associativity of sized vector appends. We first need to define sized vectors and the append operation:

variable n m o : Nat
// Definitions
open data Vec (n : Nat) (A : Type)
| 0, A ⇒ nil
| suc n, A ⇒ infixr :< A (Vec n A)
overlap def infixr ++ (Vec n A) (Vec m A) : Vec (n + m) A
| nil, ys ⇒ ys
| ys, nil ⇒ ys
| x :< xs, ys ⇒ x :< xs ++ ys
tighter :< =

It is tempting to use the below definition:

overlap def ++-assoc (xs : Vec n A) (ys : Vec m A) (zs : Vec o A)
  : (xs ++ ys) ++ zs = xs ++ (ys ++ zs)
| nil, ys, zs => refl
| x :< xs, ys, zs => pmap (x :<) (++-assoc xs ys zs)

overlap def ++-assoc (xs : Vec n A) (ys : Vec m A) (zs : Vec o A)
  : (xs ++ ys) ++ zs = xs ++ (ys ++ zs)
| nil, ys, zs => refl
| x :< xs, ys, zs => pmap (x :<) (++-assoc xs ys zs)

However, this definition is not well-typed:

• (xs ++ ys) ++ zs is of type Vec ((n + m) + o) A
• xs ++ (ys ++ zs) is of type Vec (n + (m + o)) A.

They are not the same! Fortunately, we can prove that they are propositionally equal. We need to show that natural number addition is associative, which is the key lemma of this propositional equality:

def +-assoc {a b c : Nat} : (a + b) + c = a + (b + c)
| {0} ⇒ refl
| {suc _} ⇒ pmap suc +-assoc

Now we can work on the proof of ++-assoc. Here's a lame definition that is well-typed in pre-cubical type theory, and is also hard to prove -- we cast one side of the equation to be other side:

example def ++-assoc-ty (xs : Vec n A) (ys : Vec m A) (zs : Vec o A)
  ⇒ cast (↑ pmap (fn n ⇒ Vec n A) +-assoc) ((xs ++ ys) ++ zs) = xs ++ (ys ++ zs)

It is harder to prove because in the induction step, one need to show that cast (\ i ⇒ pmap (\ n ⇒ (Vec n A)) (\ i ⇒ +-assoc i) i) is equivalent to the identity function in order to use the induction hypothesis. For the record, here's the proof:

def castRefl (a : A) : cast ↑ refl a = a ⇒ fn i ⇒ coe i 1 (fn j ⇒ A) a

But still, with this lemma it is still hard. Cubical provides a pleasant way of working with heterogeneous equality:

def Path (A : I → Type) (a : A 0) (b : A 1) ⇒ ⟦ i ⟧ A i { i := b | ~ i := a }

So if we have X : A = B and a : A, b : B, then Path (\i => X i) a b expresses the heterogeneous equality between a and b nicely.

We may then use the following type signature:

def ++-assoc (xs : Vec n A) (ys : Vec m A) (zs : Vec o A)
  ⇒ Path (fn i ⇒ Vec (+-assoc i) A) ((xs ++ ys) ++ zs) (xs ++ (ys ++ zs))

The proof is omitted (try yourself!).