# Calculus of Constructions

As an additional extension to the hierarchy of logical systems we’ve discussed, we’ll consider type theory. In type theory, every “term” has a “type” and operations are restricted to terms of a certain type.

Type theory was created to avoid paradoxes in a variety of formal logics and rewrite systems. For example, Russell’s paradox is not applicable since every sentence has its own type.

The most powerful concept in Coq is not about Coq; it’s the calculus of constructions. Coq is just a lot of awesome automation machinery on top of that.

Calculus of Constructions is a type theory based on intuitionistic logic and simply typed lambda calculus. Together with Dependent Types, Inductive Types, Recursion, they provide a neat way to do proofs.

To demonstrate why they are powerful, let’s consider one of the inference rules as an example:
${\displaystyle {\Gamma ,x:A\vdash B:K\qquad \qquad \Gamma ,x:A\vdash N:B \over {\Gamma \vdash (\lambda x:A.N):(\forall x:A.B):K}}}$

What this means is that if above the line holds, then so does everything below the line.

In this case, K is either a Prop or a Type. For simplicity, assume K is Type. To translate it, it means that:
1. If we have a judgment x of type A, from which it follows that B : K (B is a Type)
and
2. If we have a judgment x of type A, from which it follows that N : B (N has type of B)
then we can deduce that
(\x : A . N) has type (\forall (x : A) . B), and the term (\forall (x : A) . B) has type Type.

In other words, this is the rule that ties together lambda abstraction and universal quantification. It also says that a universal quantification is a type.

This rule is the reason why this flows nicely:

(* Make sure you have enabled "Display all basic low-level contents" *)
Definition test (s : Set) : bool := true.

Check test.                 (* test : forall _ : Set, bool *)

Check forall _ : Set, bool. (* forall _ : Set, bool : Type *)


Other data types are encoded similarly as with Church Numerals.

The interesting thing about this system is that it has the strong normalization property, which means that every sequence of rewrites eventually terminates.

While this is not useful for IO, it is very useful for proofs.

In any case, we can take advantage of IO by extracting code to other languages as seen in Coq to Haskell.

Lately I spent some time learning Coq. It is a very useful programming language, but guess what? It has no IO!

So, once we prove our theorems, what do we do? We extract our code!

Let’s pick a language that has IO, for example Haskell.

For fun, let’s try to prove the following and extract the definitions to Haskell:
The sum of a list containing natural numbers is always greater than or equal to 0.

But, in order to prove this, we must define everything on a core level, that is:
1. We need to define what a list is
2. We need to define what a natural number is

We can, but we won’t do that since they are already defined in Coq itself. So we’ll just go ahead and use “list” and “nat” in our code.

Here’s the definition of our recursive function that will calculate the sum of all the elements in a list of natural numbers:

Fixpoint my_sum (l : list nat) : nat :=
match l with
| nil     => 0
| x :: l' => x + my_sum l'
end.


It’s a pretty straight-forward definition. Let’s write some examples:

Example my_sum1 : my_sum [1 ; 2 ; 3] = 6.
Proof.
simpl. reflexivity.
Qed.

Example my_sum2 : my_sum [] = 0.
Proof.
simpl. reflexivity.
Qed.

Example my_sum3 : my_sum [1] = 1.
Proof.
simpl. reflexivity.
Qed.


And now, for the proof of our theorem:

(* Sum of a list of naturals will always produce a result greater than or equal to 0 *)
Theorem my_sum_nat_gt_0 : forall l : list nat, 0 <= my_sum l.
Proof.
intros l.
induction l.
- (* base *) simpl. reflexivity.
- (* i.h. *) simpl. case a.
+ (* zero *) simpl. exact IHl.
+ (* S n  *) intros n. rewrite IHl. exact (Plus.le_plus_r (S n) (my_sum l)).
Qed.


Cool! Now for the final part, which is the extraction:

Require Extraction.
Recursive Extraction my_sum.


And finally, our Haskell code generated by the commands above:

data Nat =
O
| S Nat

data List a =
Nil
| Cons a (List a)

add :: Nat -> Nat -> Nat
case n of {
O -> m;
S p -> S (add p m)}

my_sum :: (List Nat) -> Nat
my_sum l =
case l of {
Nil -> O;
Cons x l' -> add x (my_sum l')}


We can now use our function my_sum for which we know our theorem to be true, and we can use it in a programming language that supports IO.

Pretty cool stuff, right? 🙂

Bonus: You can’t do these kind of proofs in Haskell itself, since it has no support for dependent types (we’ll cover that in a future post).

# MU puzzle

The MU puzzle is an example of a Formal system we will take a look at, as described by Douglas Hofstadter in his book Gödel, Escher, Bach.

We’re given a starting string MI, and some transformation rules.

Here are the transformation rules:

 Nr. Formal rule Informal explanation Example 1. xI → xIU Add a U to the end of any string ending in I MI to MIU 2. Mx → Mxx Double the string after the M MIU to MIUIU 3. xIIIy → xUy Replace any III with a U MUIIIU to MUUU 4. xUUy → xy Remove any UU MUUU to MU

Here is an example usage to derive MIIU from MII for some of the transformation rules:

1. MI (an axiom)
2. MII (by rule 2)
3. MIIII (by rule 2)
4. MIIIIIIII (by rule 2)
5. MUIIIII (by rule 3)
6. MUUII (by rule 3)
7. MII (by rule 4)
8. MIIU (by rule 1)

More formally, we can represent the MU puzzle as follows:

1. Language
1. The set of symbols is { M, I, U }.
2. A string is a grammatical sentence (wff) if its first word is M, and no other words are Ms. E.g.: The following strings are grammatical sentences: M, MIUIU, MUUUIII.
2. MI is the starting string, thus an axiom.
3. The inference rules are the transformation rules we’ve just defined above.

The question now is, can we convert MI to MU using the transformation rules?

In order to answer this, we will use invariant (a property that holds true whenever we apply some of the rules) with induction to prove our claim.

What would be a good invariant here?

Let’s suppose we can somehow get to MUIIIU. Now, if we apply rule 3 to it, we get MUUU. Apply rule 4, and pure win!

Note that, in order to be able to apply rule 3, we need to have the number of subsequent I’s to be divisible by 3. So let’s have our invariant say that “There are no subsequent I’s in the string that are divisible by 3”.

1. For the starting axiom, we have one I. Invariant OK.
2. Applying rule 2 will be doubling the number of I’s, so we can have: I, II, IIII, IIIIIII (in particular, 2^n I’s). Invariant OK.
3. Applying rule 3 will be reducing the number of I’s by 3. But note that 2^n – 3 is still not divisible by 3. Invariant OK.

So we’ve shown that with the starting axiom MI it is not possible to get to MU. But, what if we have a different starting axiom? Or even, different rules? 🙂

But if we look carefully, we’ve used a different formal system to reason about MU (i.e. divisibility by 3). This is because the puzzle cannot be solved in its own system. Otherwise, an algorithm would keep trying different inference rules of MU indifinitely (not knowing that MU is impossible).

In general, for any formal system there’s this limitation. Gödel’s theorem shows that there’s no formal system that can contain all possible truths, because it cannot prove some truths about its own structure.

So, having experience with different formal systems and combining them as needed can be useful 🙂