# Decidable orderings in Idris

Posted on November 11, 2016

The goal of this post is to explore ordering relations in Idris. We will discuss the notion of decision procedures, seeing how to decide equality of natural numbers. We will define the notion of minimality of natural numbers and we will show that minimality is a decidable property. Finally, we will generalize the notion of minimality, and implement a verified lexicographic total ordering for vectors.

This post is a literate Idris file, so here is the module declaration and necessary imports.

module Data.Order.Total
import Data.Vect
%default total
%access public export

## Decidable equality

In Idris, decision procedures are really useful. A decision procedure lets you promote value-level information to the type-level, even at runtime. For example, a decision procedure for equality produces a decision of the equality of two things: either they are equal, in which case the procedure has constructed a proof that they are equal; or they are not, in which case the procedure has constructed a proof that the equality leads to a contradiction. (In constructive logic, this is what is understood by a proposition being false.)

We can represent decisions with a datatype.

data Dec : Type -> Type where
Yes : prop -> Dec prop
No : (prop -> Void) -> Dec prop

Let’s implement a decision for the equality of natural numbers. Recall that natural numbers are defined like this.

||| Natural numbers.
data Nat : Type where
||| Zero is a natural number.
Z : Nat
||| Every natural number has a successor.
S : Nat -> Nat

Obviously, zero is equal to zero, and zero is not equal to the successor of any number, but what if we have two successors? Induction! Let’s formalize this argument with a function that computes the decision, i.e. a decision procedure for the equality of Nat.

decEqNat : (n : Nat) -> (m : Nat) -> Dec (n = m)

This reads “for all natural numbers $$n$$, and for all natural numbers $$m$$, we can decide whether $$n$$ is equal to $$m$$”.

decEqNat Z Z = Yes Refl

Of course zero is equal to zero, and Idris recognizes this. The reason is simple: when we match Z for n and m, type refinement occurs. This means that in the type signature that is specialized for this branch of the function, n and m are replaced by Z as well. Consequently, the goal of the function has changed: we now have to produce Dec (Z = Z). This is easy: we have Refl : x = x, so unifying x with Z gives us a proof Z = Z. Then, wrapping with Yes gives us a value of type Dec (Z = Z), as required.

decEqNat (S _) Z = No (\Refl impossible)
decEqNat Z (S _) = No (\Refl impossible)

The syntax here is a bit tricky, but the idea is that the successor of a number cannot be zero: because the constructors are different, Idris recognizes that they are not equal. However, we need to show that them being equal is a contradiction.

Suppose we wrote No (\x => ?a) and asked Idris the type of the hole, it would say that x : S _t = Z. Matching with Refl gives the left-hand side type error, because Idris cannot unify S _ with Z. We explicitly mark the case as impossible. But this is the only case for the function! What happens when a function has no valid cases? The intuition is that the function has been proven to be uncallable. When this happens, the result type of the function is arbitrary. Because the function can never be called, it may as well return whatever you like; it’s not like you could ever call it anyway. In particular, we can even produce values of type Void, the type with no value, which is exactly what No requires.

The same reasoning applies to the symmetric case decEqNat Z (S _) = ?a.

At last, we have the inductive case. We are doing induction on both numbers at once. If the two smaller numbers are equal, then it suffices to apply S to both sides of the equality to show that the bigger numbers are equal. Equality is a congruence, meaning that applying a function to both sides of an equality produces another equality. This notion is captured by the function cong : (a = b) -> f a = f b.

decEqNat (S n) (S m) with (decEqNat n m)
| Yes eqPrf = Yes (cong eqPrf)

The No case is a bit trickier. Let’s proceed the same way as in the previous case to produce the contradiction.

1. Writing No (\x => ?a) gives us a goal of Void with the assumptions eqContra : (n = m) -> Void and x : S n = S m.
2. Matching Refl for x causes type refinement to occur: S n must unify with S m. However, Idris can make an additional inference. Since data constructors are injective, Idris infers also that n must unify with m, so the variable m is replaced by n in all our assumptions, giving us that eqContra : (n = n) -> Void.
3. Of course, Refl : n = n, so we can apply Refl to eqContra and get Void, as required.

Using this reasoning gives the following equation.

  | No eqContra = No (\Refl => eqContra Refl)

Now suppose we have a natural number coming from runtime - maybe we parsed a number on standard input. We can check that the number is equal to, say, 3 and get either a proof that this is so or that this is a contradiction. This proof can then be used later in the program, helping us on the type level.

Equality is just one kind of relation. Can we generalize the theory we developed for equality to other types of relations? The next simplest relation to study is orderings.

## Decidable orderings

There are, very broadly speaking, two kinds of orders: partial orders and total orders. In a partial order, it is possible to have two elements that are not ordered, and we say that are incomparable in that case. In a total order, any two elements are ordered in some way: either the first is “less than” the second, or the second is “less than” the first. I use scare quotes because the ordering can be other than numeric inequality, especially for partial orders. (Also, the strict “less than” on numbers is not a total or partial order, but it’s shorter than writing “no greater than” all the time.)

For example, consider the powerset boolean algebra of the set $$A = \{a, b, c\}$$. The elements in the boolean algebra are all possible subsets of this set. The ordering relation is set containment: one set is contained in an other (not to be confused with “is an element of another”) if all its elements are elements of the other set. So $$\{a, b\} \subseteq \{a, b, c\}$$. However, neither $$\{a, b\} \subseteq \{b, c\}$$ nor $$\{b, c\} \subseteq \{a, b\}$$, so this ordering is partial.

To be a partial order, a few properties must be satisfied. 1. The order must be reflexive: every element is “less than” itself 2. The order must be transitive: if $$A$$ is “less than” $$B$$, and $$B$$ is “less than” $$C$$, then we can infer that $$A$$ is “less than” $$C$$. 3. The order must be antisymmetric: if $$A$$ is “less than” $$B$$ and $$B$$ is “less than” $$A$$, then we can infer that $$A = B$$. You can see for yourself that these properties are satisfied in the powerset boolean algebra.

Total orders have all these same requirements, plus the additional requirement that any two elements can be ordered. Intuitively, the natural numbers are totally ordered by the less-than-or-equal relation.

Intuition isn’t good enough though. Let’s prove it. Here’s the less-than-or-equal relation for Nat. It is based on two ideas: zero is less than any natural, and if we have that a natural is less than another, then we can add one to “both sides”.

data LTE : Nat -> Nat -> Type where
LTEZero : LTE Z n
LTESucc : LTE n m -> LTE (S n) (S m)

Now let’s write a decision procedure for whether a given natural is less than another one according this relation.

decLte : (n : Nat) -> (m : Nat) -> Dec (LTE n m)

The zero case is easy, as usual. Since zero is less than any other natural, we can use the LTEZero proof immediately.

decLte Z _ = Yes LTEZero

On the other hand, if we have a successor in the first argument and a zero in the second, we have a contradiction. How can a successor of some number be less than or equal to zero? We just match both cases and dispel them as impossible, which Idris confirms because the matches each produce a type error in the left-hand side.

decLte (S _) Z
= No (\x => case x of LTEZero impossible ; LTESucc _ impossible)

(Write down the unification constraints produced by the matches to see how the contradiction arises in both cases.)

Finally, we have the inductive case, with successor in both arguments. We set up the induction hypothesis with a recursive call in a with-clause, and then see whether the smaller numbers are ordered or not. If they are, then we can apply the LTESucc rule to infer that the larger ones are too.

decLte (S n) (S m) with (decLte n m)
| Yes ltePrf = Yes (LTESucc ltePrf)

In the No case, we must prove that LTE (S n) (S m) -> Void. So we use a lambda to introduce the assumption LTE (S n) (S m). We can match this assumption with LTESucc p to get p : LTE n m. The No match gives us a proof that the smaller numbers being ordered is a contradiction, so we also have the assumption lteContra : LTE n m -> Void available to us. It will suffice to apply p to lteContra to get Void, as required.

  | No lteContra = No (\(LTESucc p) => lteContra p)

With decLte, we can decide whether one number is less than another. However, it turns out that this function cannot be used to decide which of two naturals is minimal! Before continuing to explain why, let’s see what exactly is meant by “minimal”.

||| A shorthand for inequality.
(/=) : a -> b -> Type
x /= y = Not (x = y)
symNeq : (x /= y) -> (y /= x)
symNeq neq eq = neq (sym eq)
||| Minimality among naturals.
data Minimal : Nat -> Nat -> Type where
||| The first natural is minimal: it is strictly smaller than the second.
MinLT : LTE n m -> (n /= m) -> Minimal n m
||| Both are minimal: they are equal.
MinEQ : (n = m) -> Minimal n m
||| The second natural is minimal: it is strictly smaller than the first.
MinGT : LTE m n -> (n /= m) -> Minimal n m

Now we would like to write a function to decide, given two numbers, which is minimal. This is the total order property for LTE. Here’s the signature of our decision procedure.

decMin : (n : Nat) -> (m : Nat) -> Minimal n m

It might seem surprising that decLte will be useless in implementing this function. Using decLte, an implementation strategy would be to compute a decision on whether the first argument is less than the second, and then compute a decision on whether the second is less than the first. Then four cases arise: Yes/Yes, Yes/No, No/Yes, and No/No. Intuitively, the last case is a contradiction: how can neither number be minimal? Even so, we will be unable to convince Idris that this case is in fact impossible. We would have in that case the assumptions notNM : LTE n m -> Void and notMN : LTE m n -> Void. Since we cannot match on function types, we cannot use matching to tease the contradiction out of the data. The other option then is to build either LTE n m or LTE m n so that we can invoke one our assumptions to produce Void. (If ever you produce Void in a case, you can dispel it with absurd : Void -> a which lets you derive anything.) However we are implementing the function that determines either LTE n m or LTE m n for any n and m; intuitively we can’t build either such proof just like that.

Instead, we will repeat more or less the same implementation as for decLte and decEqNat and combine them.

decMin Z Z = MinEQ Refl
decMin Z (S _) = MinLT LTEZero (\Refl impossible)
decMin (S _) Z = MinGT LTEZero (\Refl impossible)
decMin (S n) (S m) = l (decMin n m) where
l : Minimal n m -> Minimal (S n) (S m)
l (MinLT lt neq) = MinLT (LTESucc lt) (\Refl => neq Refl)
l (MinEQ eq) = MinEQ (cong eq)
l (MinGT gt neq) = MinGT (LTESucc gt) (\Refl => neq Refl)

This formalism of minimality is quite powerful. Thanks to the embedded in/equality proofs, we can use it to easily implement decEqNat

decEqNat' : (n : Nat) -> (m : Nat) -> Dec (n = m)
decEqNat' n m with (decMin n m)
| MinLT _ neq = No neq
| MinEQ eq = Yes eq
| MinGT _ neq = No neq

In fact, decMin is so strong that we can use it to implement decLte as well, but we will need to use the fact that the LTE relation is antisymmetric.

lteAntiSym : LTE n m -> LTE m n -> n = m
lteAntiSym LTEZero LTEZero = Refl
lteAntiSym (LTESucc p) (LTESucc q) = rewrite lteAntiSym p q in Refl
decLte' : (n : Nat) -> (m : Nat) -> Dec (LTE n m)
decLte' n m with (decMin n m)
| MinLT lte _ = Yes lte
| MinEQ eq = case eq of Refl => Yes lteRefl
| MinGT gte neq = No (\lte => neq (lteAntiSym lte gte))

### Generalizing minimality

We can generalize the proposition Minimal by refactoring the ordering relation into a type parameter.

data OrderedBy : {a : Type} -> (order : a -> a -> Type) -> (x : a) -> (y : a) -> Type where
LT : order x y -> (x /= y) -> OrderedBy order x y
EQ : (x = y) -> OrderedBy order x y
GT : order y x -> (x /= y) -> OrderedBy order x y

Of course, our Minimal is isomorphic to OrderedBy LTE.

minimalityIsOrderedByLte : Minimal x y -> OrderedBy LTE x y
minimalityIsOrderedByLte (MinLT lte neq) = LT lte neq
minimalityIsOrderedByLte (MinEQ eq) = EQ eq
minimalityIsOrderedByLte (MinGT gte neq) = GT gte neq
orderedByLteIsMinimality : OrderedBy LTE x y -> Minimal x y
orderedByLteIsMinimality (LT lte neq) = MinLT lte neq
orderedByLteIsMinimality (EQ eq) = MinEQ eq
orderedByLteIsMinimality (GT gte neq) = MinGT gte neq

I leave it as a (very boring) exercise to demonstrate that these functions are inverses, and to tie this in with Control.Isomorphism.

We can collect all the axioms of a total ordering on some type into an interface.

interface TotalOrder (a : Type) (order : a -> a -> Type) | order where
constructor MkTotalOrder
orderRefl : order x x
orderTrans : order x y -> order y z -> order x z
orderAntiSym : order x y -> order y x -> x = y
orderTotal : (x : a) -> (y : a) -> OrderedBy order x y

Of course, our theory of natural numbers with LTE can be made into an implementation of this interface.

TotalOrder Nat LTE where
orderRefl = lteRefl
orderTrans = lteTransitive
orderAntiSym = lteAntiSym
orderTotal x y = minimalityIsOrderedByLte (decMin x y)

## Ordering vectors

Consider the words “cat” and “dog”: the former comes before the latter, since “c” comes before “d” in alphabetical order. What happens if the first letters are equal? We defer the ordering decision to the substring made by removing the first letter. For example, consider “mouse” and “moose”: since the first two letters are the same, we compare “u” with “o” to conclude the “moose” comes before “mouse”.

The lexicographic order is a generalization of this way of applying alphabetical order. We will show that any two vectors of the same length and containing totally ordered elements (according to any total ordering) are totally ordered by the lexicographic order. First, we need to define the lexicographic order predicate for equal-length vectors.

data Lexicographic : (sub : a -> a -> Type) -> Vect n a -> Vect n a -> Type where
LexEqZ : Lexicographic sub [] []
LexEqS
: (x = y)
-> Lexicographic sub xs ys
-> Lexicographic sub (x::xs) (y::ys)
LexLT : sub x y -> (x /= y) -> Lexicographic sub (x::xs) (y::ys)

The constructors LexEqZ and LexEqS are there to handle the fact that equal vectors are lexicographically ordered. The LexLT constructor captures the idea that if we find two unequal elements ordered by the underlying relation, and those elements are the heads of arbitrary vectors, then those vectors are lexicographically ordered.

It goes without saying that the proofs for Lexicographic are a bit trickier than for LTE, so let’s look at some examples.

||| The first elements are unequal and ordered by LTE, so the vectors are
||| lexicographically ordered.
eg1 : Lexicographic LTE [1, 2, 3] [2, 1, 1]
eg1 = LexLT (LTESucc LTEZero) (\Refl impossible)
||| The vectors are equal, so they are lexicographically ordered for any
||| choice of underlying order.
eg2 : Lexicographic order ['c', 'a', 't'] ['c', 'a', 't']
eg2 = LexEqS Refl (LexEqS Refl (LexEqS Refl (LexEqZ)))
||| The vectors are equal up to the second element, and are strictly ordered
||| in the third element.
eg3 : Lexicographic LTE [1, 2, 3, 1] [1, 2, 5, 1]
eg3 = LexEqS Refl (LexEqS Refl (LexLT (LTESucc (LTESucc (LTESucc (LTEZero)))) (\Refl impossible)))

Now let’s show that lexicographic ordering over totally ordered elements is a total order. To do so, we will show that it is reflexive, transitive, antisymmetric, and totally decidable.

lexRefl : {n : Nat} -> {v : Vect n a} -> Lexicographic o v v
lexRefl {n=Z} {v=Nil} = LexEqZ
lexRefl {n=S n} {v=x::xs} = LexEqS Refl lexRefl

Reflexivity is easy, since we built it into Lexicographic via the LexEqZ and LexEqS constructors. Notice that reflexivity does not depend on any properties of the underlying ordering on the elements in the vectors.

Transitivity is much harder. It will require the transitivity and antisymmetry properties of the underlying ordering. Rather than require OrderTotal on the underlying ordering right away, we can instead just take the transitive law and antisymmetric law as arguments and use them when necessary.

lexTrans'
: {a : Type}
-> {v1, v2, v3 : Vect n a}
-> {order : a -> a -> Type}
-> ({x, y, z : a} -> order x y -> order y z -> order x z)
-> ({x, y : a} -> order x y -> order y x -> x = y)
-> Lexicographic order v1 v2
-> Lexicographic order v2 v3
-> Lexicographic order v1 v3
lexTrans' _ _ LexEqZ LexEqZ = LexEqZ

As usual, these base cases are very easy. Here, the matches cause type refinement on the vectors making them empty. Since v1 ~ [] and v3 ~ [], we can prove that they’re ordered with LexEqZ.

lexTrans' _ _ (LexLT ord neq) (LexEqS Refl _) = LexLT ord neq
lexTrans' _ _ (LexEqS Refl p) (LexLT ord neq) = LexLT ord neq

I’ll focus on the first equation. The first pair of vectors have their first elements ordered and unequal. The second pair of vectors have their first elements identically the same. So the first element of the first vector is also ordered with the first component of the third vector, and is also not equal to it. That’s enough to use LexLT to prove that the first vector is lexicographically ordered with the third one, as required.

lexTrans' trans antisym (LexEqS Refl p) (LexEqS Refl q)
= LexEqS Refl (lexTrans' trans antisym p q)

The first elements of the first and second pair of vectors in this case are equal. So the first element of the first vector is equal to the first element of the third. This is almost enough to use LexEqS to show the first vector is lexicographically ordered with the third. We make a recursive call to prove that the tails of the vectors are lexicographically ordered.

lexTrans' trans antisym (LexLT {x} {y} ordl neql) (LexLT {x=y} {y=z} ordr neqr)
= LexLT (trans ordl ordr) (\Refl => case antisym ordl ordr of Refl => neql Refl)

Here is where we need to use the properties of the underlying ordering. What we know is that the heads of the first pair of vectors are ordered, and the heads of the second pair of vectors are ordered. Using the transitivity of the underlying ordering, we can prove that the head of the first vector is ordered with the head of the third vector.

Next, we need to prove that the head of the first vector is not equal to the head of the third vector. Let x, y, and z refer to the heads of the first, second, and third vectors respectively. Supposing that x = z, some type refinement occurs, and we get that ordl : order x y and ordr : y x. Using antisymmetry, we can conclude that x = y. Matching on that equality proof unifies x with y, so all occurrences of x in types are replaced with y. Therefore, neql : (x = y) -> Void becomes neql : (y = y) -> Void. We can prove y = y with Refl, so we get produce Void, as required.

Now we can write a version of lexTrans where the properties on the underlying ordering are obtained from TotalOrder instance.

lexTrans
: (TotalOrder a order)
=> {v1, v2, v3 : Vect n a}
-> Lexicographic order v1 v2
-> Lexicographic order v2 v3
-> Lexicographic order v1 v3
lexTrans l1 l2 = lexTrans' orderTrans orderAntiSym l1 l2

For antisymmetry of lexicographic ordering, we will proceed in a similar way. We’ll write a general function that explicitly marks which properties of the underlying ordering are needed. Then we can write an easy-to-use version that fills in these functions by requiring a TotalOrder instance on the underlying order.

The antisymmetry proof isn’t very interesting, so I won’t go into as much detail as I did for transitivity. Due to how Lexicographic is designed, the only way we could possibly have both Lexicographic order v w and Lexicographic order w v at the same time is if both such proofs are LexEqS chains. Consequently, the bulk of the antisymmetry proof is showing that anything but chains of LexEqS are impossible, by deriving contradictions. The key difference with earlier “contradiction fishing” is that now we will generally construct Void on the right-hand side and dispel the case by using the function absurd : Void -> a, which lets us derive anything.

lexAntiSym'
: {v, w : Vect n a}
-> ({x, y : a} -> order x y -> order y x -> x = y)
-> Lexicographic order v w
-> Lexicographic order w v
-> v = w
lexAntiSym' _ LexEqZ LexEqZ = Refl
lexAntiSym' anti (LexEqS Refl p) (LexEqS Refl q) with (lexAntiSym' anti p q)
| Refl = Refl
lexAntiSym' _ (LexLT ord neq) (LexEqS Refl q) = absurd (neq Refl)
lexAntiSym' _ (LexEqS Refl p) (LexLT ord neq) = absurd (neq Refl)
lexAntiSym' anti (LexLT ordl neql) (LexLT ordr neqr)
= absurd (neql (anti ordl ordr))
lexAntiSym
: TotalOrder a order => {v, w : Vect n a}
-> Lexicographic order v w
-> Lexicographic order w v
-> v = w
lexAntiSym = lexAntiSym' orderAntiSym

Last but not least, we need to show that lexicographic ordering is a decidable ordering. Of course, in order to decide the lexicographic ordering of two vectors, we will need to decide the ordering of individual elements according to the underlying order; a function to do just that will be required as an argument yet again, and in a revised version of the lexicographic ordering decision procedure, this function will be supplied by a TotalOrder constraint on the underlying ordering.

decLex'
: {a : Type}
-> {order : a -> a -> Type}
-> ((x : a) -> (y : a) -> OrderedBy order x y)
-> (v : Vect n a)
-> (w : Vect n a)
-> OrderedBy (Lexicographic order) v w
decLex' _ Nil Nil = EQ Refl
decLex' decOrder (x::xs) (y::ys) with (decOrder x y)
| LT lt neq = LT (LexLT lt neq) (\Refl => neq Refl)
| GT gt neq = GT (LexLT gt (symNeq neq)) (\Refl => neq Refl)
| EQ eq with (decLex' decOrder xs ys)
| LT lt neq = LT (LexEqS eq lt) (\Refl => neq Refl)
| GT gt neq = GT (LexEqS (sym eq) gt) (\Refl => neq Refl)
| EQ eq' = case (eq, eq') of (Refl, Refl) => EQ Refl
decLex
: TotalOrder a order
=> (v : Vect n a)
-> (w : Vect n a)
-> OrderedBy (Lexicographic order) v w
decLex = decLex' orderTotal

Finally, we can collect our four proofs into an instance of TotalOrder for Lexicographic order, provided that order is also an instance of TotalOrder.

lexTotal : TotalOrder a order -> TotalOrder (Vect n a) (Lexicographic order)
lexTotal _
= MkTotalOrder
(\x => lexRefl {v=x})
(\v1, v2, v3 => lexTrans {v1} {v2} {v3})
(\v, w => lexAntiSym {v} {w})
decLex

(I used the explicit form for defining a named implementation, because when using the Haskell-style anonymous form, I would get a type error that made no sense at all.)

## Conclusion

We saw how to implement some simple decision procedures on Nat, for equality and for minimality according to LTE. We extended the notion of minimality to describe general total orderings. We captured the notion of total orders in the TotalOrder interface. Finally, we defined the lexicographic ordering for vectors, and proved that it is a total order.

The takeaway from this post is not that lexicographic ordering is total or that the minimal element of a set of two natural numbers can be decided machanically. Instead, I hope that you were able to learn some new strategies for proving things in Idris, and how powerful this language is.

There are a number of ways that the work here can be extended. The Lexicographic datatype can be first generalized to work on lists. It can also be generalized to work on vectors of possibly different lengths. Both such lexicographic orders can be shown to be total.