# 3. Why3 by Examples¶

This chapter describes the WhyML specification and programming language. A WhyML source file has suffix .mlw. It contains a list of modules. Each module contains a list of declarations. These include

• Logical declarations:

• types (abstract, record, or algebraic data types);

• functions and predicates;

• axioms, lemmas, and goals.

• Program data types. In a record type declaration, some fields can be declared mutable and/or ghost. Additionally, a record type can be declared abstract (its fields are only visible in ghost code / specification).

• Program declarations and definitions. Programs include many constructs with no counterpart in the logic:

• mutable field assignment;

• sequence;

• loops;

• exceptions;

• local and anonymous functions;

• ghost parameters and ghost code;

• annotations: pre- and postconditions, assertions, loop invariants.

A program may be non-terminating. (But termination can be proved if we wish.)

Command-line tools described in the previous chapter also apply to files containing programs. For instance

why3 prove myfile.mlw

displays the verification conditions for programs contained in file myfile.mlw, and

why3 prove -P alt-ergo myfile.mlw

runs the SMT solver Alt-Ergo on these verification conditions. All this can be performed within the GUI tool why3 ide as well. See Section 6 for more details regarding command lines.

As an introduction to WhyML, we use a small logical puzzle (Section 3.1) and then the five problems from the VSTTE 2010 verification competition [SM10]. The source code for all these examples is contained in Why3’s distribution, in sub-directory examples/. Look for files logic/einstein.why and vstte10_xxx.mlw.

## 3.1. Problem 0: Einstein’s Problem¶

Let us use Why3 to solve a little puzzle known as “Einstein’s logic problem”. (This Why3 example was contributed by Stéphane Lescuyer.) The problem is stated as follows. Five persons, of five different nationalities, live in five houses in a row, all painted with different colors. These five persons own different pets, drink different beverages, and smoke different brands of cigars. We are given the following information:

• The Englishman lives in a red house;

• The Swede has dogs;

• The Dane drinks tea;

• The green house is on the left of the white one;

• The green house’s owner drinks coffee;

• The person who smokes Pall Mall has birds;

• The yellow house’s owner smokes Dunhill;

• In the house in the center lives someone who drinks milk;

• The Norwegian lives in the first house;

• The man who smokes Blends lives next to the one who has cats;

• The man who owns a horse lives next to the one who smokes Dunhills;

• The man who smokes Blue Masters drinks beer;

• The German smokes Prince;

• The Norwegian lives next to the blue house;

• The man who smokes Blends has a neighbour who drinks water.

The question is: What is the nationality of the fish’s owner?

We start by introducing a general-purpose theory defining the notion of bijection, as two abstract types together with two functions from one to the other and two axioms stating that these functions are inverse of each other.

theory Bijection
type t
type u

function of t : u
function to_ u : t

axiom To_of : forall x : t. to_ (of x) = x
axiom Of_to : forall y : u. of (to_ y) = y
end

We now start a new theory, Einstein, which will contain all the individuals of the problem.

theory Einstein

First, we introduce enumeration types for houses, colors, persons, drinks, cigars, and pets.

type house  = H1 | H2 | H3 | H4 | H5
type color  = Blue | Green | Red | White | Yellow
type person = Dane | Englishman | German | Norwegian | Swede
type drink  = Beer | Coffee | Milk | Tea | Water
type cigar  = Blend | BlueMaster | Dunhill | PallMall | Prince
type pet    = Birds | Cats | Dogs | Fish | Horse

We now express that each house is associated bijectively to a color, by cloning the Bijection theory appropriately.

clone Bijection as Color with type t = house, type u = color

Cloning a theory makes a copy of all its declarations, possibly in combination with a user-provided substitution (see Section 7.5.6). Here we substitute type house for type t and type color for type u. As a result, we get two new functions, namely Color.of and Color.to_, from houses to colors and colors to houses, respectively, and two new axioms relating them. Similarly, we express that each house is associated bijectively to a person

clone Bijection as Owner with type t = house, type u = person

and that drinks, cigars, and pets are all associated bijectively to persons:

clone Bijection as Drink with type t = person, type u = drink
clone Bijection as Cigar with type t = person, type u = cigar
clone Bijection as Pet   with type t = person, type u = pet

Next, we need a way to state that a person lives next to another. We first define a predicate leftof over two houses.

predicate leftof (h1 h2 : house) =
match h1, h2 with
| H1, H2
| H2, H3
| H3, H4
| H4, H5 -> true
| _      -> false
end

Note how we advantageously used pattern matching, with an or-pattern for the four positive cases and a universal pattern for the remaining 21 cases. It is then immediate to define a neighbour predicate over two houses, which completes theory Einstein.

predicate rightof (h1 h2 : house) =
leftof h2 h1
predicate neighbour (h1 h2 : house) =
leftof h1 h2 \/ rightof h1 h2
end

The next theory contains the 15 hypotheses. It starts by importing theory Einstein.

theory EinsteinHints
use Einstein

Then each hypothesis is stated in terms of to_ and of functions. For instance, the hypothesis “The Englishman lives in a red house” is declared as the following axiom.

axiom Hint1: Color.of (Owner.to_ Englishman) = Red

And so on for all other hypotheses, up to “The man who smokes Blends has a neighbour who drinks water”, which completes this theory.

...
axiom Hint15:
neighbour (Owner.to_ (Cigar.to_ Blend)) (Owner.to_ (Drink.to_ Water))
end

Finally, we declare the goal in a fourth theory:

theory Problem
use Einstein
use EinsteinHints

goal G: Pet.to_ Fish = German
end

and we can use Why3 to discharge this goal with any prover of our choice.

> why3 prove -P alt-ergo einstein.why
einstein.why Goals G: Valid (1.27s, 989 steps)

The source code for this puzzle is available in the source distribution of Why3, in file examples/logic/einstein.why.

## 3.2. Problem 1: Sum and Maximum¶

Let us now move to the problems of the VSTTE 2010 verification competition [SM10]. The first problem is stated as follows:

Given an $$N$$-element array of natural numbers, write a program to compute the sum and the maximum of the elements in the array.

We assume $$N \ge 0$$ and $$a[i] \ge 0$$ for $$0 \le i < N$$, as precondition, and we have to prove the following postcondition:

$sum \le N \times max.$

In a file max_sum.mlw, we start a new module:

module MaxAndSum

We are obviously needing arithmetic, so we import the corresponding theory, exactly as we would do within a theory definition:

use int.Int

We are also going to use references and arrays from Why3 standard library, so we import the corresponding modules:

use ref.Ref
use array.Array

Modules Ref and Array respectively provide a type ref ’a for references and a type array ’a for arrays, together with useful operations and traditional syntax. They are loaded from the WhyML files ref.mlw and array.mlw in the standard library.

We are now in position to define a program function max_sum. A function definition is introduced with the keyword let. In our case, it introduces a function with two arguments, an array a and its size n:

let max_sum (a: array int) (n: int) : (int, int) = ...

(There is a function length to get the size of an array but we add this extra parameter n to stay close to the original problem statement.) The function body is a Hoare triple, that is a precondition, a program expression, and a postcondition.

let max_sum (a: array int) (n: int) : (int, int)
requires { n = length a }
requires { forall i. 0 <= i < n -> a[i] >= 0 }
ensures  { let (sum, max) = result in sum <= n * max }
= ... expression ...

The first precondition expresses that n is equal to the length of a (this will be needed for verification conditions related to array bound checking). The second precondition expresses that all elements of a are non-negative. The postcondition decomposes the value returned by the function as a pair of integers (sum, max) and states the required property.

returns { sum, max -> sum <= n * max }

We are now left with the function body itself, that is a code computing the sum and the maximum of all elements in a. With no surprise, it is as simple as introducing two local references

let sum = ref 0 in
let max = ref 0 in

scanning the array with a for loop, updating max and sum

for i = 0 to n - 1 do
if !max < a[i] then max := a[i];
sum := !sum + a[i]
done;

and finally returning the pair of the values contained in sum and max:

!sum, !max

This completes the code for function max_sum. As such, it cannot be proved correct, since the loop is still lacking a loop invariant. In this case, the loop invariant is as simple as !sum <= i * !max, since the postcondition only requires us to prove sum <= n * max. The loop invariant is introduced with the keyword invariant, immediately after the keyword do:

for i = 0 to n - 1 do
invariant { !sum <= i * !max }
...
done

There is no need to introduce a variant, as the termination of a for loop is automatically guaranteed. This completes module MaxAndSum, shown below.

module MaxAndSum

use int.Int
use ref.Ref
use array.Array

let max_sum (a: array int) (n: int) : (int, int)
requires { n = length a }
requires { forall i. 0 <= i < n -> a[i] >= 0 }
returns  { sum, max -> sum <= n * max }
= let sum = ref 0 in
let max = ref 0 in
for i = 0 to n - 1 do
invariant { !sum <= i * !max }
if !max < a[i] then max := a[i];
sum := !sum + a[i]
done;
!sum, !max

end

We can now proceed to its verification. Running why3, or better why3 ide, on file max_sum.mlw shows a single verification condition with name WP max_sum. Discharging this verification condition requires a little bit of non-linear arithmetic. Thus some SMT solvers may fail at proving it, but other succeed, e.g., CVC4.

Note: It is of course possible to execute the code to test it, before or after you prove it correct. This is detailed in Section 10.1.

Auto-dereference

Why3 features an auto-dereferencing mechanism, which simplifies the use of references. When a reference is introduced using let ref x (instead of let x = ref), the use of operator ! to access its value is not needed anymore. For instance, we can rewrite the program above as follows (the contract being unchanged and omitted):

let max_sum (a: array int) (n: int) : (sum: int, max: int)
= let ref sum = 0 in
let ref max = 0 in
for i = 0 to n - 1 do
invariant { sum <= i * max }
if max < a[i] then max <- a[i];
sum <- sum + a[i]
done;
sum, max

Note that use of operator <- for assignment (instead of :=) and the absence of ! both in the loop invariant and in the program. See Section 13.1 for more details about the auto-dereferencing mechanism.

## 3.3. Problem 2: Inverting an Injection¶

The second problem is stated as follows:

Invert an injective array $$A$$ on $$N$$ elements in the subrange from $$0$$ to $$N - 1$$, the output array $$B$$ must be such that $$B[A[i]] = i$$ for $$0 \le i < N$$.

The code is immediate, since it is as simple as

for i = 0 to n - 1 do b[a[i]] <- i done

so it is more a matter of specification and of getting the proof done with as much automation as possible. In a new file, we start a new module and we import arithmetic and arrays:

module InvertingAnInjection
use int.Int
use array.Array

It is convenient to introduce predicate definitions for the properties of being injective and surjective. These are purely logical declarations:

predicate injective (a: array int) (n: int) =
forall i j. 0 <= i < n -> 0 <= j < n -> i <> j -> a[i] <> a[j]

predicate surjective (a: array int) (n: int) =
forall i. 0 <= i < n -> exists j: int. (0 <= j < n /\ a[j] = i)

It is also convenient to introduce the predicate “being in the subrange from 0 to $$n-1$$”:

predicate range (a: array int) (n: int) =
forall i. 0 <= i < n -> 0 <= a[i] < n

Using these predicates, we can formulate the assumption that any injective array of size $$n$$ within the range $$0..n-1$$ is also surjective:

lemma injective_surjective:
forall a: array int, n: int.
injective a n -> range a n -> surjective a n

We declare it as a lemma rather than as an axiom, since it is actually provable. It requires induction and can be proved using the Coq proof assistant for instance. Finally we can give the code a specification, with a loop invariant which simply expresses the values assigned to array b so far:

let inverting (a: array int) (b: array int) (n: int)
requires { n = length a = length b }
requires { injective a n /\ range a n }
ensures  { injective b n }
= for i = 0 to n - 1 do
invariant { forall j. 0 <= j < i -> b[a[j]] = j }
b[a[i]] <- i
done

Here we chose to have array b as argument; returning a freshly allocated array would be equally simple. The whole module is given below. The verification conditions for function inverting are easily discharged automatically, thanks to the lemma.

module InvertingAnInjection

use int.Int
use array.Array

predicate injective (a: array int) (n: int) =
forall i j. 0 <= i < n -> 0 <= j < n -> i <> j -> a[i] <> a[j]

predicate surjective (a: array int) (n: int) =
forall i. 0 <= i < n -> exists j: int. (0 <= j < n /\ a[j] = i)

predicate range (a: array int) (n: int) =
forall i. 0 <= i < n -> 0 <= a[i] < n

lemma injective_surjective:
forall a: array int, n: int.
injective a n -> range a n -> surjective a n

let inverting (a: array int) (b: array int) (n: int)
requires { n = length a = length b }
requires { injective a n /\ range a n }
ensures  { injective b n }
= for i = 0 to n - 1 do
invariant { forall j. 0 <= j < i -> b[a[j]] = j }
b[a[i]] <- i
done

end

## 3.4. Problem 3: Searching a Linked List¶

The third problem is stated as follows:

Given a linked list representation of a list of integers, find the index of the first element that is equal to 0.

More precisely, the specification says

You have to show that the program returns an index i equal to the length of the list if there is no such element. Otherwise, the i-th element of the list must be equal to 0, and all the preceding elements must be non-zero.

Since the list is not mutated, we can use the algebraic data type of polymorphic lists from Why3’s standard library, defined in theory list.List. It comes with other handy theories: list.Length, which provides a function length, and list.Nth, which provides a function nth for the nth element of a list. The latter returns an option type, depending on whether the index is meaningful or not.

use int.Int
use option.Option
use export list.List
use export list.Length
use export list.Nth

It is helpful to introduce two predicates: a first one for a successful search,

predicate zero_at (l: list int) (i: int) =
nth i l = Some 0 /\ forall j. 0 <= j < i -> nth j l <> Some 0

and a second one for a non-successful search,

predicate no_zero (l: list int) =
forall j. 0 <= j < length l -> nth j l <> Some 0

We are now in position to give the code for the search function. We write it as a recursive function search that scans a list for the first zero value:

let rec search (i: int) (l: list int) : int =
match l with
| Nil      -> i
| Cons x r -> if x = 0 then i else search (i+1) r
end

Passing an index i as first argument allows to perform a tail call. A simpler code (yet less efficient) would return 0 in the first branch and 1 + search ... in the second one, avoiding the extra argument i.

We first prove the termination of this recursive function. It amounts to giving it a variant, that is a value that strictly decreases at each recursive call with respect to some well-founded ordering. Here it is as simple as the list l itself:

let rec search (i: int) (l: list int) : int variant { l } = ...

It is worth pointing out that variants are not limited to values of algebraic types. A non-negative integer term (for example, length l) can be used, or a term of any other type equipped with a well-founded order relation. Several terms can be given, separated with commas, for lexicographic ordering.

There is no precondition for function search. The postcondition expresses that either a zero value is found, and consequently the value returned is bounded accordingly,

i <= result < i + length l /\ zero_at l (result - i)

or no zero value was found, and thus the returned value is exactly i plus the length of l:

result = i + length l /\ no_zero l

Solving the problem is simply a matter of calling search with 0 as first argument. The code is given below. The verification conditions are all discharged automatically.

use int.Int
use export list.List
use export list.Length
use export list.Nth

predicate zero_at (l: list int) (i: int) =
nth i l = Some 0 /\ forall j. 0 <= j < i -> nth j l <> Some 0

predicate no_zero (l: list int) =
forall j. 0 <= j < length l -> nth j l <> Some 0

let rec search (i: int) (l: list int) : int variant { l }
ensures { (i <= result < i + length l /\ zero_at l (result - i))
\/ (result = i + length l /\ no_zero l) }
= match l with
| Nil -> i
| Cons x r -> if x = 0 then i else search (i+1) r
end

let search_list (l: list int) : int
ensures { (0 <= result < length l /\ zero_at l result)
\/ (result = length l /\ no_zero l) }
= search 0 l

end

Alternatively, we can implement the search with a while loop. To do this, we need to import references from the standard library, together with theory list.HdTl which defines functions hd and tl over lists.

use ref.Ref
use list.HdTl

Being partial functions, hd and tl return options. For the purpose of our code, though, it is simpler to have functions which do not return options, but have preconditions instead. Such a function head is defined as follows:

let head (l: list 'a) : 'a
requires { l <> Nil } ensures { hd l = Some result }
= match l with Nil -> absurd | Cons h _ -> h end

The program construct absurd denotes an unreachable piece of code. It generates the verification condition false, which is here provable using the precondition (the list cannot be Nil). Function tail is defined similarly:

let tail (l: list 'a) : list 'a
requires { l <> Nil } ensures { tl l = Some result }
= match l with Nil -> absurd | Cons _ t -> t end

Using head and tail, it is straightforward to implement the search as a while loop. It uses a local reference i to store the index and another local reference s to store the list being scanned. As long as s is not empty and its head is not zero, it increments i and advances in s using function tail.

let search_loop (l: list int) : int
ensures { ... same postcondition as in search_list ... }
= let i = ref 0 in
let s = ref l in
while !s <> Nil && head !s <> 0 do
invariant { ... }
variant   { !s }
i := !i + 1;
s := tail !s
done;
!i

The postcondition is exactly the same as for function search_list. The termination of the while loop is ensured using a variant, exactly as for a recursive function. Such a variant must strictly decrease at each execution of the loop body. The reader is invited to figure out the loop invariant.

## 3.5. Problem 4: N-Queens¶

The fourth problem is probably the most challenging one. We have to verify the implementation of a program which solves the N-queens puzzle: place N queens on an N*×*N chess board so that no queen can capture another one with a legal move. The program should return a placement if there is a solution and indicates that there is no solution otherwise. A placement is a N-element array which assigns the queen on row i to its column. Thus we start our module by importing arithmetic and arrays:

module NQueens
use int.Int
use array.Array

The code is a simple backtracking algorithm, which tries to put a queen on each row of the chess board, one by one (there is basically no better way to solve the N-queens puzzle). A building block is a function which checks whether the queen on a given row may attack another queen on a previous row. To verify this function, we first define a more elementary predicate, which expresses that queens on row pos and q do no attack each other:

predicate consistent_row (board: array int) (pos: int) (q: int) =
board[q] <> board[pos] /\
board[q] - board[pos] <> pos - q /\
board[pos] - board[q] <> pos - q

Then it is possible to define the consistency of row pos with respect to all previous rows:

predicate is_consistent (board: array int) (pos: int) =
forall q. 0 <= q < pos -> consistent_row board pos q

Implementing a function which decides this predicate is another matter. In order for it to be efficient, we want to return False as soon as a queen attacks the queen on row pos. We use an exception for this purpose and it carries the row of the attacking queen:

exception Inconsistent int

The check is implemented by a function check_is_consistent, which takes the board and the row pos as arguments, and scans rows from 0 to pos - 1 looking for an attacking queen. As soon as one is found, the exception is raised. It is caught immediately outside the loop and False is returned. Whenever the end of the loop is reached, True is returned.

let check_is_consistent (board: array int) (pos: int) : bool
requires { 0 <= pos < length board }
ensures  { result <-> is_consistent board pos }
= try
for q = 0 to pos - 1 do
invariant {
forall j:int. 0 <= j < q -> consistent_row board pos j
}
let bq   = board[q]   in
let bpos = board[pos] in
if bq        = bpos    then raise (Inconsistent q);
if bq - bpos = pos - q then raise (Inconsistent q);
if bpos - bq = pos - q then raise (Inconsistent q)
done;
True
with Inconsistent q ->
assert { not (consistent_row board pos q) };
False
end

The assertion in the exception handler is a cut for SMT solvers. This first part of the solution is given below.

module NQueens
use int.Int
use array.Array

predicate consistent_row (board: array int) (pos: int) (q: int) =
board[q] <> board[pos] /\
board[q] - board[pos] <> pos - q /\
board[pos] - board[q] <> pos - q

predicate is_consistent (board: array int) (pos: int) =
forall q. 0 <= q < pos -> consistent_row board pos q

exception Inconsistent int

let check_is_consistent (board: array int) (pos: int)
requires { 0 <= pos < length board }
ensures  { result <-> is_consistent board pos }
= try
for q = 0 to pos - 1 do
invariant {
forall j:int. 0 <= j < q -> consistent_row board pos j
}
let bq   = board[q]   in
let bpos = board[pos] in
if bq        = bpos    then raise (Inconsistent q);
if bq - bpos = pos - q then raise (Inconsistent q);
if bpos - bq = pos - q then raise (Inconsistent q)
done;
True
with Inconsistent q ->
assert { not (consistent_row board pos q) };
False
end

We now proceed with the verification of the backtracking algorithm. The specification requires us to define the notion of solution, which is straightforward using the predicate is_consistent above. However, since the algorithm will try to complete a given partial solution, it is more convenient to define the notion of partial solution, up to a given row. It is even more convenient to split it in two predicates, one related to legal column values and another to consistency of rows:

predicate is_board (board: array int) (pos: int) =
forall q. 0 <= q < pos -> 0 <= board[q] < length board

predicate solution (board: array int) (pos: int) =
is_board board pos /\
forall q. 0 <= q < pos -> is_consistent board q

The algorithm will not mutate the partial solution it is given and, in case of a search failure, will claim that there is no solution extending this prefix. For this reason, we introduce a predicate comparing two chess boards for equality up to a given row:

predicate eq_board (b1 b2: array int) (pos: int) =
forall q. 0 <= q < pos -> b1[q] = b2[q]

The search itself makes use of an exception to signal a successful search:

exception Solution

The backtracking code is a recursive function bt_queens which takes the chess board, its size, and the starting row for the search. The termination is ensured by the obvious variant n - pos.

let rec bt_queens (board: array int) (n: int) (pos: int) : unit
variant  { n - pos }

The precondition relates board, pos, and n and requires board to be a solution up to pos:

requires { 0 <= pos <= n = length board }
requires { solution board pos }

The postcondition is twofold: either the function exits normally and then there is no solution extending the prefix in board, which has not been modified; or the function raises Solution and we have a solution in board.

ensures  { eq_board board (old board) pos }
ensures  { forall b:array int. length b = n -> is_board b n ->
eq_board board b pos -> not (solution b n) }
raises   { Solution -> solution board n }
=

Whenever we reach the end of the chess board, we have found a solution and we signal it using exception Solution:

if pos = n then raise Solution;

Otherwise we scan all possible positions for the queen on row pos with a for loop:

for i = 0 to n - 1 do

The loop invariant states that we have not modified the solution prefix so far, and that we have not found any solution that would extend this prefix with a queen on row pos at a column below i:

invariant { eq_board board (old board) pos }
invariant { forall b:array int.  length b = n -> is_board b n ->
eq_board board b pos -> 0 <= b[pos] < i -> not (solution b n) }

Then we assign column i to the queen on row pos and we check for a possible attack with check_is_consistent. If not, we call bt_queens recursively on the next row.

board[pos] <- i;
if check_is_consistent board pos then bt_queens board n (pos + 1)
done

This completes the loop and function bt_queens as well. Solving the puzzle is a simple call to bt_queens, starting the search on row 0. The postcondition is also twofold, as for bt_queens, yet slightly simpler.

let queens (board: array int) (n: int) : unit
requires { length board = n }
ensures  { forall b:array int.
length b = n -> is_board b n -> not (solution b n) }
raises   { Solution -> solution board n }
= bt_queens board n 0

This second part of the solution is given below. With the help of a few auxiliary lemmas — not given here but available from Why3’s sources — the verification conditions are all discharged automatically, including the verification of the lemmas themselves.

predicate is_board (board: array int) (pos: int) =
forall q. 0 <= q < pos -> 0 <= board[q] < length board

predicate solution (board: array int) (pos: int) =
is_board board pos /\
forall q. 0 <= q < pos -> is_consistent board q

predicate eq_board (b1 b2: array int) (pos: int) =
forall q. 0 <= q < pos -> b1[q] = b2[q]

exception Solution

let rec bt_queens (board: array int) (n: int) (pos: int) : unit
variant  { n - pos }
requires { 0 <= pos <= n = length board }
requires { solution board pos }
ensures  { eq_board board (old board) pos }
ensures  { forall b:array int. length b = n -> is_board b n ->
eq_board board b pos -> not (solution b n) }
raises   { Solution -> solution board n }
= if pos = n then raise Solution;
for i = 0 to n - 1 do
invariant { eq_board board (old board) pos }
invariant { forall b:array int. length b = n -> is_board b n ->
eq_board board b pos -> 0 <= b[pos] < i -> not (solution b n) }
board[pos] <- i;
if check_is_consistent board pos then bt_queens board n (pos + 1)
done

let queens (board: array int) (n: int) : unit
requires { length board = n }
ensures  { forall b:array int.
length b = n -> is_board b n -> not (solution b n) }
raises   { Solution -> solution board n }
= bt_queens board n 0

end

## 3.6. Problem 5: Amortized Queue¶

The last problem consists in verifying the implementation of a well-known purely applicative data structure for queues. A queue is composed of two lists, front and rear. We push elements at the head of list rear and pop them off the head of list front. We maintain that the length of front is always greater or equal to the length of rear. (See for instance Okasaki’s Purely Functional Data Structures [Oka98] for more details.)

We have to implement operations empty, head, tail, and enqueue over this data type, to show that the invariant over lengths is maintained, and finally to show that a client invoking these operations observes an abstract queue given by a sequence.

In a new module, we import arithmetic and theory list.ListRich, a combo theory that imports all list operations we will require: length, reversal, and concatenation.

module AmortizedQueue
use int.Int
use option.Option
use export list.ListRich

The queue data type is naturally introduced as a polymorphic record type. The two list lengths are explicitly stored, for greater efficiency.

type queue 'a = { front: list 'a; lenf: int;
rear : list 'a; lenr: int; }
invariant { length front = lenf >= length rear = lenr }
by { front = Nil; lenf = 0; rear = Nil; lenr = 0 }

The type definition is accompanied with an invariant — a logical property imposed on any value of the type. Why3 assumes that any queue satisfies the invariant at any function entry and it requires that any queue satisfies the invariant at function exit (be the queue created or modified). The by clause ensures the non-vacuity of this type with invariant. If you omit it, a goal with an existential statement is generated. (See Section 7.5.1 for more details about record types with invariants.)

For the purpose of the specification, it is convenient to introduce a function sequence which builds the sequence of elements of a queue, that is the front list concatenated to the reversed rear list.

function sequence (q: queue 'a) : list 'a = q.front ++ reverse q.rear

It is worth pointing out that this function can only be used in specifications. We start with the easiest operation: building the empty queue.

let empty () : queue 'a
ensures { sequence result = Nil }
= { front = Nil; lenf = 0; rear = Nil; lenr = 0 }

The postcondition states that the returned queue represents the empty sequence. Another postcondition, saying that the returned queue satisfies the type invariant, is implicit. Note the cast to type queue 'a. It is required, for the type checker not to complain about an undefined type variable.

The next operation is head, which returns the first element from a given queue q. It naturally requires the queue to be non empty, which is conveniently expressed as sequence q not being Nil.

let head (q: queue 'a) : 'a
requires { sequence q <> Nil }
ensures  { hd (sequence q) = Some result }
= let Cons x _ = q.front in x

The fact that the argument q satisfies the type invariant is implicitly assumed. The type invariant is required to prove the absurdity of q.front being Nil (otherwise, sequence q would be Nil as well).

The next operation is tail, which removes the first element from a given queue. This is more subtle than head, since we may have to re-structure the queue to maintain the invariant. Since we will have to perform a similar operation when implementing operation enqueue later, it is a good idea to introduce a smart constructor create that builds a queue from two lists while ensuring the invariant. The list lengths are also passed as arguments, to avoid unnecessary computations.

let create (f: list 'a) (lf: int) (r: list 'a) (lr: int) : queue 'a
requires { lf = length f /\ lr = length r }
ensures  { sequence result = f ++ reverse r }
= if lf >= lr then
{ front = f; lenf = lf; rear = r; lenr = lr }
else
let f = f ++ reverse r in
{ front = f; lenf = lf + lr; rear = Nil; lenr = 0 }

If the invariant already holds, it is simply a matter of building the record. Otherwise, we empty the rear list and build a new front list as the concatenation of list f and the reversal of list r. The principle of this implementation is that the cost of this reversal will be amortized over all queue operations. Implementing function tail is now straightforward and follows the structure of function head.

let tail (q: queue 'a) : queue 'a
requires { sequence q <> Nil }
ensures  { tl (sequence q) = Some (sequence result) }
= let Cons _ r = q.front in
create r (q.lenf - 1) q.rear q.lenr

The last operation is enqueue, which pushes a new element in a given queue. Reusing the smart constructor create makes it a one line code.

let enqueue (x: 'a) (q: queue 'a) : queue 'a
ensures { sequence result = sequence q ++ Cons x Nil }
= create q.front q.lenf (Cons x q.rear) (q.lenr + 1)

The code is given below. The verification conditions are all discharged automatically.

module AmortizedQueue

use int.Int
use option.Option
use list.ListRich

type queue 'a = { front: list 'a; lenf: int;
rear : list 'a; lenr: int; }
invariant { length front = lenf >= length rear = lenr }
by { front = Nil; lenf = 0; rear = Nil; lenr = 0 }

function sequence (q: queue 'a) : list 'a =
q.front ++ reverse q.rear

let empty () : queue 'a
ensures { sequence result = Nil }
= { front = Nil; lenf = 0; rear = Nil; lenr = 0 }

let head (q: queue 'a) : 'a
requires { sequence q <> Nil }
ensures  { hd (sequence q) = Some result }
= let Cons x _ = q.front in x

let create (f: list 'a) (lf: int) (r: list 'a) (lr: int) : queue 'a
requires { lf = length f /\ lr = length r }
ensures  { sequence result = f ++ reverse r }
= if lf >= lr then
{ front = f; lenf = lf; rear = r; lenr = lr }
else
let f = f ++ reverse r in
{ front = f; lenf = lf + lr; rear = Nil; lenr = 0 }

let tail (q: queue 'a) : queue 'a
requires { sequence q <> Nil }
ensures  { tl (sequence q) = Some (sequence result) }
= let Cons _ r = q.front in
create r (q.lenf - 1) q.rear q.lenr

let enqueue (x: 'a) (q: queue 'a) : queue 'a
ensures { sequence result = sequence q ++ Cons x Nil }
= create q.front q.lenf (Cons x q.rear) (q.lenr + 1)

end