BasicsFunctional Programming in Coq


The functional programming style brings programming closer to mathematics: If a procedure or method has no side effects, then pretty much all you need to understand about it is how it maps inputs to outputs — that is, you can think of its behavior as just computing a mathematical function. This is one reason for the word "functional" in "functional programming." This direct connection between programs and simple mathematical objects supports both sound informal reasoning and formal proofs of correctness.
The other sense in which functional programming is "functional" is that it emphasizes the use of functions (or methods) as first-class values — i.e., values that can be passed as arguments to other functions, returned as results, stored in data structures, etc. The recognition that functions can be treated as data in this way enables a host of useful idioms, as we will see.
Other common features of functional languages include algebraic data types and pattern matching, which make it easy to construct and manipulate rich data structures, and sophisticated polymorphic type systems that support abstraction and code reuse. Coq shares all of these features.

Enumerated Types

One unusual aspect of Coq is that its set of built-in features is extremely small. For example, instead of providing the usual palette of atomic data types (booleans, integers, strings, etc.), Coq offers an extremely powerful mechanism for defining new data types from scratch — so powerful that all these familiar types arise as instances.
Naturally, the Coq distribution comes with an extensive standard library providing definitions of booleans, numbers, and many common data structures like lists and hash tables. But there is nothing magic or primitive about these library definitions: they are ordinary user code.
To see how this works, let's start with a very simple example.

Days of the Week

The following declaration tells Coq that we are defining a new set of data values — a type.

Inductive day : Type :=
  | monday : day
  | tuesday : day
  | wednesday : day
  | thursday : day
  | friday : day
  | saturday : day
  | sunday : day.

The type is called day, and its members are monday, tuesday, etc. The second through eighth lines of the definition can be read "monday is a day, tuesday is a day, etc."
Having defined day, we can write functions that operate on days.

Definition next_weekday (d:day) : day :=
  match d with
  | mondaytuesday
  | tuesdaywednesday
  | wednesdaythursday
  | thursdayfriday
  | fridaymonday
  | saturdaymonday
  | sundaymonday

One thing to note is that the argument and return types of this function are explicitly declared. Like most functional programming languages, Coq can often work out these types even if they are not given explicitly — i.e., it performs some type inference — but we'll always include them to make reading easier.
Having defined a function, we should check that it works on some examples. There are actually three different ways to do this in Coq. First, we can use the command Eval compute to evaluate a compound expression involving next_weekday.

Eval compute in (next_weekday friday).
   (* ==> monday : day *)
Eval compute in (next_weekday (next_weekday saturday)).
   (* ==> tuesday : day *)

If you have a computer handy, now would be an excellent moment to fire up the Coq interpreter under your favorite IDE — either CoqIde or Proof General — and try this for yourself. Load this file (Basics.v) from the book's accompanying Coq sources, find the above example, submit it to Coq, and observe the result.
The keyword compute tells Coq precisely how to evaluate the expression we give it. For the moment, compute is the only one we'll need; later on we'll see some alternatives that are sometimes useful.
Second, we can record what we expect the result to be in the form of a Coq example:

Example test_next_weekday:
  (next_weekday (next_weekday saturday)) = tuesday.

This declaration does two things: it makes an assertion (that the second weekday after saturday is tuesday), and it gives the assertion a name that can be used to refer to it later. Having made the assertion, we can also ask Coq to verify it, like this:

Proof. simpl. reflexivity. Qed.

The details are not important for now (we'll come back to them in a bit), but essentially this can be read as "The assertion we've just made can be proved by observing that both sides of the equality evaluate to the same thing, after some simplification."
Third, we can ask Coq to "extract," from a Definition, a program in some other, more conventional, programming language (OCaml, Scheme, or Haskell) with a high-performance compiler. This facility is very interesting, since it gives us a way to construct fully certified programs in mainstream languages. Indeed, this is one of the main uses for which Coq was developed. We'll come back to this topic in later chapters. More information can also be found in the Coq'Art book by Bertot and Casteran, as well as the Coq reference manual.


In a similar way, we can define the type bool of booleans, with members true and false.

Inductive bool : Type :=
  | true : bool
  | false : bool.

Although we are rolling our own booleans here for the sake of building up everything from scratch, Coq does, of course, provide a default implementation of the booleans in its standard library, together with a multitude of useful functions and lemmas. (Take a look at Coq.Init.Datatypes in the Coq library documentation if you're interested.) Whenever possible, we'll name our own definitions and theorems so that they exactly coincide with the ones in the standard library.
Functions over booleans can be defined in the same way as above:

Definition negb (b:bool) : bool :=
  match b with
  | truefalse
  | falsetrue

Definition andb (b1:bool) (b2:bool) : bool :=
  match b1 with
  | trueb2
  | falsefalse

Definition orb (b1:bool) (b2:bool) : bool :=
  match b1 with
  | truetrue
  | falseb2

The last two illustrate the syntax for multi-argument function definitions.
The following four "unit tests" constitute a complete specification — a truth table — for the orb function:

Example test_orb1: (orb true false) = true.
Proof. reflexivity. Qed.
Example test_orb2: (orb false false) = false.
Proof. reflexivity. Qed.
Example test_orb3: (orb false true) = true.
Proof. reflexivity. Qed.
Example test_orb4: (orb true true) = true.
Proof. reflexivity. Qed.

(Note that we've dropped the simpl in the proofs. It's not actually needed because reflexivity will automatically perform simplification.)
A note on notation: We use square brackets to delimit fragments of Coq code in comments in .v files; this convention, also used by the coqdoc documentation tool, keeps them visually separate from the surrounding text. In the html version of the files, these pieces of text appear in a different font.
The values Admitted and admit can be used to fill a hole in an incomplete definition or proof. We'll use them in the following exercises. In general, your job in the exercises is to replace admit or Admitted with real definitions or proofs.

Exercise: 1 star (nandb)

Complete the definition of the following function, then make sure that the Example assertions below can each be verified by Coq.
This function should return true if either or both of its inputs are false.

Definition nandb (b1:bool) (b2:bool) : bool :=
  (* FILL IN HERE *) admit.

Remove "Admitted." and fill in each proof with "Proof. reflexivity. Qed."

Example test_nandb1: (nandb true false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb2: (nandb false false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb3: (nandb false true) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb4: (nandb true true) = false.
(* FILL IN HERE *) Admitted.

Exercise: 1 star (andb3)

Do the same for the andb3 function below. This function should return true when all of its inputs are true, and false otherwise.

Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool :=
  (* FILL IN HERE *) admit.

Example test_andb31: (andb3 true true true) = true.
(* FILL IN HERE *) Admitted.
Example test_andb32: (andb3 false true true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb33: (andb3 true false true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb34: (andb3 true true false) = false.
(* FILL IN HERE *) Admitted.

Function Types

The Check command causes Coq to print the type of an expression. For example, the type of negb true is bool.

Check true.
(* ===> true : bool *)
Check (negb true).
(* ===> negb true : bool *)

Functions like negb itself are also data values, just like true and false. Their types are called function types, and they are written with arrows.

Check negb.
(* ===> negb : bool -> bool *)

The type of negb, written bool bool and pronounced "bool arrow bool," can be read, "Given an input of type bool, this function produces an output of type bool." Similarly, the type of andb, written bool bool bool, can be read, "Given two inputs, both of type bool, this function produces an output of type bool."


Technical digression: Coq provides a fairly sophisticated module system, to aid in organizing large developments. In this course we won't need most of its features, but one is useful: If we enclose a collection of declarations between Module X and End X markers, then, in the remainder of the file after the End, these definitions will be referred to by names like instead of just foo. Here, we use this feature to introduce the definition of the type nat in an inner module so that it does not shadow the one from the standard library.

Module Playground1.

The types we have defined so far are examples of "enumerated types": their definitions explicitly enumerate a finite set of elements. A more interesting way of defining a type is to give a collection of "inductive rules" describing its elements. For example, we can define the natural numbers as follows:

Inductive nat : Type :=
  | O : nat
  | S : nat nat.

The clauses of this definition can be read:
  • O is a natural number (note that this is the letter "O," not the numeral "0").
  • S is a "constructor" that takes a natural number and yields another one — that is, if n is a natural number, then S n is too.
Let's look at this in a little more detail.
Every inductively defined set (day, nat, bool, etc.) is actually a set of expressions. The definition of nat says how expressions in the set nat can be constructed:
  • the expression O belongs to the set nat;
  • if n is an expression belonging to the set nat, then S n is also an expression belonging to the set nat; and
  • expressions formed in these two ways are the only ones belonging to the set nat.
The same rules apply for our definitions of day and bool. The annotations we used for their constructors are analogous to the one for the O constructor, and indicate that each of those constructors doesn't take any arguments.
These three conditions are the precise force of the Inductive declaration. They imply that the expression O, the expression S O, the expression S (S O), the expression S (S (S O)), and so on all belong to the set nat, while other expressions like true, andb true false, and S (S false) do not.
We can write simple functions that pattern match on natural numbers just as we did above — for example, the predecessor function:

Definition pred (n : nat) : nat :=
  match n with
    | OO
    | S n'n'

The second branch can be read: "if n has the form S n' for some n', then return n'."

End Playground1.

Definition minustwo (n : nat) : nat :=
  match n with
    | OO
    | S OO
    | S (S n') ⇒ n'

Because natural numbers are such a pervasive form of data, Coq provides a tiny bit of built-in magic for parsing and printing them: ordinary arabic numerals can be used as an alternative to the "unary" notation defined by the constructors S and O. Coq prints numbers in arabic form by default:

Check (S (S (S (S O)))).
Eval compute in (minustwo 4).

The constructor S has the type nat nat, just like the functions minustwo and pred:

Check S.
Check pred.
Check minustwo.

These are all things that can be applied to a number to yield a number. However, there is a fundamental difference: functions like pred and minustwo come with computation rules — e.g., the definition of pred says that pred 2 can be simplified to 1 — while the definition of S has no such behavior attached. Although it is like a function in the sense that it can be applied to an argument, it does not do anything at all!
For most function definitions over numbers, pure pattern matching is not enough: we also need recursion. For example, to check that a number n is even, we may need to recursively check whether n-2 is even. To write such functions, we use the keyword Fixpoint.

Fixpoint evenb (n:nat) : bool :=
  match n with
  | Otrue
  | S Ofalse
  | S (S n') ⇒ evenb n'

We can define oddb by a similar Fixpoint declaration, but here is a simpler definition that will be a bit easier to work with:

Definition oddb (n:nat) : bool := negb (evenb n).

Example test_oddb1: (oddb (S O)) = true.
Proof. reflexivity. Qed.
Example test_oddb2: (oddb (S (S (S (S O))))) = false.
Proof. reflexivity. Qed.

Naturally, we can also define multi-argument functions by recursion. (Once again, we use a module to avoid polluting the namespace.)

Module Playground2.

Fixpoint plus (n : nat) (m : nat) : nat :=
  match n with
    | Om
    | S n'S (plus n' m)

Adding three to two now gives us five, as we'd expect.

Eval compute in (plus (S (S (S O))) (S (S O))).

The simplification that Coq performs to reach this conclusion can be visualized as follows:

(*  plus (S (S (S O))) (S (S O))    
==> S (plus (S (S O)) (S (S O))) by the second clause of the match
==> S (S (plus (S O) (S (S O)))) by the second clause of the match
==> S (S (S (plus O (S (S O))))) by the second clause of the match
==> S (S (S (S (S O))))          by the first clause of the match

As a notational convenience, if two or more arguments have the same type, they can be written together. In the following definition, (n m : nat) means just the same as if we had written (n : nat) (m : nat).

Fixpoint mult (n m : nat) : nat :=
  match n with
    | OO
    | S n'plus m (mult n' m)

Example test_mult1: (mult 3 3) = 9.
Proof. reflexivity. Qed.

You can match two expressions at once by putting a comma between them:

Fixpoint minus (n m:nat) : nat :=
  match n, m with
  | O , _O
  | S _ , On
  | S n', S m'minus n' m'

The _ in the first line is a wildcard pattern. Writing _ in a pattern is the same as writing some variable that doesn't get used on the right-hand side. This avoids the need to invent a bogus variable name.

End Playground2.

Fixpoint exp (base power : nat) : nat :=
  match power with
    | OS O
    | S pmult base (exp base p)

Exercise: 1 star (factorial)

Recall the standard factorial function:
    factorial(0)  =  1 
    factorial(n)  =  n * factorial(n-1)     (if n>0)
Translate this into Coq.

Fixpoint factorial (n:nat) : nat :=
(* FILL IN HERE *) admit.

Example test_factorial1: (factorial 3) = 6.
(* FILL IN HERE *) Admitted.
Example test_factorial2: (factorial 5) = (mult 10 12).
(* FILL IN HERE *) Admitted.
We can make numerical expressions a little easier to read and write by introducing "notations" for addition, multiplication, and subtraction.

                       (at level 50, left associativity)
                       : nat_scope.
Notation "x - y" := (minus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x × y" := (mult x y)
                       (at level 40, left associativity)
                       : nat_scope.

Check ((0 + 1) + 1).

(The level, associativity, and nat_scope annotations control how these notations are treated by Coq's parser. The details are not important, but interested readers can refer to the "More on Notation" subsection in the "Optional Material" section at the end of this chapter.)
Note that these do not change the definitions we've already made: they are simply instructions to the Coq parser to accept x + y in place of plus x y and, conversely, to the Coq pretty-printer to display plus x y as x + y.
When we say that Coq comes with nothing built-in, we really mean it: even equality testing for numbers is a user-defined operation! The beq_nat function tests natural numbers for equality, yielding a boolean. Note the use of nested matches (we could also have used a simultaneous match, as we did in minus.)

Fixpoint beq_nat (n m : nat) : bool :=
  match n with
  | Omatch m with
         | Otrue
         | S m'false
  | S n'match m with
            | Ofalse
            | S m'beq_nat n' m'

Similarly, the ble_nat function tests natural numbers for less-or-equal, yielding a boolean.

Fixpoint ble_nat (n m : nat) : bool :=
  match n with
  | Otrue
  | S n'
      match m with
      | Ofalse
      | S m'ble_nat n' m'

Example test_ble_nat1: (ble_nat 2 2) = true.
Proof. reflexivity. Qed.
Example test_ble_nat2: (ble_nat 2 4) = true.
Proof. reflexivity. Qed.
Example test_ble_nat3: (ble_nat 4 2) = false.
Proof. reflexivity. Qed.

Exercise: 2 stars (blt_nat)

The blt_nat function tests natural numbers for less-than, yielding a boolean. Instead of making up a new Fixpoint for this one, define it in terms of a previously defined function.
Note: If you have trouble with the simpl tactic, try using compute, which is like simpl on steroids. However, there is a simple, elegant solution for which simpl suffices.

Definition blt_nat (n m : nat) : bool :=
  (* FILL IN HERE *) admit.

Example test_blt_nat1: (blt_nat 2 2) = false.
(* FILL IN HERE *) Admitted.
Example test_blt_nat2: (blt_nat 2 4) = true.
(* FILL IN HERE *) Admitted.
Example test_blt_nat3: (blt_nat 4 2) = false.
(* FILL IN HERE *) Admitted.

Proof by Simplification

Now that we've defined a few datatypes and functions, let's turn to the question of how to state and prove properties of their behavior. Actually, in a sense, we've already started doing this: each Example in the previous sections makes a precise claim about the behavior of some function on some particular inputs. The proofs of these claims were always the same: use reflexivity to check that both sides of the = simplify to identical values.
(By the way, it will be useful later to know that reflexivity actually does somewhat more simplification than simpl does — for example, it tries "unfolding" defined terms, replacing them with their right-hand sides. The reason for this difference is that, when reflexivity succeeds, the whole goal is finished and we don't need to look at whatever expanded expressions reflexivity has found; by contrast, simpl is used in situations where we may have to read and understand the new goal, so we would not want it blindly expanding definitions.)
The same sort of "proof by simplification" can be used to prove more interesting properties as well. For example, the fact that 0 is a "neutral element" for + on the left can be proved just by observing that 0 + n reduces to n no matter what n is, a fact that can be read directly off the definition of plus.

Theorem plus_O_n : n : nat, 0 + n = n.
  intros n. reflexivity. Qed.

(Note: You may notice that the above statement looks different in the original source file and the final html output. In Coq files, we write the universal quantifier using the "forall" reserved identifier. This gets printed as an upside-down "A", the familiar symbol used in logic.)
The form of this theorem and proof are almost exactly the same as the examples above; there are just a few differences.
First, we've used the keyword Theorem instead of Example. Indeed, the difference is purely a matter of style; the keywords Example and Theorem (and a few others, including Lemma, Fact, and Remark) mean exactly the same thing to Coq.
Secondly, we've added the quantifier n:nat, so that our theorem talks about all natural numbers n. In order to prove theorems of this form, we need to to be able to reason by assuming the existence of an arbitrary natural number n. This is achieved in the proof by intros n, which moves the quantifier from the goal to a "context" of current assumptions. In effect, we start the proof by saying "OK, suppose n is some arbitrary number."
The keywords intros, simpl, and reflexivity are examples of tactics. A tactic is a command that is used between Proof and Qed to tell Coq how it should check the correctness of some claim we are making. We will see several more tactics in the rest of this lecture, and yet more in future lectures.
Step through these proofs in Coq and notice how the goal and context change.

Theorem plus_1_l : n:nat, 1 + n = S n.
  intros n. reflexivity. Qed.

Theorem mult_0_l : n:nat, 0 × n = 0.
  intros n. reflexivity. Qed.

The _l suffix in the names of these theorems is pronounced "on the left."

Proof by Rewriting

Here is a slightly more interesting theorem:

Theorem plus_id_example : n m:nat,
  n = m
  n + n = m + m.

Instead of making a completely universal claim about all numbers n and m, this theorem talks about a more specialized property that only holds when n = m. The arrow symbol is pronounced "implies."
As before, we need to be able to reason by assuming the existence of some numbers n and m. We also need to assume the hypothesis n = m. The intros tactic will serve to move all three of these from the goal into assumptions in the current context.
Since n and m are arbitrary numbers, we can't just use simplification to prove this theorem. Instead, we prove it by observing that, if we are assuming n = m, then we can replace n with m in the goal statement and obtain an equality with the same expression on both sides. The tactic that tells Coq to perform this replacement is called rewrite.

  intros n m. (* move both quantifiers into the context *)
  intros H. (* move the hypothesis into the context *)
  rewrite H. (* Rewrite the goal using the hypothesis *)
  reflexivity. Qed.

The first line of the proof moves the universally quantified variables n and m into the context. The second moves the hypothesis n = m into the context and gives it the (arbitrary) name H. The third tells Coq to rewrite the current goal (n + n = m + m) by replacing the left side of the equality hypothesis H with the right side.
(The arrow symbol in the rewrite has nothing to do with implication: it tells Coq to apply the rewrite from left to right. To rewrite from right to left, you can use rewrite . Try making this change in the above proof and see what difference it makes in Coq's behavior.)

Exercise: 1 star (plus_id_exercise)

Remove "Admitted." and fill in the proof.

Theorem plus_id_exercise : n m o : nat,
  n = m m = o n + m = m + o.
  (* FILL IN HERE *) Admitted.
As we've seen in earlier examples, the Admitted command tells Coq that we want to skip trying to prove this theorem and just accept it as a given. This can be useful for developing longer proofs, since we can state subsidiary facts that we believe will be useful for making some larger argument, use Admitted to accept them on faith for the moment, and continue thinking about the larger argument until we are sure it makes sense; then we can go back and fill in the proofs we skipped. Be careful, though: every time you say Admitted (or admit) you are leaving a door open for total nonsense to enter Coq's nice, rigorous, formally checked world!
We can also use the rewrite tactic with a previously proved theorem instead of a hypothesis from the context.

Theorem mult_0_plus : n m : nat,
  (0 + n) × m = n × m.
  intros n m.
  rewrite plus_O_n.
  reflexivity. Qed.

Exercise: 2 stars (mult_S_1)

Theorem mult_S_1 : n m : nat,
  m = S n
  m × (1 + n) = m × m.
  (* FILL IN HERE *) Admitted.

Proof by Case Analysis

Of course, not everything can be proved by simple calculation: In general, unknown, hypothetical values (arbitrary numbers, booleans, lists, etc.) can block the calculation. For example, if we try to prove the following fact using the simpl tactic as above, we get stuck.

Theorem plus_1_neq_0_firsttry : n : nat,
  beq_nat (n + 1) 0 = false.
  intros n.
  simpl. (* does nothing! *)

The reason for this is that the definitions of both beq_nat and + begin by performing a match on their first argument. But here, the first argument to + is the unknown number n and the argument to beq_nat is the compound expression n + 1; neither can be simplified.
What we need is to be able to consider the possible forms of n separately. If n is O, then we can calculate the final result of beq_nat (n + 1) 0 and check that it is, indeed, false. And if n = S n' for some n', then, although we don't know exactly what number n + 1 yields, we can calculate that, at least, it will begin with one S, and this is enough to calculate that, again, beq_nat (n + 1) 0 will yield false.
The tactic that tells Coq to consider, separately, the cases where n = O and where n = S n' is called destruct.

Theorem plus_1_neq_0 : n : nat,
  beq_nat (n + 1) 0 = false.
  intros n. destruct n as [| n'].
    reflexivity. Qed.

The destruct generates two subgoals, which we must then prove, separately, in order to get Coq to accept the theorem as proved. (No special command is needed for moving from one subgoal to the other. When the first subgoal has been proved, it just disappears and we are left with the other "in focus.") In this proof, each of the subgoals is easily proved by a single use of reflexivity.
The annotation "as [| n']" is called an intro pattern. It tells Coq what variable names to introduce in each subgoal. In general, what goes between the square brackets is a list of lists of names, separated by |. Here, the first component is empty, since the O constructor is nullary (it doesn't carry any data). The second component gives a single name, n', since S is a unary constructor.
The destruct tactic can be used with any inductively defined datatype. For example, we use it here to prove that boolean negation is involutive — i.e., that negation is its own inverse.

Theorem negb_involutive : b : bool,
  negb (negb b) = b.
  intros b. destruct b.
    reflexivity. Qed.

Note that the destruct here has no as clause because none of the subcases of the destruct need to bind any variables, so there is no need to specify any names. (We could also have written as [|], or as [].) In fact, we can omit the as clause from any destruct and Coq will fill in variable names automatically. Although this is convenient, it is arguably bad style, since Coq often makes confusing choices of names when left to its own devices.

Exercise: 1 star (zero_nbeq_plus_1)

Theorem zero_nbeq_plus_1 : n : nat,
  beq_nat 0 (n + 1) = false.
  (* FILL IN HERE *) Admitted.

More Exercises

Exercise: 2 stars (boolean functions)

Use the tactics you have learned so far to prove the following theorem about boolean functions.

Theorem identity_fn_applied_twice :
  (f : bool bool),
  ((x : bool), f x = x)
  (b : bool), f (f b) = b.
  (* FILL IN HERE *) Admitted.

Now state and prove a theorem negation_fn_applied_twice similar to the previous one but where the second hypothesis says that the function f has the property that f x = negb x.


Exercise: 2 stars (andb_eq_orb)

Prove the following theorem. (You may want to first prove a subsidiary lemma or two. Alternatively, remember that you do not have to introduce all hypotheses at the same time.)

Theorem andb_eq_orb :
  (b c : bool),
  (andb b c = orb b c)
  b = c.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars (binary)

Consider a different, more efficient representation of natural numbers using a binary rather than unary system. That is, instead of saying that each natural number is either zero or the successor of a natural number, we can say that each binary number is either
  • zero,
  • twice a binary number, or
  • one more than twice a binary number.
(a) First, write an inductive definition of the type bin corresponding to this description of binary numbers.
(Hint: Recall that the definition of nat from class,
    Inductive nat : Type :=
      | O : nat
      | S : nat  nat.
says nothing about what O and S "mean." It just says "O is in the set called nat, and if n is in the set then so is S n." The interpretation of O as zero and S as successor/plus one comes from the way that we use nat values, by writing functions to do things with them, proving things about them, and so on. Your definition of bin should be correspondingly simple; it is the functions you will write next that will give it mathematical meaning.)
(b) Next, write an increment function for binary numbers, and a function to convert binary numbers to unary numbers.
(c) Write some unit tests for your increment and binary-to-unary functions. Notice that incrementing a binary number and then converting it to unary should yield the same result as first converting it to unary and then incrementing.


Optional Material

More on Notation

                       (at level 50, left associativity)
                       : nat_scope.
                       (at level 40, left associativity)
                       : nat_scope.

For each notation-symbol in Coq we can specify its precedence level and its associativity. The precedence level n can be specified by the keywords at level n and it is helpful to disambiguate expressions containing different symbols. The associativity is helpful to disambiguate expressions containing more occurrences of the same symbol. For example, the parameters specified above for + and × say that the expression 1+2×3×4 is a shorthand for the expression (1+((2×3)×4)). Coq uses precedence levels from 0 to 100, and left, right, or no associativity.
Each notation-symbol in Coq is also active in a notation scope. Coq tries to guess what scope you mean, so when you write S(O×O) it guesses nat_scope, but when you write the cartesian product (tuple) type bool×bool it guesses type_scope. Occasionally you have to help it out with percent-notation by writing (x×y)%nat, and sometimes in Coq's feedback to you it will use %nat to indicate what scope a notation is in.
Notation scopes also apply to numeral notation (3,4,5, etc.), so you may sometimes see 0%nat which means O, or 0%Z which means the Integer zero.

Fixpoints and Structural Recursion

Fixpoint plus' (n : nat) (m : nat) : nat :=
  match n with
    | Om
    | S n'S (plus' n' m)

When Coq checks this definition, it notes that plus' is "decreasing on 1st argument." What this means is that we are performing a structural recursion over the argument n — i.e., that we make recursive calls only on strictly smaller values of n. This implies that all calls to plus' will eventually terminate. Coq demands that some argument of every Fixpoint definition is "decreasing".
This requirement is a fundamental feature of Coq's design: In particular, it guarantees that every function that can be defined in Coq will terminate on all inputs. However, because Coq's "decreasing analysis" is not very sophisticated, it is sometimes necessary to write functions in slightly unnatural ways.

Exercise: 2 stars, optional (decreasing)

To get a concrete sense of this, find a way to write a sensible Fixpoint definition (of a simple function on numbers, say) that does terminate on all inputs, but that Coq will not accept because of this restriction.


(* $Date: 2013-12-03 06:45:41 -0600 (Tue, 03 Dec 2013) $ *)