# BasicsFunctional Programming

(* \$Date: 2012-07-21 13:38:33 -0400 (Sat, 21 Jul 2012) \$ *)

# Enumerated Types

In Coq's programming language, almost nothing is built in — not even booleans or numbers! Instead, it provides powerful tools for defining new types of data and functions that process and transform them.

## Days of the Week

Let's start with a very simple example. The following declaration tells Coq that we are defining a new set of data values — a "type." The type is called day, and its members are monday, tuesday, etc. The lines of the definition can be read "monday is a day, tuesday is a day, etc."

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

Having defined day, we can write functions that operate on days.

Definition next_weekday (d:day) : day :=
match d with
| monday => tuesday
| tuesday => wednesday
| wednesday => thursday
| thursday => friday
| friday => monday
| saturday => monday
| sunday => monday
end.

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 simpl to evaluate a compound expression involving next_weekday.

Eval simpl in (next_weekday friday).
(* ==> monday : day *)
Eval simpl 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 simpl ("simplify") tells Coq precisely how to evaluate the expression we give it. For the moment, simpl 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 are the same after 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 won't have space to dig further into this topic, but more information can be found in the Coq'Art book by Bertot and CastÃ©ran, as well as the Coq reference manual.

## Booleans

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
| true => false
| false => true
end.

Definition andb (b1:bool) (b2:bool) : bool :=
match b1 with
| true => b2
| false => false
end.

Definition orb (b1:bool) (b2:bool) : bool :=
match b1 with
| true => true
| false => b2
end.

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

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 following bit of Coq hackery defines a magic value called admit that can fill a hole in an incomplete definition or proof. We'll use it in the definition of nandb in the following exercise. In general, your job in the exercises is to replace admit or Admitted with real definitions or proofs.

Definition admit {T: Type} : T. Admitted.

#### Exercise: 1 star (nandb)

Complete the definition of the following function, then make sure that the Example assertions below each can 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. simpl. 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)

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."

## Numbers

Technical digression: Coq provides a fairly fancy module system, to aid in organizing large developments. In this course, we won't need most of its features, but one of them 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, all these definitions will be referred to by names like X.foo instead of just foo. This means that the new definition will not clash with the unqualified name foo later, which would otherwise be an error (a name can only be defined once in a given scope). 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 (weekday, 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.
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, predecessor:

Definition pred (n : nat) : nat :=
match n with
| O => O
| S n' => n'
end.

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
| O => O
| S O => O
| S (S n') => n'
end.

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 simpl 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 n can be simplified to match n with | O => O | S m' => m' end — while the definition of S has no such behavior attached. Although it is 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
| O => true
| S O => false
| S (S n') => evenb n'
end.

When Coq checks this definition, it notes that evenb 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 evenb will eventually terminate. Coq demands that some argument of every Fixpoint definition is decreasing.
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. simpl. reflexivity. Qed.
Example test_oddb2: (oddb (S (S (S (S O))))) = false.
Proof. simpl. 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
| O => m
| S n' => S (plus n' m)
end.

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

Eval simpl 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
| O => O
| S n' => plus m (mult n' m)
end.

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 _ , O => n
| S n', S m' => minus n' m'
end.

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
| O => S O
| S p => mult base (exp base p)
end.

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

#### 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.

Notation "x + y" := (plus x y)
(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).

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.
Each notation-symbol in Coq is 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.
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
| O => match m with
| O => true
| S m' => false
end
| S n' => match m with
| O => false
| S m' => beq_nat n' m'
end
end.

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
| O => true
| S n' =>
match m with
| O => false
| S m' => ble_nat n' m'
end
end.

Example test_ble_nat1: (ble_nat 2 2) = true.
Proof. simpl. reflexivity. Qed.
Example test_ble_nat2: (ble_nat 2 4) = true.
Proof. simpl. reflexivity. Qed.
Example test_ble_nat3: (ble_nat 4 2) = false.
Proof. simpl. 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 the function's definition to simplify the expressions on both sides of the = and notice that they become identical.
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, since the definition of + is recursive in its first argument.

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

The reflexivity command implicitly simplifies both sides of the equality before testing to see if they are the same, so we can shorten the proof a little. (It will be useful later to know that reflexivity actually does somwhat more than simpl — 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.)

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

The form of this theorem and proof are almost exactly the same as the examples above: the only differences are that we've added the quantifier n:nat and that we've used the keyword Theorem instead of Example. Indeed, the latter 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.
The keywords 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.

#### Exercise: 1 star, optional (simpl_plus)

What will Coq print in response to this query?

(* Eval simpl in (forall n:nat, n + 0 = n). *)

(* Eval simpl in (forall n:nat, 0 + n = n). *)

Explain the difference.

# The intros Tactic

Aside from unit tests, which apply functions to particular arguments, most of the properties we will be interested in proving about programs will begin with some quantifiers (e.g., "for all numbers n, ...") and/or hypothesis ("assuming m=n, ..."). In such situations, we will need to be able to reason by assuming the hypothesis — i.e., we start by saying "OK, suppose n is some arbitrary number," or "OK, suppose m=n."
The intros tactic permits us to do this by moving one or more quantifiers or hypotheses from the goal to a "context" of current assumptions.
For example, here is a slightly different proof of the same theorem.

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

Step through this proof in Coq and notice how the goal and context change.

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

Theorem mult_0_l : n:nat, 0 * n = 0.
Proof.
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."
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.

Proof.
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 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.
Proof.
(* FILL IN HERE *) Admitted.
The Admitted command tells Coq that we want to give up 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 admit or Admitted 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.
Proof.
intros n m.
rewrite plus_O_n.
reflexivity. Qed.

#### Exercise: 2 stars, recommended (mult_1_plus)

Theorem mult_1_plus : n m : nat,
(1 + n) * m = m + (n * m).
Proof.
(* FILL IN HERE *) Admitted.

# Case Analysis

Of course, not everything can be proved by simple calculation: In general, unknown, hypothetical values (arbitrary numbers, booleans, lists, etc.) can show up in the "head position" of functions that we want to reason about, blocking simplification. 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.
Proof.
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.
Proof.
intros n. destruct n as [| n'].
reflexivity.
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.
Proof.
intros b. destruct b.
reflexivity.
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.
Proof.
(* FILL IN HERE *) Admitted.

# Naming Cases

The fact that there is no explicit command for moving from one branch of a case analysis to the next can make proof scripts rather hard to read. In larger proofs, with nested case analyses, it can even become hard to stay oriented when you're sitting with Coq and stepping through the proof. (Imagine trying to remember that the first five subgoals belong to the inner case analysis and the remaining seven cases are what remains of the outer one...) Disciplined use of indentation and comments can help, but a better way is to use the Case tactic.
Case is not built into Coq: we need to define it ourselves. There is no need to understand how it works — just skip over the definition to the example that follows. It uses some facilities of Coq that we have not discussed — the string library (just for the concrete syntax of quoted strings) and the Ltac command, which allows us to declare custom tactics. Kudos to Aaron Bohannon for this nice hack!

Require String. Open Scope string_scope.

Ltac move_to_top x :=
match reverse goal with
| H : _ _ => try move x after H
end.

Tactic Notation "assert_eq" ident(x) constr(v) :=
let H := fresh in
assert (x = v) as H by reflexivity;
clear H.

Tactic Notation "Case_aux" ident(x) constr(name) :=
first [
set (x := name); move_to_top x
| assert_eq x name; move_to_top x
| fail 1 "because we are working on a different case" ].

Tactic Notation "Case" constr(name) := Case_aux Case name.
Tactic Notation "SCase" constr(name) := Case_aux SCase name.
Tactic Notation "SSCase" constr(name) := Case_aux SSCase name.
Tactic Notation "SSSCase" constr(name) := Case_aux SSSCase name.
Tactic Notation "SSSSCase" constr(name) := Case_aux SSSSCase name.
Tactic Notation "SSSSSCase" constr(name) := Case_aux SSSSSCase name.
Tactic Notation "SSSSSSCase" constr(name) := Case_aux SSSSSSCase name.
Tactic Notation "SSSSSSSCase" constr(name) := Case_aux SSSSSSSCase name.

Here's an example of how Case is used. Step through the following proof and observe how the context changes.

Theorem andb_true_elim1 : b c : bool,
andb b c = true b = true.
Proof.
intros b c H.
destruct b.
Case "b = true".
reflexivity.
Case "b = false".
rewrite H. reflexivity. Qed.

Case does something very trivial: It simply adds a string that we choose (tagged with the identifier "Case") to the context for the current goal. When subgoals are generated, this string is carried over into their contexts. When the last of these subgoals is finally proved and the next top-level goal (a sibling of the current one) becomes active, this string will no longer appear in the context and we will be able to see that the case where we introduced it is complete. Also, as a sanity check, if we try to execute a new Case tactic while the string left by the previous one is still in the context, we get a nice clear error message.
For nested case analyses (i.e., when we want to use a destruct to solve a goal that has itself been generated by a destruct), there is an SCase ("subcase") tactic.

#### Exercise: 2 stars (andb_true_elim2)

Prove andb_true_elim2, marking cases (and subcases) when you use destruct.

Theorem andb_true_elim2 : b c : bool,
andb b c = true c = true.
Proof.
(* FILL IN HERE *) Admitted.
There are no hard and fast rules for how proofs should be formatted in Coq — in particular, where lines should be broken and how sections of the proof should be indented to indicate their nested structure. However, if the places where multiple subgoals are generated are marked with explicit Case tactics placed at the beginning of lines, then the proof will be readable almost no matter what choices are made about other aspects of layout.
This is a good place to mention one other piece of (possibly obvious) advice about line lengths. Beginning Coq users sometimes tend to the extremes, either writing each tactic on its own line or entire proofs on one line. Good style lies somewhere in the middle. In particular, one reasonable convention is to limit yourself to 80-character lines. Lines longer than this are hard to read and can be inconvenient to display and print. Many editors have features that help enforce this.

# Induction

We proved above that 0 is a neutral element for + on the left using a simple partial evaluation argument. The fact that it is also a neutral element on the right...

Theorem plus_0_r_firsttry : n:nat,
n + 0 = n.

... cannot be proved in the same simple way. Just applying reflexivity doesn't work: the n in n + 0 is an arbitrary unknown number, so the match in the definition of + can't be simplified. And reasoning by cases using destruct n doesn't get us much further: the branch of the case analysis where we assume n = 0 goes through, but in the branch where n = S n' for some n' we get stuck in exactly the same way. We could use destruct n' to get one step further, but since n can be arbitrarily large, if we try to keep on going this way we'll never be done.

Proof.
intros n.
simpl. (* Does nothing! *)

Case analysis gets us a little further, but not all the way:

Theorem plus_0_r_secondtry : n:nat,
n + 0 = n.
Proof.
intros n. destruct n as [| n'].
Case "n = 0".
reflexivity. (* so far so good... *)
Case "n = S n'".
simpl. (* ...but here we are stuck again *)

To prove such facts — indeed, to prove most interesting facts about numbers, lists, and other inductively defined sets — we need a more powerful reasoning principle: induction.
Recall (from high school) the principle of induction over natural numbers: If P(n) is some proposition involving a natural number n and we want to show that P holds for all numbers n, we can reason like this:
• show that P(O) holds;
• show that, for any n', if P(n') holds, then so does P(S n');
• conclude that P(n) holds for all n.
In Coq, the steps are the same but the order is backwards: we begin with the goal of proving P(n) for all n and break it down (by applying the induction tactic) into two separate subgoals: first showing P(O) and then showing P(n') P(S n'). Here's how this works for the theorem we are trying to prove at the moment:

Theorem plus_0_r : n:nat, n + 0 = n.
Proof.
intros n. induction n as [| n'].
Case "n = 0". reflexivity.
Case "n = S n'". simpl. rewrite IHn'. reflexivity. Qed.

Like destruct, the induction tactic takes an as... clause that specifies the names of the variables to be introduced in the subgoals. In the first branch, n is replaced by 0 and the goal becomes 0 + 0 = 0, which follows by simplification. In the second, n is replaced by S n' and the assumption n' + 0 = n' is added to the context (with the name IHn', i.e., the Induction Hypothesis for n'). The goal in this case becomes (S n') + 0 = S n', which simplifies to S (n' + 0) = S n', which in turn follows from the induction hypothesis.

Theorem minus_diag : n,
minus n n = 0.
Proof.
(* WORKED IN CLASS *)
intros n. induction n as [| n'].
Case "n = 0".
simpl. reflexivity.
Case "n = S n'".
simpl. rewrite IHn'. reflexivity. Qed.

#### Exercise: 2 stars, recommended (basic_induction)

Theorem mult_0_r : n:nat,
n * 0 = 0.
Proof.
(* FILL IN HERE *) Admitted.

Theorem plus_n_Sm : n m : nat,
S (n + m) = n + (S m).
Proof.
(* FILL IN HERE *) Admitted.

Theorem plus_comm : n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.

Fixpoint double (n:nat) :=
match n with
| O => O
| S n' => S (S (double n'))
end.

#### Exercise: 2 stars (double_plus)

Lemma double_plus : n, double n = n + n .
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 1 star (destruct_induction)

Briefly explain the difference between the tactics destruct and induction.
(* FILL IN HERE *)

# Formal vs. Informal Proof

"Informal proofs are algorithms; formal proofs are code."
The question of what, exactly, constitutes a "proof" of a mathematical claim has challenged philosophers for millenia. A rough and ready definition, though, could be this: a proof of a mathematical proposition P is a written (or spoken) text that instills in the reader or hearer the certainty that P is true. That is, a proof is an act of communication.
Now, acts of communication may involve different sorts of readers. On one hand, the "reader" can be a program like Coq, in which case the "belief" that is instilled is a simple mechanical check that P can be derived from a certain set of formal logical rules, and the proof is a recipe that guides the program in performing this check. Such recipes are formal proofs.
Alternatively, the reader can be a human being, in which case the proof will be written in English or some other natural language, thus necessarily informal. Here, the criteria for success are less clearly specified. A "good" proof is one that makes the reader believe P. But the same proof may be read by many different readers, some of whom may be convinced by a particular way of phrasing the argument, while others may not be. One reader may be particularly pedantic, inexperienced, or just plain thick-headed; the only way to convince them will be to make the argument in painstaking detail. But another reader, more familiar in the area, may find all this detail so overwhelming that they lose the overall thread. All they want is to be told the main ideas, because it is easier to fill in the details for themselves. Ultimately, there is no universal standard, because there is no single way of writing an informal proof that is guaranteed to convince every conceivable reader. In practice, however, mathematicians have developed a rich set of conventions and idioms for writing about complex mathematical objects that, within a certain community, make communication fairly reliable. The conventions of this stylized form of communication give a fairly clear standard for judging proofs good or bad.
Because we are using Coq in this course, we will be working heavily with formal proofs. But this doesn't mean we can ignore the informal ones! Formal proofs are useful in many ways, but they are not very efficient ways of communicating ideas between human beings.
For example, here is a proof that addition is associative:

Theorem plus_assoc' : n m p : nat,
n + (m + p) = (n + m) + p.
Proof. intros n m p. induction n as [| n']. reflexivity.
simpl. rewrite IHn'. reflexivity. Qed.

Coq is perfectly happy with this as a proof. For a human, however, it is difficult to make much sense of it. If you're used to Coq you can probably step through the tactics one after the other in your mind and imagine the state of the context and goal stack at each point, but if the proof were even a little bit more complicated this would be next to impossible. Instead, a mathematician mighty write it like this:
• Theorem: For any n, m and p,
n + (m + p) = (n + m) + p.
Proof: By induction on n.
• First, suppose n = 0. We must show
0 + (m + p) = (0 + m) + p.
This follows directly from the definition of +.
• Next, suppose n = S n', where
n' + (m + p) = (n' + m) + p.
We must show
(S n') + (m + p) = ((S n') + m) + p.
By the definition of +, this follows from
S (n' + (m + p)) = S ((n' + m) + p),
which is immediate from the induction hypothesis.
The overall form of the proof is basically similar. This is no accident, of course: Coq has been designed so that its induction tactic generates the same sub-goals, in the same order, as the bullet points that a mathematician would write. But there are significant differences of detail: the formal proof is much more explicit in some ways (e.g., the use of reflexivity) but much less explicit in others; in particular, the "proof state" at any given point in the Coq proof is completely implicit, whereas the informal proof reminds the reader several times where things stand.
Here is a formal proof that shows the structure more clearly:

Theorem plus_assoc : n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
intros n m p. induction n as [| n'].
Case "n = 0".
reflexivity.
Case "n = S n'".
simpl. rewrite IHn'. reflexivity. Qed.

#### Exercise: 2 stars (plus_comm_informal)

Translate your solution for plus_comm into an informal proof.
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)

#### Exercise: 2 stars, optional (beq_nat_refl_informal)

Write an informal proof of the following theorem, using the informal proof of plus_assoc as a model. Don't just paraphrase the Coq tactics into English!
Theorem: true = beq_nat n n for any n.
Proof: (* FILL IN HERE *)

#### Exercise: 1 star, optional (beq_nat_refl)

Theorem beq_nat_refl : n : nat,
true = beq_nat n n.
Proof.
(* FILL IN HERE *) Admitted.

# Proofs Within Proofs

In Coq, as in informal mathematics, large proofs are very often broken into a sequence of theorems, with later proofs referring to earlier theorems. Occasionally, however, a proof will need some miscellaneous fact that is too trivial (and of too little general interest) to bother giving it its own top-level name. In such cases, it is convenient to be able to simply state and prove the needed "sub-theorem" right at the point where it is used. The assert tactic allows us to do this. For example, our earlier proof of the mult_0_plus theorem referred to a previous theorem named plus_O_n. We can also use assert to state and prove plus_O_n in-line:

Theorem mult_0_plus' : n m : nat,
(0 + n) * m = n * m.
Proof.
intros n m.
assert (H: 0 + n = n).
Case "Proof of assertion". reflexivity.
rewrite H.
reflexivity. Qed.

The assert tactic introduces two sub-goals. The first is the assertion itself; by prefixing it with H: we name the assertion H. (Note that we could also name the assertion with as just as we did above with destruct and induction, i.e., assert (0 + n = n) as H. Also note that we mark the proof of this assertion with a Case, both for readability and so that, when using Coq interactively, we can see when we're finished proving the assertion by observing when the "Proof of assertion" string disappears from the context.) The second goal is the same as the one at the point where we invoke assert, except that, in the context, we have the assumption H that 0 + n = n. That is, assert generates one subgoal where we must prove the asserted fact and a second subgoal where we can use the asserted fact to make progress on whatever we were trying to prove in the first place.
Actually, assert will turn out to be handy in many sorts of situations. For example, suppose we want to prove that (n + m) + (p + q) = (m + n) + (p + q). The only difference between the two sides of the = is that the arguments m and n to the first inner + are swapped, so it seems we should be able to use the commutativity of addition (plus_comm) to rewrite one into the other. However, the rewrite tactic is a little stupid about where it applies the rewrite. There are three uses of + here, and it turns out that doing rewrite plus_comm will affect only the outer one.

Theorem plus_rearrange_firsttry : n m p q : nat,
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
(* We just need to swap (n + m) for (m + n)...
it seems like plus_comm should do the trick! *)

rewrite plus_comm.
(* Doesn't work...Coq rewrote the wrong plus! *)

To get plus_comm to apply at the point where we want it, we can introduce a local lemma stating that n + m = m + n (for the particular m and n that we are talking about here), prove this lemma using plus_comm, and then use this lemma to do the desired rewrite.

Theorem plus_rearrange : n m p q : nat,
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
assert (H: n + m = m + n).
Case "Proof of assertion".
rewrite plus_comm. reflexivity.
rewrite H. reflexivity. Qed.

#### Exercise: 4 stars, recommended (mult_comm)

Use assert to help prove this theorem. You shouldn't need to use induction.

Theorem plus_swap : n m p : nat,
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.

Now prove commutativity of multiplication. (You will probably need to define and prove a separate subsidiary theorem to be used in the proof of this one.) You may find that plus_swap comes in handy.

Theorem mult_comm : m n : nat,
m * n = n * m.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, optional (evenb_n__oddb_Sn)

Theorem evenb_n__oddb_Sn : n : nat,
evenb n = negb (evenb (S n)).
Proof.
(* FILL IN HERE *) Admitted.

# More Exercises

#### Exercise: 3 stars, optional (more_exercises)

Take a piece of paper. For each of the following theorems, first think about whether (a) it can be proved using only simplification and rewriting, (b) it also requires case analysis (destruct), or (c) it also requires induction. Write down your prediction. Then fill in the proof. (There is no need to turn in your piece of paper; this is just to encourage you to reflect before hacking!)

Theorem ble_nat_refl : n:nat,
true = ble_nat n n.
Proof.
(* FILL IN HERE *) Admitted.

Theorem zero_nbeq_S : n:nat,
beq_nat 0 (S n) = false.
Proof.
(* FILL IN HERE *) Admitted.

Theorem andb_false_r : b : bool,
andb b false = false.
Proof.
(* FILL IN HERE *) Admitted.

Theorem plus_ble_compat_l : n m p : nat,
ble_nat n m = true ble_nat (p + n) (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.

Theorem S_nbeq_0 : n:nat,
beq_nat (S n) 0 = false.
Proof.
(* FILL IN HERE *) Admitted.

Theorem mult_1_l : n:nat, 1 * n = n.
Proof.
(* FILL IN HERE *) Admitted.

Theorem all3_spec : b c : bool,
orb
(andb b c)
(orb (negb b)
(negb c))
= true.
Proof.
(* FILL IN HERE *) Admitted.

Theorem mult_plus_distr_r : n m p : nat,
(n + m) * p = (n * p) + (m * p).
Proof.
(* FILL IN HERE *) Admitted.

Theorem mult_assoc : n m p : nat,
n * (m * p) = (n * m) * p.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, optional (plus_swap')

The replace tactic allows you to specify a particular subterm to rewrite and what you want it rewritten to. More precisely, replace (t) with (u) replaces (all copies of) expression t in the goal by expression u, and generates t = u as an additional subgoal. This is often useful when a plain rewrite acts on the wrong part of the goal.
Use the replace tactic to do a proof of plus_swap', just like plus_swap but without needing assert (n + m = m + n).

Theorem plus_swap' : n m p : nat,
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, optional

Theorem bool_fn_applied_thrice :
(f : bool bool) (b : bool),
f (f (f b)) = f b.
Proof.
intros f b.
destruct b.
Case "b = true".
remember (f true) as ftrue.
destruct ftrue.
SCase "f true = true".
rewrite Heqftrue.
symmetry.
apply Heqftrue.
SCase "f true = false".
remember (f false) as ffalse.
destruct ffalse.
SSCase "f false = true".
symmetry.
apply Heqftrue.
SSCase "f false = false".
symmetry.
apply Heqffalse.
remember (f false) as ffalse.
destruct ffalse.
SCase "f false = true".
remember (f true) as ftrue.
destruct ftrue.
SSCase "f true = true".
symmetry.
apply Heqftrue.
SSCase "f true = false".
symmetry.
apply Heqffalse.
SCase "f false = false".
rewrite Heqffalse.
symmetry.
apply Heqffalse.
Qed.

#### Exercise: 4 stars, recommended (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 a nat (whatever that is), and if n is a nat 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) Finally, prove that your increment and binary-to-unary functions commute: that is, incrementing a binary number and then converting it to unary yields the same result as first converting it to unary and then incrementing.

(* FILL IN HERE *)

#### Exercise: 5 stars (binary_inverse)

This exercise is a continuation of the previous exercise about binary numbers. You will need your definitions and theorems from the previous exercise to complete this one.
(a) First, write a function to convert natural numbers to binary numbers. Then prove that starting with any natural number, converting to binary, then converting back yields the same natural number you started with.
(b) You might naturally think that we should also prove the opposite direction: that starting with a binary number, converting to a natural, and then back to binary yields the same number we started with. However, it is not true! Explain what the problem is.
(c) Define a function normalize from binary numbers to binary numbers such that for any binary number b, converting to a natural and then back to binary yields (normalize b). Prove it.

(* FILL IN HERE *)

#### Exercise: 2 stars, optional (decreasing)

The requirement that some argument to each function be "decreasing" 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.
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.

(* FILL IN HERE *)