# Introduction

The functional programming style is founded on simple, everyday mathematical intuition: If a procedure or method has no side effects, then (ignoring efficiency) all we need to understand about it is how it maps inputs to outputs — that is, we can think of it as just a concrete method for computing a mathematical function. This is one sense of the word "functional" in "functional programming." The direct connection between programs and simple mathematical objects supports both formal correctness proofs and sound informal reasoning about program behavior.
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, included in data structures, etc. The recognition that functions can be treated as data gives rise to a host of useful and powerful programming idioms.
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 supporting abstraction and code reuse. Coq offers all of these features.
The first half of this chapter introduces the most essential elements of Coq's functional programming language, called Gallina. The second half introduces some basic tactics that can be used to prove properties of Coq programs.

# Data and Functions

## Enumerated Types

One notable 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 a powerful mechanism for defining new data types from scratch, with all these familiar types as instances.
Naturally, the Coq distribution comes preloaded 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. To illustrate this, we will explicitly recapitulate all the definitions we need in this course, rather than just getting them implicitly from the library.

## Days of the Week

To see how this definition mechanism works, 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.
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 and following 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
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 figure out these types for itself when they are not given explicitly — i.e., it can do type inference — but we'll generally 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 Compute to evaluate a compound expression involving next_weekday.
Compute (next_weekday friday).
(* ==> monday : day *)

Compute (next_weekday (next_weekday saturday)).
(* ==> tuesday : day *)
(We show Coq's responses in comments, but, if you have a computer handy, this 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 Coq sources, find the above example, submit it to Coq, and observe the result.)
Second, we can record what we expect the result to be in the form of a Coq example:
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 our 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 go from proved-correct algorithms written in Gallina to efficient machine code. (Of course, we are trusting the correctness of the OCaml/Haskell/Scheme compiler, and of Coq's extraction facility itself, but this is still a big step forward from the way most software is developed today.) Indeed, this is one of the main uses for which Coq was developed. We won't really talk about extraction more in this course.

## Homework Submission Guidelines

If you are using Software Foundations in a course, your instructor may use automatic scripts to help grade your homework assignments. In order for these scripts to work correctly (so that you get full credit for your work!), please be careful to follow these rules:
• The grading scripts work by extracting marked regions of the .v files that you submit. It is therefore important that you do not alter the "markup" that delimits exercises: the Exercise header, the name of the exercise, the "empty square bracket" marker at the end, etc. Please leave this markup exactly as you find it.
• Do not delete exercises. If you skip an exercise (e.g., because it is marked Optional, or because you can't solve it), it is OK to leave a partial proof in your .v file, but in this case please make sure it ends with Admitted (not, for example Abort).
• It is fine to use additional definitions (of helper functions, useful lemmas, etc.) in your solutions. You can put these between the exercise header and the theorem you are asked to prove.

## Booleans

In a similar way, we can define the standard 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, 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. First, we can use booleans to define a predicate, a function that identifies some subset of a given set:
Definition is_weekday (d:day) : bool :=
match d with
| mondaytrue
| tuesdaytrue
| wednesdaytrue
| thursdaytrue
| fridaytrue
| saturdayfalse
| sundayfalse
end.
We can also define some of the usual operations on booleans. First comes not or negation, which is often written as the operator !.
Definition negb (b:bool) : bool :=
match b with
| truefalse
| falsetrue
end.
Another common way of expressing functions from booleans to booleans is with a truth table.
```    |---|--------|
| b | negb b |
|---|--------|
| T |    F   |
| F |    T   |
|---|--------|
```
Each column of the truth table represents an expression of type bool. Here the first column represents an arbitrary input b, which can be true (written T) or false (written F). It's typical to consider the initial columns of a truth table as representing inputs and the final column as representing an output.
Each row of the truth table gives a possible assignment: you can read the first row as saying that if b = true, then negb b = false; the second row says that if b = false, then negb b = true.
Definition andb (b1:bool) (b2:bool) : bool :=
match b1 with
| trueb2
| falsefalse
end.
When constructing a truth table with more than one input, it's important to make sure your truth table has every possible input configuration accounted for. People have different ways of doing so, but I tend to like the following format, where we exhaust all of the possibilities for the first column to be true, and then we consider the cases where the first column is false. Electrical engineers, however, like to do it the opposite way: when false is 0 and true is 1, it makes sense to count "up".
It doesn't particularly matter which method you choose, but it's important to be consistent!
```    |----|----|------------|
| b1 | b2 | andb b1 b2 |
|----|----|------------|
| T  | T  |      T     |
| T  | F  |      F     |
| F  | T  |      F     |
| F  | F  |      F     |
|----|----|------------|
```
Definition orb (b1:bool) (b2:bool) : bool :=
match b1 with
| truetrue
| falseb2
end.
```    |----|----|-----------|
| b1 | b2 | orb b1 b2 |
|----|----|-----------|
| T  | T  |     T     |
| T  | F  |     T     |
| F  | T  |     T     |
| F  | F  |     F     |
|----|----|-----------|
```
The last two of these definitions illustrate Coq's syntax for multi-argument function definitions. The corresponding multi-argument application syntax is illustrated by the following "unit tests," which 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.
We can also introduce some familiar syntax for the boolean operations we have just defined. The Notation command defines a new symbolic notation for an existing definition.
Notation "x && y" := (andb x y).
Notation "x || y" := (orb x y).

Example test_orb5: false || false || true = true.
Proof. simpl. reflexivity. Qed.
A note on notation: In .v files, we use square brackets to delimit fragments of Coq code within comments; 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 command Admitted can be used as a placeholder for an incomplete proof. We'll use it in exercises, to indicate the parts that we're leaving for you — i.e., your job is to replace Admitteds with real proofs.

#### Exercise: 1 star (nandb)

Remove "Admitted." and complete the definition of the following function; then make sure that the Example assertions below can each be verified by Coq. (Remove "Admitted." and fill in each proof, following the model of the orb tests above.) The function should return true if either or both of its inputs are false.
Definition nandb (b1:bool) (b2:bool) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

#### Exercise: 1 star (nandb_truth_table)

Write the truth table for nandb. Make sure to leave it in a comment, or your file won't compile!
Note that nandb gets its name because its the Negation of AND.
(* FILL IN HERE *)
Check your truth table with the tests below.
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.

#### Exercise: 1 star (test_nandb4)

Example test_nandb4: (nandb true true) = false.
(* FILL IN HERE *) Admitted.
Truth tables are a particularly nice way of calculating compound expressions involving booleans. In addition to having input and output columns, we can have intermediate columns representing subexpressions of the boolean we're interested in. When building such a truth table, every subexpression of the final result should show up as a column.
```    |---|--------|-----------------|
| b | negb b | orb b (negb b)  |
|---|--------|-----------------|
| T |    F   |        T        |
| F |    T   |        T        |
|---|--------|-----------------|
```

#### Exercise: 1 star (compound_truthtable)

Write the truth table for the expression orb (negb b1) b2.
(* FILL IN HERE *)

#### Exercise: 1 star (impb)

Write a function impb such that impb b1 b2 has the same truth table as orb (negb b1) b2. Don't just trivially define it as orb (negb b1) b2, though! Try using a match.
Definition impb (b1:bool) (b2:bool) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

#### Exercise: 1 star (andb3)

Do the same for the andb3 function below: write the definition and the truth table. This function should return true when all of its inputs are true, and false otherwise.
Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

(* FILL IN HERE *)
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

Every expression in Coq has a type, describing what sort of thing it computes. The Check command asks Coq to print the type of an expression.
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."

## Compound Types

The types we have defined so far are examples of "enumerated types": their definitions explicitly enumerate a finite set of elements, each of which is just a bare constructor. Here is a more interesting type definition, where one of the constructors takes an argument:
Inductive rgb : Type :=
| red : rgb
| green : rgb
| blue : rgb.

Inductive color : Type :=
| black : color
| white : color
| primary : rgbcolor.
Let's look at this in a little more detail.
Every inductively defined type (day, bool, rgb, color, etc.) contains a set of constructor expressions built from constructors like red, primary, true, false, monday, etc. The definitions of rgb and color say how expressions in the sets rgb and color can be built:
• red, green, and blue are the constructors of rgb;
• black, white, and primary are the constructors of color;
• the expression red belongs to the set rgb, as do the expressions green and blue;
• the expressions black and white belong to the set color;
• if p is an expression belonging to the set rgb, then primary p (pronounced "the constructor primary applied to the argument p") is an expression belonging to the set color; and
• expressions formed in these ways are the only ones belonging to the sets rgb and color.
We can define functions on colors using pattern matching just as we have done for day and bool.
Definition monochrome (c : color) : bool :=
match c with
| blacktrue
| whitetrue
| primary pfalse
end.
Since the primary constructor takes an argument, a pattern matching primary should include either a variable (as above) or a constant of appropriate type (as below). How does monochrome white evaluate? We can write out the computation like so, simulating the way Coq works:
monochrome white
```    computes to
```
match white with
| black ⇒ true
| white ⇒ true
| primary p ⇒ false
end
Coq considers each case in turn, comparing the scrutinee (here, white) against the patterns (here, black, white, and primary p).
The scrutinee will always be a value; the pattern will be a mix of constructors and variable names. The match construct tries to match the scrutinee to each pattern, where a match is when every constructor is matched with an identical constructor or a variable.
Concretely, the scrutinee white doesn't match the pattern black, because white and black are different constructors. So Coq will skip that case, and keep evaluating:
match white with
| black ⇒ true
| white ⇒ true
| primary p ⇒ false
end
```    computes to
```
match white with
| white ⇒ true
| primary p ⇒ false
end
We're being particularly careful with our computation steps—-as we gain facility with Coq, we won't need to step each case. But here it helps us see that each pattern is considered in turn, from top to bottom.
The next pattern is white, which matches our scrutinee white, so our pattern matches and we can execute the code in that branch of the match:
match white with
| white ⇒ true
| primary p ⇒ false
end
```    computes to
```
true
In this case, there were no variables in the match to keep track of, but in general, Coq will bind each variable in a pattern to the corresponding part of the scrutinee. For example, suppose we had run monochrome (primary blue):
monochrome (primary blue)
```    computes to
```
match (primary bluewith
| black ⇒ true
| white ⇒ true
| primary p ⇒ false
end
```    computes to
```
match (primary bluewith
| white ⇒ true
| primary p ⇒ false
end
```    computes to
```
match (primary bluewith
| primary p ⇒ false
end
At this point, the argument to the primary constructor, blue, in the scrutinee corresponds to the variable p in the pattern primary p. So what Coq will do is evaluate the corresponding branch of the match, remember that p is equal to blue. (In this case, p doesn't occur at all on the right-hand side of , so it doesn't matter. But it could!)
match (primary bluewith
| primary p ⇒ false
end
```     computes to
```
false (* with p bound to blue *)
Definition isred (c : color) : bool :=
match c with
| blackfalse
| whitefalse
| primary redtrue
| primary _false
end.
The pattern primary _ here is shorthand for "primary applied to any rgb constructor except red." Recall again that Coq applies patterns in order. For example:
isred (primary green)
computes to
match (primary greenwith
| black ⇒ false
| white ⇒ false
| primary red ⇒ true
| primary _ ⇒ false
computes to
match (primary greenwith
| white ⇒ false
| primary red ⇒ true
| primary _ ⇒ false
computes to
match (primary greenwith
| primary red ⇒ true
| primary _ ⇒ false
At this point, our scrutinee primary green doesn't match the pattern primary red: while the outermost constructors are both primary, the arguments differ: green and red are different constructors of color. So Coq will skip this pattern:
match (primary greenwith
| primary red ⇒ true
| primary _ ⇒ false
computes to
match (primary greenwith
| primary _ ⇒ false
computes to
false
In this case, the wildcard pattern _ has the same effect as the dummy pattern variable p in the definition of monochrome. Both functions don't actually use the argument to primary. We could have written isred a different way:
Definition isred' (c : color) : bool :=
match c with
| blackfalse
| whitefalse
| primary p
match p with
| redtrue
| _false
end
end.
In this definition, we explicitly name the primary color we're working with and do a nested pattern match. We can compute isred' (primary green) as follows:
isred' (primary green)
computes to
match (primary greenwith
| black ⇒ false
| white ⇒ false
| primary p ⇒
match p with
| red ⇒ true
| _ ⇒ false
end
end
computes to
match (primary greenwith
| white ⇒ false
| primary p ⇒
match p with
| red ⇒ true
| _ ⇒ false
end
end
computes to
match (primary greenwith
| primary p ⇒
match p with
| red ⇒ true
| _ ⇒ false
end
end
computes to
match green with
| red ⇒ true
| _ ⇒ false
end
computes to
match green with
| _ ⇒ false
end
computes to
false
It's good practice to use wildcards when you don't need to name the variable—-it helps prevent mistakes, like referring to the wrong variable.
It's also good practice to try to condense pattern matching: the definition of isred is cleaner than that of isred'.

## Modules

Coq provides a 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 are referred to by names like X.foo instead of just foo. We will use this feature to introduce the definition of the type nat in an inner module so that it does not interfere with the one from the standard library (which we want to use in the rest because it comes with a tiny bit of convenient special notation).
Module NatPlayground.

## Numbers

In this section, we'll define the natural numbers, that is, the numbers:
0, 1, 2, 3, ...
You may have heard in a math class that 0 isn't a natural number—-for our purposes, it will be. Computer scientists almost always start counting from 0!
In other programming languages, it's easy to take numbers for granted: int is simply a built-in type, represented in some low-level way on the computer (typically binary).
But in Coq we're building everything from scratch. The representation we'll be using is called unary, which is an odd sort of way of writing numbers. You can think of it like tick marks, so the number zero is represented by no tick marks, the number one is one tick mark, the number two is two tick marks, etc.
That is, a unary natural number is either:
• zero (i.e., no tick marks), or
• a natural number followed by another tick mark.
Notice that our definition of natural numbers is in terms of natural numbers themselves-—that is, to allow the rules describing the natural numbers are inductive.
Here's the above definition translated into Coq:
Inductive nat : Type :=
| O : nat
| S : natnat.
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 can be put in front of a natural number to yield another one — if n is a natural number, then S n is too.
Again, let's look at this in a little more detail. The definition of nat says how expressions in the set nat can be built:
• O and S are constructors;
• 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, bool, color, etc.
The above 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 built from data constructors, like true, andb true false, S (S false), and O (O (O S)) do not.
A critical point here is that what we've done so far is just to define a representation of numbers: a way of writing them down. The names O and S are arbitrary, and at this point they have no special meaning — they are just two different marks that we can use to write down numbers (together with a rule that says any nat will be written as some string of S marks followed by an O). If we like, we can write essentially the same definition this way:
Inductive nat' : Type :=
| stop : nat'
| tick : nat'nat'.
The interpretation of these marks comes from how we use them to compute. We use O and S rather than these longer names because (a) concision is nice, (b) we'll be using nat quite a bit, and (c) that's how the Coq standard library defines them.
We interpret by writing functions that pattern match on representations of natural numbers just as we did above with booleans and days — for example, here is the predecessor function, which takes a given number and returns one less than that number, i.e.,
pred (S n) should compute to n
For example, pred (S O) computes to O, i.e., 0 is the predecessor of 1.
Definition pred (n : nat) : nat :=
match n with
| OO
| S n'n'
end.
Our definition has an unfortunate property: pred O computes to O, i.e., 0 is its own predecessor. A mathematician might take a different course of action, saying that pred O is undefined—-there really is no natural number that comes before 0, since 0 is the first one! Coq won't let us leave anything undefined, though, so we'll simply have to be careful to remember that pred O = O.
The second branch can be read: "if n has the form S n' for some n', then return 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)))).
(* ===> 4 : nat *)

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

Compute (minustwo 4).
(* ===> 2 : nat *)
The constructor S has the type nat nat, just like pred and functions like minustwo:
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 between the first one and the other two: 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! It is just a way of writing down numbers. (Think about standard arabic numerals: the numeral 1 is not a computation; it's a piece of data. When we write 111 to mean the number one hundred and eleven, we are using 1, three times, to write down a concrete representation of a number.)
For most function definitions over numbers, just pattern matching is not enough: we also need recursion, i.e., functions that refer to themselves. 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.
In your prior programming experience, you may not have spent a lot of thought on recursion. Many languages use loops (using words like while or for) rather than recursion. Coq doesn't have loops—-only recursion. Don't worry if you're not particularly familiar with recursion... you'll get lots of practice!
Fixpoint evenb (n:nat) : bool :=
match n with
| Otrue
| S Ofalse
| S (S n') ⇒ evenb n'
end.
The call to evenb n' is a recursive call. One might worry: is it okay to define a function in terms of itself? How do we know that evenb terminates?
We can check an example: is 3 even?
evenb 3
is the same as
evenb (S (S (S O)))
computes to
match (S (S (S O))) with
| O        ⇒ true
| S O      ⇒ false
| S (S n') ⇒ evenb n'
end
computes to
match (S (S (S O))) with
| S O      ⇒ false
| S (S n') ⇒ evenb n'
end
computes to
match (S (S (S O))) with
| S (S n') ⇒ evenb n'
end
computes to
evenb n' (* with n' bound to S O *)
is the same as
evenb (S 0)
computes to
match (S Owith
| O        ⇒ true
| S O      ⇒ false
| S (S n') ⇒ evenb n'
end
computes to
match (S Owith
| S O      ⇒ false
| S (S n') ⇒ evenb n'
end
computes to
false
More generally, evenb only ever makes recursive calls on smaller inputs: if we put in the number n, we'll make a recursive call on two less than n.
A surprising fact: Coq will only let us write functions that are guaranteed to terminate on all inputs. A good rule of thumb is that every recursive function should make a recursive call only on subparts of its input.
We can define oddb by a similar Fixpoint declaration, but here is a simpler definition:
Definition oddb (n:nat) : bool := negb (evenb n).

Example test_oddb1: oddb 1 = true.
Proof. simpl. reflexivity. Qed.
Example test_oddb2: oddb 4 = false.
Proof. simpl. reflexivity. Qed.
(You will notice if you step through these proofs that simpl actually has no effect on the goal — all of the work is done by reflexivity. We'll see more about why that is shortly.)
Naturally, we can also define multi-argument functions by recursion.
Module NatPlayground2.

Fixpoint plus (n : nat) (m : nat) : nat :=
match n with
| Om
| S n'S (plus n' m)
end.
Adding three to two now gives us five, as we'd expect.
Compute (plus 3 2).
The simplification that Coq performs to reach this conclusion can be visualized as follows, a little more succinctly than what we wrote above:
(*  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)
end.

Example test_mult1: (mult 3 3) = 9.
Proof. simpl. 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 _ , O) ⇒ n
| (S n', S m') ⇒ minus n' m'
end.
Our minus function has the same problem pred does: minus 0 n results in 0! This 'truncating' subtraction is necessary because we're working with natural numbers, i.e., there are no negative numbers.
Keep in mind: these definitions have suggestive names, but it's up to us as humans to agree that minus corresponds to our notion of subtraction. I would wager that it does: if we pick an n and an m such that m n, then minus n m yields n - m. We don't have the tools to prove this... yet!
Again, 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 variable name.
The exp function defines exponentiation: exp n m should yield n to the mth power.
Fixpoint exp (base power : nat) : nat :=
match power with
| OS O
| S pmult base (exp base p)
end.

#### Exercise: 1 star (factorial)

Here's the standard mathematical factorial function:
```       factorial(0)  =  1
factorial(n)  =  n * factorial(n-1)     (if n>0)
```
We'll meet this function again later in the course. For now, translate this into Coq.
Fixpoint factorial (n:nat) : nat
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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).
(The level, associativity, and nat_scope annotations control how these notations are treated by Coq's parser. The details are not important for our purposes, but interested readers can refer to the optional "More on Notation" 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 almost nothing built-in, we really mean it: even equality testing for numbers is a user-defined operation! We now define a function beq_nat, which 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
end
| S n'match m with
| Ofalse
| S m'beq_nat n' m'
end
end.
The leb function tests whether its first argument is less than or equal to its second argument, yielding a boolean.
Fixpoint leb (n m : nat) : bool :=
match n with
| Otrue
| S n'
match m with
| Ofalse
| S m'leb n' m'
end
end.

Example test_leb1: (leb 2 2) = true.
Proof. simpl. reflexivity. Qed.
Example test_leb2: (leb 2 4) = true.
Proof. simpl. reflexivity. Qed.
Example test_leb3: (leb 4 2) = false.
Proof. simpl. reflexivity. Qed.

#### Exercise: 1 star (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.
Definition blt_nat (n m : nat) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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 stating and proving properties of their behavior. Actually, 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 simpl to simplify both sides of the equation, then use reflexivity to check that both sides contain identical values.
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.
Proof.
intros n. simpl. reflexivity. Qed.
What's happening here? It's worth stepping through the proof slowly, looking at the various views: there's the main script screen, the screen with the goal and context, and a response screen.
We write Theorem and then a name for our theorem—-here plus_O_n. Then we write a proposition, in this case n : nat, 0 + n = n, i.e., for any possible natural number n, it is the case that 0 + n is equal to n. We'll talk more about what counts as a proposition later in the course.
After we type the Proof. keyword, we're shown a screen like the following:
subgoalsubgoal 1 (ID 50)

============================
n : nat, 0 + n = n
The first line says how many cases we're considering in our proof at present: there's 1 subgoal, and we're currently working on it. That is: our proof has only one case.
Then there's a line. Beneath the line is our *goal*—-that's what we're trying to prove. Above the line is our *context*, which is currently empty.
The way a proof works is that we try to show that given our context, our goal *holds*, that is, that our goal is a true proposition.
In Coq, we use tactics to manipulate the goal and context, where Coq will keep track of the goal and context for us. On paper (or a chalkboard or in person), we'll use natural language (for CS 54, English) to to manipulate the goal and context, which we'll keep track of ourselves.
Throughout the course, we'll try to keep parallel tracks in mind: how does proof work in Coq and how does it work on paper? After this course, you'll probably only use paper proofs. We'll be using Coq as a tool to help you learn the ropes of paper proof. One of the hardest things about paper proof is that it can be very easy to get confused and make mistakes, breaking the "rules". Coq enforces the rules! Using Coq will help you internalize the rules.
After the Proof. keyword comes our *proof script*, a series of *tactics* telling Coq how to manipulate the proof state. Each tactic is followed by a period.
Before explaining the proof script, let's see the above proof in English.
• Theorem: For any natural number n,
0 + n = n.
Proof: Let n be given. We must show that
0 + n = n.
By the definition of plus, we know that 0+n reduces to n. Finally, we have
n = n
immediately. Qed.
How does the above proof work? To start out, we needed to show that n : nat, 0 + n = n. The line Let n be given phrases the proof as a sort of a game. To show that 0 + n = n for any n, we let our imaginary "opponent" give us any n they please. Now we have that n in hand—-that is, in our proof's context—-we want to show that 0 + n = n, our new goal. Showing this goal is as simple as looking at the definition of plus: the first case of the pattern match says that when the first argument to plus is 0, plus just returns its second argument—-in this case n. So 0 + n reduces to n; finally, we observe that n = n immediately—-an inarguable fact about natural numbers. How does the proof in Coq work? Our first tactic is intros with the argument n. Our goal is n : nat, 0 + n = n; in English, our goal is to show that for every possible n that is a natural number 0 + n is equal to n. The intros tactic *introduces* things into the context: our goal was a , so intros will try to introduce whatever's been *quantified* by the for all. Here give a name explicitly: intros n says to name the introduced variable n and put it into our context:
subgoalsubgoal 1 (ID 51)

n : nat
============================
0 + n = n
We could have chosen a different name—-go back and change it to read intros m and see what goal you get! (It's generally good style to use the name in the quantifier, since it's a little clearer.)
Our next tactic is simpl, which asks Coq to simplify our goal, running a few steps of computation:
subgoalsubgoal 1 (ID 53)

n : nat
============================
n = n
Our context remains the same, but our new goal is to show that n = n. To do so, we use the reflexivity tactic, named after the property of equality: all things are equal to themselves.
When we run the tactic, we get a new readout:
No more subgoals.
(dependent evars: (printing disabled) )
Coq is being a little too lowkey here: we proved it! To celebrate (and tell Coq that we're satisfied with our proof), we say Qed., which is short for quod erat demonstrandum, which is Latin for "that which was to be proved" which is perhaps better said as "and we've proved what we want to" or "and that's the proof" or "I told you so!" or "mic drop". Then Coq acknowledges that we're truly done:
plus_O_n is defined
The paper and Coq proofs are very much in parallel. Here's the paper proof annotated with Coq tactics:
• Theorem: For any natural number n,
0 + n = n.
Proof: Let n be given (intros n.). We must show that
0 + n = n.
By the definition of plus, we know that 0+n reduces to n (simpl.). Finally, we have
n = n
immediately (reflexivity.). Qed.
Throughout the course, we'll have to work hard to write good proofs. At first our proofs will follow the Coq tactics quite closely. But just like any other kind of writing, proof writing is a craft to be honed and practiced.
(You may notice that the above statement looks different in the .v file in your IDE than it does in the HTML rendition in your browser, if you are viewing both. In .v files, we write the universal quantifier using the reserved identifier "forall." When the .v files are converted to HTML, this gets transformed into an upside-down-A symbol.)
This is a good place to mention that reflexivity is a bit more powerful than we have admitted. In the examples we have seen, the calls to simpl were actually not needed, because reflexivity can perform some simplification automatically when checking that two sides are equal; simpl was just added so that we could see the intermediate state — after simplification but before finishing the proof. Here is a shorter proof of the theorem:
Theorem plus_O_n' : n : nat, 0 + n = n.
Proof.
intros n. reflexivity. Qed.
Moreover, it will be useful later to know that reflexivity 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, if reflexivity succeeds, the whole goal is finished and we don't need to look at whatever expanded expressions reflexivity has created by all this simplification and unfolding; by contrast, simpl is used in situations where we may have to read and understand the new goal that it creates, so we would not want it blindly expanding definitions and leaving the goal in a messy state.
The form of the theorem we just stated and its proof are almost exactly the same as the simpler examples we saw earlier; there are just a few differences.
First, we've used the keyword Theorem instead of Example. This difference is mostly a matter of style; the keywords Example and Theorem (and a few others, including Lemma, Fact, and Remark) mean pretty much the same thing to Coq.
Second, we've added the quantifier n:nat, so that our theorem talks about all natural numbers n. Informally, to prove theorems of this form, we generally start by saying "Suppose n is some number..." Formally, this is achieved in the proof by intros n, which moves n from the quantifier in the goal to a context of current assumptions.
The keywords intros, simpl, and reflexivity are examples of tactics. A tactic is a command that is used between Proof and Qed to guide the process of checking some claim we are making. We will see several more tactics in the rest of this chapter and yet more in future chapters.
Other similar theorems can be proved with the same pattern.
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."
It is worth stepping through these proofs to observe how the context and the goal change. You may want to add calls to simpl before reflexivity to see the simplifications that Coq performs on the terms before checking that they are equal.

# Proof by Rewriting

This theorem is a bit more interesting than the others we've seen:
Theorem plus_id_example : n m:nat,
n = m
n + n = m + m.
Instead of making a universal claim about all numbers n and m, it talks about a more specialized property that only holds when n = m. The arrow symbol is pronounced "implies." We've already defined a notion of implies on booleans impb; now we have a notion of implies on propositions.
The way a proof with implies works is: you have to prove what's to the right of the arrow... but you may assume what's to the left. That is, to show that n + n = m + m, we know (in our context) not only that n and m are natural numbers, but in fact it is the case that n = m.
As before, we need to be able to reason by assuming we are given such 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.
Proof.
(* move both quantifiers into the context: *)
intros n m.
(* move the hypothesis into the context: *)
intros H.
(* rewrite the goal using the hypothesis: *)
rewriteH.
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.) Here's a paper proof of the same theorem, with the tactics interwoven.
• Theorem: For any natural numbers n and m, if
n = m
then
n + n = m + m.
Proof: Let n and m be given (intros n m.) such that n = m (intros H). We must show:
n + n = m + m.
Since we've assumed that n = m, we can replace each reference to n on the left-hand side with a reference to m, and we have
m + m = m + m
immediately (reflexivity.). Qed.

#### Exercise: 1 star (plus_id_exercise)

Remove "Admitted." and fill in the proof.
Theorem plus_id_exercise : n m o : nat,
n = mm = on + m = m + o.
Proof.
(* FILL IN HERE *) Admitted.
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 lemmas that we believe will be useful for making some larger argument, use Admitted to accept them on faith for the moment, and continue working on the main 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 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. If the statement of the previously proved theorem involves quantified variables, as in the example below, Coq tries to instantiate them by matching with the current goal.
Theorem mult_0_plus : n m : nat,
(0 + n) * m = n * m.
Proof.
intros n m.
rewriteplus_O_n.
reflexivity. Qed.
Note that we've just rewritten by an existing theorem: the proof of mult_0_plus uses the proof of plus_O_n.
In an informal proof, we might use language by "by plus_0_n, we know...". For example:
• Theorem: For any natural number n and m,
(0 + n) * m = n * m.
Proof: Let n and m be given (intros n m.). We must show:
(0 + n) * m = n * m.
By plus_0_n, we know that 0 + n = n, so we can replace 0 + n on the left-hand side with just n, and we have
n * m = n * m
immediately (reflexivity.). Qed.

#### Exercise: 2 stars (mult_S_1)

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

(* (N.b. This proof can actually be completed without using rewrite,
but please do use rewrite for the sake of the exercise.) *)

# Proof by Case Analysis

Of course, not everything can be proved by simple calculation and rewriting: In general, unknown, hypothetical values (arbitrary numbers, booleans, lists, etc.) can block simplification. For example, if we try to prove the following fact using the simpl tactic as above, we get stuck. (We then use the Abort command to give up on it for the moment.)
Theorem plus_1_neq_0_firsttry : n : nat,
beq_nat (n + 1) 0 = false.
Proof.
intros n.
simpl. (* does nothing! *)
Abort.
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.
To make progress, we need 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. 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 |. In this case, the first component is empty, since the O constructor is nullary (it doesn't have any arguments). The second component gives a single name, n', since S is a unary constructor.
The - signs on the second and third lines are called bullets, and they mark the parts of the proof that correspond to each generated subgoal. The proof script that comes after a bullet is the entire proof for a subgoal. In this example, each of the subgoals is easily proved by a single use of reflexivity, which itself performs some simplification — e.g., the first one simplifies beq_nat (S n' + 1) 0 to false by first rewriting (S n' + 1) to S (n' + 1), then unfolding beq_nat, and then simplifying the match.
Marking cases with bullets is entirely optional: if bullets are not present, Coq simply asks you to prove each subgoal in sequence, one at a time. But it is a good idea to use bullets. For one thing, they make the structure of a proof apparent, making it more readable. Also, bullets instruct Coq to ensure that a subgoal is complete before trying to verify the next one, preventing proofs for different subgoals from getting mixed up. These issues become especially important in large developments, where fragile proofs lead to long debugging sessions.
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 bullets 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 also a good place to mention one other piece of somewhat obvious advice about line lengths. Beginning Coq users sometimes tend to the extremes, either writing each tactic on its own line or writing entire proofs on one line. Good style lies somewhere in the middle. One reasonable convention is to limit yourself to 80-character lines.
The destruct tactic can be used with any inductively defined datatype. For example, we use it next to prove that boolean negation is involutive — i.e., that negation is its own inverse.
• Theorem: For any natural number n,

beq_nat (n + 1) 0 = false.
Proof: Let n be given (intros n). We go by cases on n (destruct n as [| n']):
• First, suppose n=0. We must show that beq_nat (0 + 1) 0 = false, which we have by definition (reflexivity).
• Next, suppose n=S n'. We must show that beq_nat ((S n') + 1) 0 = false, which holds by the definition of beq_nat (reflexivity).
Qed.
Notice how the destruct lines up with what we say in the paper proof. We open with announcing our intention—-to do case analysis on n. While Coq wants to know the names of any possible subparts of n up front, in the as pattern of the destruct; in the paper proof, we find out the names when we write out which case we're in explicitly: each subcase is written as its own bullet, where we announce the case we're in.
We've encounted our first significant divergence (of many!) between how Coq and paper proofs work. Coq keeps track of the context for you, so Coq proof scripts can leave quite a bit implicit. Paper proofs are meant to communicate an idea from one human to another, so it's important to check in and make sure everyone is on the same page—-by, for example, saying what each case is up front.
Finally, note that we're not "transliterating" the Coq: the two uses of reflexivity in each branch are written slightly differently in the paper proof.
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. This is generally considered bad style, since Coq often makes confusing choices of names when left to its own devices. In a paper proof, it'd be good to announce that b=false in the first case and b=true in the second case.
It is sometimes useful to invoke destruct inside a subgoal, generating yet more proof obligations. In this case, we use different kinds of bullets to mark goals on different "levels." For example:
Theorem andb_commutative : b c, andb b c = andb c b.
Proof.
intros b c. destruct b.
- destruct c.
+ reflexivity.
+ reflexivity.
- destruct c.
+ reflexivity.
+ reflexivity.
Qed.
Each pair of calls to reflexivity corresponds to the subgoals that were generated after the execution of the destruct c line right above it.
Besides - and +, we can use * (asterisk) as a third kind of bullet. We can also enclose sub-proofs in curly braces, which is useful in case we ever encounter a proof that generates more than three levels of subgoals:
Theorem andb_commutative' : b c, andb b c = andb c b.
Proof.
intros b c. destruct b.
{ destruct c.
{ reflexivity. }
{ reflexivity. } }
{ destruct c.
{ reflexivity. }
{ reflexivity. } }
Qed.
Since curly braces mark both the beginning and the end of a proof, they can be used for multiple subgoal levels, as this example shows. Furthermore, curly braces allow us to reuse the same bullet shapes at multiple levels in a proof:
Theorem andb3_exchange :
b c d, andb (andb b c) d = andb (andb b d) c.
Proof.
intros b c d. destruct b.
- destruct c.
{ destruct d.
- reflexivity.
- reflexivity. }
{ destruct d.
- reflexivity.
- reflexivity. }
- destruct c.
{ destruct d.
- reflexivity.
- reflexivity. }
{ destruct d.
- reflexivity.
- reflexivity. }
Qed.
Before closing the chapter, let's mention one final convenience. As you may have noticed, many proofs perform case analysis on a variable right after introducing it:
intros x y. destruct y as [|y].
This pattern is so common that Coq provides a shorthand for it: we can perform case analysis on a variable when introducing it by using an intro pattern instead of a variable name. For instance, here is a shorter proof of the plus_1_neq_0 theorem above.
Theorem plus_1_neq_0' : n : nat,
beq_nat (n + 1) 0 = false.
Proof.
intros [|n].
- reflexivity.
- reflexivity. Qed.

#### Exercise: 1 star (minus_n_O)

Lemma minus_n_O : n : nat, n = n - 0.
Proof.
(* FILL IN HERE *) Admitted.
If there are no arguments to name, we can just write [].
Theorem andb_commutative'' :
b c, andb b c = andb c b.
Proof.
intros [] [].
- reflexivity.
- reflexivity.
- reflexivity.
- reflexivity.
Qed.

#### Exercise: 2 stars (andb_true_elim2)

Prove the following claim, marking cases (and subcases) with bullets when you use destruct.
Theorem andb_true_elim2 : b c : bool,
andb b c = truec = true.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars (nandb__negb_andb)

Theorem nandb__negb_andb : b1 b2 : bool,
nandb b1 b2 = negb (andb b1 b2).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars (nandb__negb_andb_paper)

(* Prove the previous theorem on paper, i.e., show that
[ b1 b2 : bool, nandb b1 b2 = negb (andb b1 b2). ]
Write your proof in a comment, following the format seen in
the paper proofs so far. Keep in mind that your paper proof can do
cases on more than one thing at once, or you can do cases inside
other cases.  *)

(* FILL IN HERE *)

#### Exercise: 2 stars (impb_spec)

Theorem impb_spec : b1 b2 : bool,
impb b1 b2 = orb (negb b1) b2.
Proof.
(* FILL IN HERE *) Admitted.

#### 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.

#### Exercise: 2 stars (zero_nbeq_plus_1_paper)

Prove the previous theorem on paper, i.e., show that
n : nat
beq_nat 0 (n + 1) = false.
Write your proof in a comment, following the format seen in the paper proofs so far.
(* FILL IN HERE *)

## More on Notation (Optional)

(In general, sections marked Optional are not needed to follow the rest of the book, except possibly other Optional sections. On a first reading, you might want to skim these sections so that you know what's there for future reference.)
Recall the notation definitions for infix plus and times, i.e., conventional mathematical operators which go between the things they're operating on:
Notation "x + y" := (plus x y)
(at level 50, left associativity)
: nat_scope.
Notation "x * y" := (mult x y)
(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 is specified by writing at level n; this helps Coq parse compound expressions. The associativity setting helps to disambiguate expressions containing multiple occurrences of the same symbol. For example, the parameters specified above for + and * say that the expression 1+2*3*4 is shorthand for (1+((2*3)*4)). Coq uses precedence levels from 0 to 100, and left, right, or no associativity. We will see more examples of this later, e.g., in the Lists chapter.
Each notation symbol is also associated with a notation scope. Coq tries to guess what scope is meant from context, so when it sees S(O*O) it guesses nat_scope, but when it sees the cartesian product (tuple) type bool*bool (which we'll see in later chapters) it guesses type_scope. Occasionally, it is necessary to help it out with percent-notation by writing (x*y)%nat, and sometimes in what Coq prints 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 (the natural number 0 that we're using in this chapter), or 0%Z, which means the Integer zero (which comes from a different part of the standard library).
Pro tip: Coq's notation mechanism is not especially powerful. Don't expect too much from it!

## Fixpoints and Structural Recursion (Optional)

Here is a copy of the definition of addition:
Fixpoint plus' (n : nat) (m : nat) : nat :=
match n with
| Om
| S n'S (plus' n' m)
end.
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 reject because of this restriction.
(* FILL IN HERE *)

# 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 : boolbool),
( (x : bool), f x = x) →
(b : bool), f (f b) = b.
Proof.
(* FILL IN HERE *) Admitted.
Now state and prove a theorem in Coq; call it negation_fn_applied_twice. It should be similar to the previous one but where the second hypothesis says that the function f has the property that f x = negb x.
(* FILL IN HERE *)

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

Prove the following theorem. (Hint: This one can be a bit tricky, depending on how you approach it. You will probably need both destruct and rewrite, but destructing everything in sight is not the best way.)
Theorem andb_eq_orb :
(b c : bool),
(andb b c = orb b c) →
b = c.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, optional (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 above,
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.)
One caveat: If you use O or S as constructor names in your definition, it will confuse the auto-grader script in BasicsTest.v. Please choose different names.
(b) Next, write an increment function incr for binary numbers, and a function bin_to_nat to convert binary numbers to unary numbers.
(c) Write five unit tests test_bin_incr1, test_bin_incr2, etc. for your increment and binary-to-unary functions. (A "unit test" in Coq is a specific Example that can be proved with just reflexivity, as we've done for several of our definitions.) 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.
(* FILL IN HERE *)