# BasicsFunctional Programming in Coq

# Introduction

*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.

*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.

*Gallina*. The second half introduces some basic

*tactics*that can be used to prove properties of Coq programs.

## Enumerated Types

*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.

## Days of the Week

*type*.

Inductive day : Type :=

| monday

| tuesday

| wednesday

| thursday

| friday

| saturday

| sunday.

| monday

| tuesday

| wednesday

| thursday

| friday

| saturday

| sunday.

The type is called day, and its members are monday,
tuesday, etc.
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.

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 figure out these types for
itself when they are not given explicitly — i.e., it can do
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.

*type inference*— but we'll generally include them to make reading easier.
Compute (next_weekday friday).

(* ==> monday : day *)

Compute (next_weekday (next_weekday saturday)).

(* ==> tuesday : day *)

(* ==> 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:
Example test_next_weekday:

(next_weekday (next_weekday saturday)) = tuesday.

(next_weekday (next_weekday saturday)) = tuesday.

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

Proof. simpl. reflexivity. Qed.

The details are not important for now (we'll come back to
them in a bit), but essentially this can be read as "The assertion
we've just made can be proved by observing that both sides of the
equality evaluate to the same thing, after some simplification."
Third, we can ask Coq to

*extract*, from 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'll come back to this topic in later chapters.## Homework Submission Guidelines

*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.

*test script*(BasicsTest.v) that automatically calculates points for the finished homework problems in the chapter. These scripts are mostly for the auto-grading infrastructure that your instructor may use to help process assignments, but you may also like to use them to double-check that your file is well formatted before handing it in. In a terminal window either type make BasicsTest.vo or do the following:

coqc -Q . LF Basics.v

coqc -Q . LF BasicsTest.v

There is no need to hand in BasicsTest.v itself (or Preface.v).
coqc -Q . LF BasicsTest.v

*If your class is using the Canvas system to hand in assignments*:

- If you submit multiple versions of the assignment, you may notice that they are given different names. This is fine: The most recent submission is the one that will be graded.
- To hand in multiple files at the same time (if more than one chapter is assigned in the same week), you need to make a single submission with all the files at once using the button "Add another file" just above the comment box.

## Booleans

Inductive bool : Type :=

| true

| false.

| true

| false.

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:

Definition negb (b:bool) : bool :=

match b with

| true ⇒ false

| false ⇒ true

end.

Definition andb (b

match b

| true ⇒ b

| false ⇒ false

end.

Definition orb (b

match b

| true ⇒ true

| false ⇒ b

end.

match b with

| true ⇒ false

| false ⇒ true

end.

Definition andb (b

_{1}:bool) (b_{2}:bool) : bool :=match b

_{1}with| true ⇒ b

_{2}| false ⇒ false

end.

Definition orb (b

_{1}:bool) (b_{2}:bool) : bool :=match b

_{1}with| true ⇒ true

| false ⇒ b

_{2}end.

The last two of these 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.

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.

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.

#### Exercise: 1 star, standard (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. (I.e., 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 (b

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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.

☐
_{1}:bool) (b_{2}:bool) : bool(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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, standard (andb3)

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

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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.

☐
_{1}:bool) (b_{2}:bool) (b_{3}:bool) : bool(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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.

## Types

Check true.

(* ===> true : bool *)

Check (negb true).

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

(* ===> 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 *)

(* ===> 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."

## New Types from Old

Inductive rgb : Type :=

| red

| green

| blue.

Inductive color : Type :=

| black

| white

| primary (p : rgb).

| red

| green

| blue.

Inductive color : Type :=

| black

| white

| primary (p : rgb).

Let's look at this in a little more detail.
Every inductively defined type (day, bool, rgb, color,
etc.) contains a set of
We can define functions on colors using pattern matching just as
we have done for day and bool.

*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.

Definition monochrome (c : color) : bool :=

match c with

| black ⇒ true

| white ⇒ true

| primary q ⇒ false

end.

match c with

| black ⇒ true

| white ⇒ true

| primary q ⇒ false

end.

Since the primary constructor takes an argument, a pattern
matching primary should include either a variable (as above —
note that we can choose its name freely) or a constant of
appropriate type (as below).

Definition isred (c : color) : bool :=

match c with

| black ⇒ false

| white ⇒ false

| primary red ⇒ true

| primary _ ⇒ false

end.

match c with

| black ⇒ false

| white ⇒ false

| primary red ⇒ true

| primary _ ⇒ false

end.

The pattern primary _ here is shorthand for "primary applied
to any rgb constructor except red." (The wildcard pattern _
has the same effect as the dummy pattern variable p in the
definition of monochrome.)

## Tuples

_{0}and B

_{1}for the two possible bit values), and then define the datatype nybble, which is essentially a tuple of four bits.

Inductive bit : Type :=

| B

| B

Inductive nybble : Type :=

| bits (b

Check (bits B

(* ==> bits B

| B

_{0}| B

_{1}.Inductive nybble : Type :=

| bits (b

_{0}b_{1}b_{2}b_{3}: bit).Check (bits B

_{1}B_{0}B_{1}B_{0}).(* ==> bits B

_{1}B_{0}B_{1}B_{0}: nybble *)
The bits constructor acts as a wrapper for its contents.
Unwrapping can be done by pattern-matching, as in the all_zero
function which tests a nybble to see if all its bits are O.
Note that we are using underscore (_) as a

*wildcard pattern*to avoid inventing variable names that will not be used.
Definition all_zero (nb : nybble) : bool :=

match nb with

| (bits B

| (bits _ _ _ _) ⇒ false

end.

Compute (all_zero (bits B

(* ===> false : bool *)

Compute (all_zero (bits B

(* ===> true : bool *)

match nb with

| (bits B

_{0}B_{0}B_{0}B_{0}) ⇒ true| (bits _ _ _ _) ⇒ false

end.

Compute (all_zero (bits B

_{1}B_{0}B_{1}B_{0})).(* ===> false : bool *)

Compute (all_zero (bits B

_{0}B_{0}B_{0}B_{0})).(* ===> true : bool *)

## Modules

*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

*proofs*simpler.

*n*, the result is the representation of

*n+1*. (S stands for "successor", or "scratch" if one is in prison.) Here is the complete datatype definition.

Inductive nat : Type :=

| O

| S (n : nat).

| O

| S (n : nat).

With this definition, 0 is represented by O, 1 by S O,
2 by S (S O), and so on.
The clauses of this definition can be read:
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:
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

- 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.

- 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.

*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

| tick (foo : nat').

| stop

| tick (foo : nat').

The
We can do this 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:

*interpretation*of these marks comes from how we use them to compute.
Definition pred (n : nat) : nat :=

match n with

| O ⇒ O

| S n' ⇒ n'

end.

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 NatPlayground.

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 decimal numerals can be used as an alternative to
the "unary" notation defined by the constructors S and O. Coq
prints numbers in decimal form by default:

Check (S (S (S (S O)))).

(* ===> 4 : nat *)

Definition minustwo (n : nat) : nat :=

match n with

| O ⇒ O

| S O ⇒ O

| S (S n') ⇒ n'

end.

Compute (minustwo 4).

(* ===> 2 : nat *)

(* ===> 4 : nat *)

Definition minustwo (n : nat) : nat :=

match n with

| O ⇒ O

| S O ⇒ O

| 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.

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
For most function definitions over numbers, just 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.

*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 decimal 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.)
Fixpoint evenb (n:nat) : bool :=

match n with

| O ⇒ true

| S O ⇒ false

| S (S n') ⇒ evenb n'

end.

match n with

| O ⇒ true

| S O ⇒ false

| S (S n') ⇒ evenb n'

end.

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.

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

| O ⇒ m

| S n' ⇒ S (plus n' m)

end.

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.

Compute (plus 3 2).

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

*)

==> 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.

Example test_mult1: (mult 3 3) = 9.

Proof. simpl. reflexivity. Qed.

match n with

| O ⇒ O

| 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.

End NatPlayground2.

Fixpoint exp (base power : nat) : nat :=

match power with

| O ⇒ S O

| S p ⇒ mult base (exp base p)

end.

match n, m with

| O , _ ⇒ O

| S _ , O ⇒ n

| S n', S m' ⇒ minus n' m'

end.

End NatPlayground2.

Fixpoint exp (base power : nat) : nat :=

match power with

| O ⇒ S O

| S p ⇒ mult base (exp base p)

end.

#### Exercise: 1 star, standard (factorial)

Recall the standard mathematical factorial function:factorial(0) = 1 factorial(n) = n * factorial(n-1) (if n>0)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.

☐
(* 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.

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).

(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 "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 is a user-defined operation!
Here is a function eqb, 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 eqb (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' ⇒ eqb n' m'

end

end.

match n with

| O ⇒ match m with

| O ⇒ true

| S m' ⇒ false

end

| S n' ⇒ match m with

| O ⇒ false

| S m' ⇒ eqb n' m'

end

end.

Similarly, 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

| O ⇒ true

| S n' ⇒

match m with

| O ⇒ false

| 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.

match n with

| O ⇒ true

| S n' ⇒

match m with

| O ⇒ false

| 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.

Since we'll be using these (especially eqb) a lot, let's give
them infix notations.

Notation "x =? y" := (eqb x y) (at level 70) : nat_scope.

Notation "x <=? y" := (leb x y) (at level 70) : nat_scope.

Example test_leb3': (4 <=? 2) = false.

Proof. simpl. reflexivity. Qed.

Notation "x <=? y" := (leb x y) (at level 70) : nat_scope.

Example test_leb3': (4 <=? 2) = false.

Proof. simpl. reflexivity. Qed.

#### Exercise: 1 star, standard (ltb)

The ltb 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. (It can be done with just one previously defined function, but you can use two if you need to.)
Definition ltb (n m : nat) : bool

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.

Example test_ltb1: (ltb 2 2) = false.

(* FILL IN HERE *) Admitted.

Example test_ltb2: (ltb 2 4) = true.

(* FILL IN HERE *) Admitted.

Example test_ltb3: (ltb 4 2) = false.

(* FILL IN HERE *) Admitted.

☐
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.

Example test_ltb1: (ltb 2 2) = false.

(* FILL IN HERE *) Admitted.

Example test_ltb2: (ltb 2 4) = true.

(* FILL IN HERE *) Admitted.

Example test_ltb3: (ltb 4 2) = false.

(* FILL IN HERE *) Admitted.

# Proof by Simplification

Theorem plus_O_n : ∀n : nat, 0 + n = n.

Proof.

intros n. simpl. reflexivity. Qed.

Proof.

intros n. simpl. reflexivity. Qed.

(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.

Proof.

intros n. reflexivity. Qed.

Moreover, it will be useful later to know that reflexivity
does somewhat
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
The keywords intros, simpl, and reflexivity are examples of
Other similar theorems can be proved with the same pattern.

*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.*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.*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.
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.

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.

Theorem plus_id_example : ∀n m:nat,

n = m →

n + n = m + m.

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."
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: *)

rewrite → H.

reflexivity. Qed.

(* move both quantifiers into the context: *)

intros n m.

(* move the hypothesis into the context: *)

intros H.

(* rewrite the goal using the hypothesis: *)

rewrite → H.

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.)

#### Exercise: 1 star, standard (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.

☐
n = m → m = o → n + m = m + o.

Proof.

(* FILL IN HERE *) Admitted.

Theorem mult_0_plus : ∀n m : nat,

(0 + n) * m = n * m.

Proof.

intros n m.

rewrite → plus_O_n.

reflexivity. Qed.

(0 + n) * m = n * m.

Proof.

intros n m.

rewrite → plus_O_n.

reflexivity. Qed.

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 with tactics other than

rewrite, but please do use rewrite for the sake of the exercise.) *)

☐
m = S n →

m * (1 + n) = m * m.

Proof.

(* FILL IN HERE *) Admitted.

(* (N.b. This proof can actually be completed with tactics other than

rewrite, but please do use rewrite for the sake of the exercise.) *)

# Proof by Case Analysis

Theorem plus_1_neq_0_firsttry : ∀n : nat,

(n + 1) =? 0 = false.

Proof.

intros n.

simpl. (* does nothing! *)

Abort.

(n + 1) =? 0 = false.

Proof.

intros n.

simpl. (* does nothing! *)

Abort.

The reason for this is that the definitions of both
eqb 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 eqb 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 (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, (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,

(n + 1) =? 0 = false.

Proof.

intros n. destruct n as [| n'] eqn:E.

- reflexivity.

- reflexivity. Qed.

(n + 1) =? 0 = false.

Proof.

intros n. destruct n as [| n'] eqn:E.

- reflexivity.

- reflexivity. Qed.

The destruct generates
The annotation "as [| n']" is called an
In each subgoal, Coq remembers the assumption about n that is
relevant for this subgoal — either n = 0 or n = S n' for some
n'. The eqn:E annotation tells destruct to give the name E to
this equation. (Leaving off the eqn:E annotation causes Coq to
elide these assumptions in the subgoals. This slightly
streamlines proofs where the assumptions are not explicitly used,
but it is better practice to keep them for the sake of
documentation, as they can help keep you oriented when working
with the subgoals.)
The - signs on the second and third lines are called
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.

*two*subgoals, which we must then prove, separately, in order to get Coq to accept the theorem.*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.*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 second one simplifies (S n' + 1) =? 0 to false by first rewriting (S n' + 1) to S (n' + 1), then unfolding eqb, and then simplifying the match.
Theorem negb_involutive : ∀b : bool,

negb (negb b) = b.

Proof.

intros b. destruct b eqn:E.

- reflexivity.

- reflexivity. Qed.

negb (negb b) = b.

Proof.

intros b. destruct b eqn:E.

- 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
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:

*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.
Theorem andb_commutative : ∀b c, andb b c = andb c b.

Proof.

intros b c. destruct b eqn:Eb.

- destruct c eqn:Ec.

+ reflexivity.

+ reflexivity.

- destruct c eqn:Ec.

+ reflexivity.

+ reflexivity.

Qed.

Proof.

intros b c. destruct b eqn:Eb.

- destruct c eqn:Ec.

+ reflexivity.

+ reflexivity.

- destruct c eqn:Ec.

+ 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 eqn:Eb.

{ destruct c eqn:Ec.

{ reflexivity. }

{ reflexivity. } }

{ destruct c eqn:Ec.

{ reflexivity. }

{ reflexivity. } }

Qed.

Proof.

intros b c. destruct b eqn:Eb.

{ destruct c eqn:Ec.

{ reflexivity. }

{ reflexivity. } }

{ destruct c eqn:Ec.

{ 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 eqn:Eb.

- destruct c eqn:Ec.

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

- destruct c eqn:Ec.

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

Qed.

∀b c d, andb (andb b c) d = andb (andb b d) c.

Proof.

intros b c d. destruct b eqn:Eb.

- destruct c eqn:Ec.

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

- destruct c eqn:Ec.

{ destruct d eqn:Ed.

- reflexivity.

- reflexivity. }

{ destruct d eqn:Ed.

- 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] eqn:E.

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. (You'll also note one downside of this shorthand: we lose
the equation recording the assumption we are making in each
subgoal, which we previously got from the eqn:E annotation.)
Theorem plus_1_neq_0' : ∀n : nat,

(n + 1) =? 0 = false.

Proof.

intros [|n].

- reflexivity.

- reflexivity. Qed.

(n + 1) =? 0 = false.

Proof.

intros [|n].

- reflexivity.

- reflexivity. Qed.

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.

∀b c, andb b c = andb c b.

Proof.

intros [] [].

- reflexivity.

- reflexivity.

- reflexivity.

- reflexivity.

Qed.

#### Exercise: 2 stars, standard (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 = true → c = true.

Proof.

(* FILL IN HERE *) Admitted.

☐
andb b c = true → c = true.

Proof.

(* FILL IN HERE *) Admitted.

Theorem zero_nbeq_plus_1 : ∀n : nat,

0 =? (n + 1) = false.

Proof.

(* FILL IN HERE *) Admitted.

☐
0 =? (n + 1) = false.

Proof.

(* FILL IN HERE *) Admitted.

## More on Notation (Optional)

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.

(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
Each notation symbol is also associated with a
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!

*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.*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.
Fixpoint plus' (n : nat) (m : nat) : nat :=

match n with

| O ⇒ m

| S n' ⇒ S (plus' n' m)

end.

match n with

| O ⇒ m

| 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
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.

*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."#### Exercise: 2 stars, standard, 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. (If you choose to turn in this optional exercise as part of a homework assignment, make sure you comment out your solution so that it doesn't cause Coq to reject the whole file!)
(* FILL IN HERE *)

☐
# More Exercises

#### Exercise: 1 star, standard (indentity_fn_applied_twice)

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

∀(f : bool → bool),

(∀(x : bool), f x = x) →

∀(b : bool), f (f b) = b.

Proof.

(* FILL IN HERE *) Admitted.

☐
∀(f : bool → bool),

(∀(x : bool), f x = x) →

∀(b : bool), f (f b) = b.

Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 1 star, standard (negation_fn_applied_twice)

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

(* The Import statement on the next line tells Coq to use the

standard library String module. We'll use strings more in later

chapters, but for the moment we just need syntax for literal

strings for the grader comments. *)

From Coq Require Export String.

(* Do not modify the following line: *)

Definition manual_grade_for_negation_fn_applied_twice : option (nat*string) := None.

☐
(* The Import statement on the next line tells Coq to use the

standard library String module. We'll use strings more in later

chapters, but for the moment we just need syntax for literal

strings for the grader comments. *)

From Coq Require Export String.

(* Do not modify the following line: *)

Definition manual_grade_for_negation_fn_applied_twice : option (nat*string) := None.

#### Exercise: 3 stars, standard, 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.

☐
∀(b c : bool),

(andb b c = orb b c) →

b = c.

Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, standard (binary)

We can generalize our unary representation of natural numbers to the more efficient binary representation by treating a binary number as a sequence of constructors A and B (representing 0s and 1s), terminated by a Z. For comparison, in the unary representation, a number is a sequence of Ss terminated by an O.
decimal binary unary

0 Z O

1 B Z S O

2 A (B Z) S (S O)

3 B (B Z) S (S (S O))

4 A (A (B Z)) S (S (S (S O)))

5 B (A (B Z)) S (S (S (S (S O))))

6 A (B (B Z)) S (S (S (S (S (S O)))))

7 B (B (B Z)) S (S (S (S (S (S (S O))))))

8 A (A (A (B Z))) S (S (S (S (S (S (S (S O)))))))

Note that the low-order bit is on the left and the high-order bit
is on the right — the opposite of the way binary numbers are
usually written. This choice makes them easier to manipulate.
0 Z O

1 B Z S O

2 A (B Z) S (S O)

3 B (B Z) S (S (S O))

4 A (A (B Z)) S (S (S (S O)))

5 B (A (B Z)) S (S (S (S (S O))))

6 A (B (B Z)) S (S (S (S (S (S O)))))

7 B (B (B Z)) S (S (S (S (S (S (S O))))))

8 A (A (A (B Z))) S (S (S (S (S (S (S (S O)))))))

Inductive bin : Type :=

| Z

| A (n : bin)

| B (n : bin).

| Z

| A (n : bin)

| B (n : bin).

Complete the definitions below of an increment function incr
for binary numbers, and a function bin_to_nat to convert
binary numbers to unary numbers.

Fixpoint incr (m:bin) : bin

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Fixpoint bin_to_nat (m:bin) : nat

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Fixpoint bin_to_nat (m:bin) : nat

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

The following "unit tests" of your increment and binary-to-unary
functions should pass after you have defined those functions correctly.
Of course, unit tests don't fully demonstrate the correctness of
your functions! We'll return to that thought at the end of the
next chapter.

Example test_bin_incr1 : (incr (B Z)) = A (B Z).

(* FILL IN HERE *) Admitted.

Example test_bin_incr2 : (incr (A (B Z))) = B (B Z).

(* FILL IN HERE *) Admitted.

Example test_bin_incr3 : (incr (B (B Z))) = A (A (B Z)).

(* FILL IN HERE *) Admitted.

Example test_bin_incr4 : bin_to_nat (A (B Z)) = 2.

(* FILL IN HERE *) Admitted.

Example test_bin_incr5 :

bin_to_nat (incr (B Z)) = 1 + bin_to_nat (B Z).

(* FILL IN HERE *) Admitted.

Example test_bin_incr6 :

bin_to_nat (incr (incr (B Z))) = 2 + bin_to_nat (B Z).

(* FILL IN HERE *) Admitted.

☐
(* FILL IN HERE *) Admitted.

Example test_bin_incr2 : (incr (A (B Z))) = B (B Z).

(* FILL IN HERE *) Admitted.

Example test_bin_incr3 : (incr (B (B Z))) = A (A (B Z)).

(* FILL IN HERE *) Admitted.

Example test_bin_incr4 : bin_to_nat (A (B Z)) = 2.

(* FILL IN HERE *) Admitted.

Example test_bin_incr5 :

bin_to_nat (incr (B Z)) = 1 + bin_to_nat (B Z).

(* FILL IN HERE *) Admitted.

Example test_bin_incr6 :

bin_to_nat (incr (incr (B Z))) = 2 + bin_to_nat (B Z).

(* FILL IN HERE *) Admitted.

(* Mon Mar 25 14:36:23 EDT 2019 *)