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Basics: Functional Programming in Coq

(* REMINDER:

          #####################################################
          ###  PLEASE DO NOT DISTRIBUTE SOLUTIONS PUBLICLY  ###
          #####################################################

   (See the [Preface] for why.)
*)

(* ################################################################# *)

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
  | 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.

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:

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.next_weekday (next_weekday saturday) = tuesday simpl.tuesday = tuesday 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

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.

You will also notice that each chapter (like Basics.v) is accompanied by a 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).

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

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

Inductive bool : Type :=
  | 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 (b1:bool) (b2:bool) : bool :=
  match b1 with
  | true => b2
  | false => false
  end.

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

The last two 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.orb true false = true
Proof.orb true false = true simpl.true = true reflexivity.  Qed.
Example test_orb2:  (orb false false) = false.orb false false = false
Proof.orb false false = false simpl.false = false reflexivity.  Qed.
Example test_orb3:  (orb false true)  = true.orb false true = true
Proof.orb false true = true simpl.true = true reflexivity.  Qed.
Example test_orb4:  (orb true  true)  = true.orb true true = true
Proof.orb true true = true simpl.true = true 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.false || false || true = true
Proof.false || false || true = true simpl.true = true 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, 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 (b1:bool) (b2:bool) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).b1, b2:bool
bool Admitted.

Example test_nandb1:               (nandb true false) = true.nandb true false = true
(* FILL IN HERE *) Admitted.
Example test_nandb2:               (nandb false false) = true.nandb false false = true
(* FILL IN HERE *) Admitted.
Example test_nandb3:               (nandb false true) = true.nandb false true = true
(* FILL IN HERE *) Admitted.
Example test_nandb4:               (nandb true true) = false.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 (b1:bool) (b2:bool) (b3:bool) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).b1, b2, b3:bool
bool Admitted.

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

☐

(* ================================================================= *)

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."

(* ================================================================= *)

New Types from Old

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
  | 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 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
  | 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.

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

A single constructor with multiple parameters can be used to create a tuple type. As an example, consider representing the four bits in a nybble (half a byte). We first define a datatype bit that resembles bool (using the constructors B0 and B1 for the two possible bit values), and then define the datatype nybble, which is essentially a tuple of four bits.

Inductive bit : Type :=
  | B0
  | B1.

Inductive nybble : Type :=
  | bits (b0 b1 b2 b3 : bit).

Check (bits B1 B0 B1 B0).bits B1 B0 B1 B0
     : 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 B0 B0 B0 B0) => true
    | (bits _ _ _ _) => false
  end.

Compute (all_zero (bits B1 B0 B1 B0)).= false
: bool

Compute (all_zero (bits B0 B0 B0 B0)).= true
: bool


(* ================================================================= *)

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

The types we have defined so far, "enumerated types" such as day, bool, and bit, and tuple types such as nybble built from them, share the property that each type has a finite set of values. The natural numbers are an infinite set, and we need to represent all of them in a datatype with a finite number of constructors. There are many representations of numbers to choose from. We are most familiar with decimal notation (base 10), using the digits 0 through 9, for example, to form the number 123. You may have encountered hexadecimal notation (base 16), in which the same number is represented as 7B, or octal (base 8), where it is 173, or binary (base 2), where it is 1111011. Using an enumerated type to represent digits, we could use any of these to represent natural numbers. There are circumstances where each of these choices can be useful.

Binary is valuable in computer hardware because it can in turn be represented with two voltage levels, resulting in simple circuitry. Analogously, we wish here to choose a representation that makes proofs simpler.

Indeed, there is a representation of numbers that is even simpler than binary, namely unary (base 1), in which only a single digit is used (as one might do while counting days in prison by scratching on the walls). To represent unary with a Coq datatype, we use two constructors. The capital-letter O constructor represents zero. When the S constructor is applied to the representation of the natural number 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).

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:

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
  | tick (foo : nat').

The interpretation of these marks comes from how we use them to compute.

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:

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

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

End 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

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

Check S.S
     : nat -> nat
Check pred.Nat.pred
     : nat -> nat
Check minustwo.minustwo
     : nat -> nat

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 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.)

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.

Fixpoint evenb (n:nat) : bool :=
  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.oddb 1 = true
Proof.oddb 1 = true simpl.oddb 1 = true reflexivity.  Qed.
Example test_oddb2:    oddb 4 = false.oddb 4 = false
Proof.oddb 4 = false simpl.oddb 4 = false 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.

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

Compute (plus 3 2).= 5
: nat

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

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

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

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

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

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_ ." *).factorial:nat -> nat
n:nat
nat Admitted.

Example test_factorial1:          (factorial 3) = 6.factorial 3 = 6
(* FILL IN HERE *) Admitted.
Example test_factorial2:          (factorial 5) = (mult 10 12).factorial 5 = 10 * 12
(* FILL IN HERE *) Admitted.

☐

Again, we can make numerical expressions 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).0 + 1 + 1
     : nat

(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.

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.leb 2 2 = true
Proof.leb 2 2 = true simpl.true = true reflexivity.  Qed.
Example test_leb2:             (leb 2 4) = true.leb 2 4 = true
Proof.leb 2 4 = true simpl.true = true reflexivity.  Qed.
Example test_leb3:             (leb 4 2) = false.leb 4 2 = false
Proof.leb 4 2 = false simpl.false = false 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.(4 <=? 2) = false
Proof.(4 <=? 2) = false simpl.false = false 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_ ." *).n, m:nat
bool Admitted.

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

Example test_ltb1:             (ltb 2 2) = false.(2 <? 2) = false
(* FILL IN HERE *) Admitted.
Example test_ltb2:             (ltb 2 4) = true.(2 <? 4) = true
(* FILL IN HERE *) Admitted.
Example test_ltb3:             (ltb 4 2) = false.(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 : forall n : nat, 0 + n = n.forall n : nat, 0 + n = n
Proof.forall n : nat, 0 + n = n
  intros n.n:nat
0 + n = n simpl.n:nat
n = n 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' : forall n : nat, 0 + n = n.forall n : nat, 0 + n = n
Proof.forall n : nat, 0 + n = n
  intros n.n:nat
0 + n = 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 : forall n:nat, 1 + n = S n.forall n : nat, 1 + n = S n
Proof.forall n : nat, 1 + n = S n
  intros n.n:nat
1 + n = S n reflexivity.  Qed.

Theorem mult_0_l : forall n:nat, 0 * n = 0.forall n : nat, 0 * n = 0
Proof.forall n : nat, 0 * n = 0
  intros n.n:nat
0 * n = 0 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 : forall n m:nat,
  n = m ->
  n + n = m + m.forall 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."

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.forall n m : nat, n = m -> n + n = m + m
  (* move both quantifiers into the context: *)
  intros n m.n, m:nat
n = m -> n + n = m + m
  (* move the hypothesis into the context: *)
  intros H.n, m:nat
H:n = m
n + n = m + m
  (* rewrite the goal using the hypothesis: *)
  rewrite -> H.n, m:nat
H:n = m
m + m = m + m
  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 : forall n m o : nat,
  n = m -> m = o -> n + m = m + o.forall n m o : nat, n = m -> m = o -> n + m = m + o
Proof.forall n m o : nat, n = m -> m = o -> n + m = m + o
  (* 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 : forall n m : nat,
  (0 + n) * m = n * m.forall n m : nat, (0 + n) * m = n * m
Proof.forall n m : nat, (0 + n) * m = n * m
  intros n m.n, m:nat
(0 + n) * m = n * m
  rewrite -> plus_O_n.n, m:nat
n * m = n * m
  reflexivity.  Qed.

Exercise: 2 stars, standard (mult_S_1)

Theorem mult_S_1 : forall n m : nat,
  m = S n ->
  m * (1 + n) = m * m.forall n m : nat, m = S n -> m * (1 + n) = m * m
Proof.forall n m : nat, m = S n -> m * (1 + n) = m * m
  (* 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

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 : forall n : nat,
  (n + 1) =? 0 = false.forall n : nat, (n + 1 =? 0) = false
Proof.forall n : nat, (n + 1 =? 0) = false
  intros n.n:nat
(n + 1 =? 0) = false
  simpl.  (* does nothing! *)n:nat
(n + 1 =? 0) = false
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 : forall n : nat,
  (n + 1) =? 0 = false.forall n : nat, (n + 1 =? 0) = false
Proof.forall n : nat, (n + 1 =? 0) = false
  intros n.n:nat
(n + 1 =? 0) = false destruct n as [| n'] eqn:E.n:nat
E:n = 0
(0 + 1 =? 0) = false
n, n':nat
E:n = S n'
(S n' + 1 =? 0) = false
  -n:nat
E:n = 0
(0 + 1 =? 0) = false reflexivity.
  -n, n':nat
E:n = S n'
(S n' + 1 =? 0) = false 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.

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 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.

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 negb_involutive : forall b : bool,
  negb (negb b) = b.forall b : bool, negb (negb b) = b
Proof.forall b : bool, negb (negb b) = b
  intros b.b:bool
negb (negb b) = b destruct b eqn:E.b:bool
E:b = true
negb (negb true) = true
b:bool
E:b = false
negb (negb false) = false
  -b:bool
E:b = true
negb (negb true) = true reflexivity.
  -b:bool
E:b = false
negb (negb false) = false 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.

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 : forall b c, andb b c = andb c b.forall b c : bool, b && c = c && b
Proof.forall b c : bool, b && c = c && b
  intros b c.b, c:bool
b && c = c && b destruct b eqn:Eb.b, c:bool
Eb:b = true
true && c = c && true
b, c:bool
Eb:b = false
false && c = c && false
  -b, c:bool
Eb:b = true
true && c = c && true destruct c eqn:Ec.b, c:bool
Eb:b = true
Ec:c = true
true && true = true && true
b, c:bool
Eb:b = true
Ec:c = false
true && false = false && true
    +b, c:bool
Eb:b = true
Ec:c = true
true && true = true && true reflexivity.
    +b, c:bool
Eb:b = true
Ec:c = false
true && false = false && true reflexivity.
  -b, c:bool
Eb:b = false
false && c = c && false destruct c eqn:Ec.b, c:bool
Eb:b = false
Ec:c = true
false && true = true && false
b, c:bool
Eb:b = false
Ec:c = false
false && false = false && false
    +b, c:bool
Eb:b = false
Ec:c = true
false && true = true && false reflexivity.
    +b, c:bool
Eb:b = false
Ec:c = false
false && false = false && false 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' : forall b c, andb b c = andb c b.forall b c : bool, b && c = c && b
Proof.forall b c : bool, b && c = c && b
  intros b c.b, c:bool
b && c = c && b destruct b eqn:Eb.b, c:bool
Eb:b = true
true && c = c && true
b, c:bool
Eb:b = false
false && c = c && false
  {b, c:bool
Eb:b = true
true && c = c && true destruct c eqn:Ec.b, c:bool
Eb:b = true
Ec:c = true
true && true = true && true
b, c:bool
Eb:b = true
Ec:c = false
true && false = false && true
    {b, c:bool
Eb:b = true
Ec:c = true
true && true = true && true reflexivity. }b, c:bool
Eb:b = true
Ec:c = false
true && false = false && true
    {b, c:bool
Eb:b = true
Ec:c = false
true && false = false && true reflexivity. } }b, c:bool
Eb:b = false
false && c = c && false
  {b, c:bool
Eb:b = false
false && c = c && false destruct c eqn:Ec.b, c:bool
Eb:b = false
Ec:c = true
false && true = true && false
b, c:bool
Eb:b = false
Ec:c = false
false && false = false && false
    {b, c:bool
Eb:b = false
Ec:c = true
false && true = true && false reflexivity. }b, c:bool
Eb:b = false
Ec:c = false
false && false = false && false
    {b, c:bool
Eb:b = false
Ec:c = false
false && false = false && false 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 :
  forall b c d, andb (andb b c) d = andb (andb b d) c.forall b c d : bool, b && c && d = b && d && c
Proof.forall b c d : bool, b && c && d = b && d && c
  intros b c d.b, c, d:bool
b && c && d = b && d && c destruct b eqn:Eb.b, c, d:bool
Eb:b = true
true && c && d = true && d && c
b, c, d:bool
Eb:b = false
false && c && d = false && d && c
  -b, c, d:bool
Eb:b = true
true && c && d = true && d && c destruct c eqn:Ec.b, c, d:bool
Eb:b = true
Ec:c = true
true && true && d = true && d && true
b, c, d:bool
Eb:b = true
Ec:c = false
true && false && d = true && d && false
    {b, c, d:bool
Eb:b = true
Ec:c = true
true && true && d = true && d && true destruct d eqn:Ed.b, c, d:bool
Eb:b = true
Ec:c = true
Ed:d = true
true && true && true = true && true && true
b, c, d:bool
Eb:b = true
Ec:c = true
Ed:d = false
true && true && false = true && false && true
      -b, c, d:bool
Eb:b = true
Ec:c = true
Ed:d = true
true && true && true = true && true && true reflexivity.
      -b, c, d:bool
Eb:b = true
Ec:c = true
Ed:d = false
true && true && false = true && false && true reflexivity. }b, c, d:bool
Eb:b = true
Ec:c = false
true && false && d = true && d && false
    {b, c, d:bool
Eb:b = true
Ec:c = false
true && false && d = true && d && false destruct d eqn:Ed.b, c, d:bool
Eb:b = true
Ec:c = false
Ed:d = true
true && false && true = true && true && false
b, c, d:bool
Eb:b = true
Ec:c = false
Ed:d = false
true && false && false = true && false && false
      -b, c, d:bool
Eb:b = true
Ec:c = false
Ed:d = true
true && false && true = true && true && false reflexivity.
      -b, c, d:bool
Eb:b = true
Ec:c = false
Ed:d = false
true && false && false = true && false && false reflexivity. }
  -b, c, d:bool
Eb:b = false
false && c && d = false && d && c destruct c eqn:Ec.b, c, d:bool
Eb:b = false
Ec:c = true
false && true && d = false && d && true
b, c, d:bool
Eb:b = false
Ec:c = false
false && false && d = false && d && false
    {b, c, d:bool
Eb:b = false
Ec:c = true
false && true && d = false && d && true destruct d eqn:Ed.b, c, d:bool
Eb:b = false
Ec:c = true
Ed:d = true
false && true && true = false && true && true
b, c, d:bool
Eb:b = false
Ec:c = true
Ed:d = false
false && true && false = false && false && true
      -b, c, d:bool
Eb:b = false
Ec:c = true
Ed:d = true
false && true && true = false && true && true reflexivity.
      -b, c, d:bool
Eb:b = false
Ec:c = true
Ed:d = false
false && true && false = false && false && true reflexivity. }b, c, d:bool
Eb:b = false
Ec:c = false
false && false && d = false && d && false
    {b, c, d:bool
Eb:b = false
Ec:c = false
false && false && d = false && d && false destruct d eqn:Ed.b, c, d:bool
Eb:b = false
Ec:c = false
Ed:d = true
false && false && true = false && true && false
b, c, d:bool
Eb:b = false
Ec:c = false
Ed:d = false
false && false && false = false && false && false
      -b, c, d:bool
Eb:b = false
Ec:c = false
Ed:d = true
false && false && true = false && true && false reflexivity.
      -b, c, d:bool
Eb:b = false
Ec:c = false
Ed:d = false
false && false && false = false && false && false 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' : forall n : nat,
  (n + 1) =? 0 = false.forall n : nat, (n + 1 =? 0) = false
Proof.forall n : nat, (n + 1 =? 0) = false
  intros [|n].(0 + 1 =? 0) = false
n:nat
(S n + 1 =? 0) = false
  -(0 + 1 =? 0) = false reflexivity.
  -n:nat
(S n + 1 =? 0) = false reflexivity.  Qed.

If there are no arguments to name, we can just write [].

Theorem andb_commutative'' :
  forall b c, andb b c = andb c b.forall b c : bool, b && c = c && b
Proof.forall b c : bool, b && c = c && b
  intros [] [].true && true = true && true
true && false = false && true
false && true = true && false
false && false = false && false
  -true && true = true && true reflexivity.
  -true && false = false && true reflexivity.
  -false && true = true && false reflexivity.
  -false && false = false && false 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 : forall b c : bool,
  andb b c = true -> c = true.forall b c : bool, b && c = true -> c = true
Proof.forall b c : bool, b && c = true -> c = true
  (* FILL IN HERE *) Admitted.

☐

Exercise: 1 star, standard (zero_nbeq_plus_1)

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

☐

(* ================================================================= *)

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:

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
  | 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 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, 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

Each SF chapter comes with a tester file (e.g. BasicsTest.v), containing scripts that check most of the exercises. You can run make BasicsTest.vo in a terminal and check its output to make sure you didn't miss anything.

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 :
  forall (f : bool -> bool),
  (forall (x : bool), f x = x) ->
  forall (b : bool), f (f b) = b.forall f : bool -> bool,
(forall x : bool, f x = x) ->
forall b : bool, f (f b) = b
Proof.forall f : bool -> bool,
(forall x : bool, f x = x) ->
forall b : bool, f (f b) = b
  (* 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.

☐

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 :
  forall (b c : bool),
  (andb b c = orb b c) ->
  b = c.forall b c : bool, b && c = b || c -> b = c
Proof.forall b c : bool, b && c = b || c -> b = c
  (* FILL IN HERE *) Admitted.

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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.

For example:

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.

Inductive bin : Type :=
  | Z
  | A (n : bin)
  | B (n : bin).

(a) 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_ ." *).incr:bin -> bin
m:bin
bin Admitted.

Fixpoint bin_to_nat (m:bin) : nat
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).bin_to_nat:bin -> nat
m:bin
nat Admitted.

(b) 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 *)

(* Do not modify the following line: *)
Definition manual_grade_for_binary : option (nat*string) := None.

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(* Wed Jan 9 12:02:44 EST 2019 *)