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Michael Zhang 3432fc448c
complete basics
2020-06-04 00:02:38 -05:00

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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.
(** 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 *)
(** 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.
Proof. simpl. reflexivity. Qed.
Example test_orb2: (orb false false) = false.
Proof. simpl. reflexivity. Qed.
Example test_orb3: (orb false true) = true.
Proof. simpl. reflexivity. Qed.
Example test_orb4: (orb true true) = true.
Proof. simpl. reflexivity. Qed.
(** We can also introduce some familiar syntax for the boolean
operations we have just defined. The [Notation] command defines a new
symbolic notation for an existing definition. *)
Notation "x && y" := (andb x y).
Notation "x || y" := (orb x y).
Example test_orb5: false || false || true = true.
Proof. simpl. reflexivity. Qed.
(** _A note on notation_: In [.v] files, we use square brackets
to delimit fragments of Coq code within comments; this convention,
also used by the [coqdoc] documentation tool, keeps them visually
separate from the surrounding text. In the HTML version of the
files, these pieces of text appear in a [different font].
The command [Admitted] can be used as a placeholder for an
incomplete proof. We'll use it in exercises, to indicate the
parts that we're leaving for you -- i.e., your job is to replace
[Admitted]s 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 :=
match b1, b2 with
| true, true => false
| _, _ => true
end.
Example test_nandb1: (nandb true false) = true.
Proof. reflexivity. Qed.
Example test_nandb2: (nandb false false) = true.
Proof. reflexivity. Qed.
Example test_nandb3: (nandb false true) = true.
Proof. reflexivity. Qed.
Example test_nandb4: (nandb true true) = false.
Proof. reflexivity. Qed.
(** [] *)
(** **** 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 :=
match b1, b2, b3 with
| true, true, true => true
| _, _, _ => false
end.
Example test_andb31: (andb3 true true true) = true.
Proof. reflexivity. Qed.
Example test_andb32: (andb3 false true true) = false.
Proof. reflexivity. Qed.
Example test_andb33: (andb3 true false true) = false.
Proof. reflexivity. Qed.
Example test_andb34: (andb3 true true false) = false.
Proof. reflexivity. Qed.
(** [] *)
(* ================================================================= *)
(** ** 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.
Check pred.
Check minustwo.
(** These are all things that can be applied to a number to yield a
number. However, there is a fundamental difference between the
first one and the other two: functions like [pred] and [minustwo]
come with _computation rules_ -- e.g., the definition of [pred]
says that [pred 2] can be simplified to [1] -- while the
definition of [S] has no such behavior attached. Although it is
like a function in the sense that it can be applied to an
argument, it does not _do_ anything at all! It is just a way of
writing down numbers. (Think about standard 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.
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.
(** 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]
*)
(** 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.
(** 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 :=
match n with
| 0 => 1
| S n => (S n) * factorial (n)
end.
Example test_factorial1: (factorial 3) = 6.
Proof. reflexivity. Qed.
Example test_factorial2: (factorial 5) = (mult 10 12).
Proof. reflexivity. Qed.
(** [] *)
(** 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).
(** (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
[eq]uality, yielding a [b]oolean. Note the use of nested
[match]es (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.
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.
(** **** Exercise: 1 star, standard (ltb)
The [ltb] function tests natural numbers for [l]ess-[t]han,
yielding a [b]oolean. 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 :=
1 <=? m - n.
Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.
Example test_ltb1: (ltb 2 2) = false.
Proof. reflexivity. Qed.
Example test_ltb2: (ltb 2 4) = true.
Proof. reflexivity. Qed.
Example test_ltb3: (ltb 4 2) = false.
Proof. reflexivity. Qed.
(** [] *)
(* ################################################################# *)
(** * 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.
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
[forall] 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.
Proof.
intros n. reflexivity. Qed.
(** Moreover, it will be useful later to know that [reflexivity]
does somewhat _more_ simplification than [simpl] does -- for
example, it tries "unfolding" defined terms, replacing them with
their right-hand sides. The reason for this difference is that,
if reflexivity succeeds, the whole goal is finished and we don't
need to look at whatever expanded expressions [reflexivity] has
created by all this simplification and unfolding; by contrast,
[simpl] is used in situations where we may have to read and
understand the new goal that it creates, so we would not want it
blindly expanding definitions and leaving the goal in a messy
state.
The form of the theorem we just stated and its proof are almost
exactly the same as the simpler examples we saw earlier; there are
just a few differences.
First, we've used the keyword [Theorem] instead of [Example].
This difference is mostly a matter of style; the keywords
[Example] and [Theorem] (and a few others, including [Lemma],
[Fact], and [Remark]) mean pretty much the same thing to Coq.
Second, we've added the quantifier [forall 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.
Proof.
intros n. reflexivity. Qed.
Theorem mult_0_l : forall n:nat, 0 * n = 0.
Proof.
intros n. reflexivity. Qed.
(** The [_l] suffix in the names of these theorems is
pronounced "on the left." *)
(** It is worth stepping through these proofs to observe how the
context and the goal change. You may want to add calls to [simpl]
before [reflexivity] to see the simplifications that Coq performs
on the terms before checking that they are equal. *)
(* ################################################################# *)
(** * Proof by Rewriting *)
(** This theorem is a bit more interesting than the others we've
seen: *)
Theorem plus_id_example : 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.
(* 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 : forall n m o : nat,
n = m -> m = o -> n + m = m + o.
Proof.
intros n m o.
intros H J.
rewrite -> H.
rewrite -> J.
reflexivity.
Qed.
(** [] *)
(** 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.
Proof.
intros n m.
rewrite -> plus_O_n.
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.
Proof.
intros n m.
intros H.
simpl.
rewrite <- H.
reflexivity.
Qed.
(* (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.
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 : forall n : nat,
(n + 1) =? 0 = false.
Proof.
intros n. destruct n as [| n'] eqn:E.
- reflexivity.
- reflexivity. Qed.
(** The [destruct] generates _two_ subgoals, which we must then
prove, separately, in order to get Coq to accept the theorem.
The annotation "[as [| n']]" is called an _intro pattern_. It
tells Coq what variable names to introduce in each subgoal. In
general, what goes between the square brackets is a _list of
lists_ of names, separated by [|]. In this case, the first
component is empty, since the [O] constructor is nullary (it
doesn't have any arguments). The second component gives a single
name, [n'], since [S] is a unary constructor.
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.
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 _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.
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' : forall 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.
(** 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.
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' : forall n : nat,
(n + 1) =? 0 = false.
Proof.
intros [|n].
- reflexivity.
- 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.
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 : forall b c : bool,
andb b c = true -> c = true.
Proof.
intros [] [].
- reflexivity.
- intros H. destruct H. reflexivity.
- reflexivity.
- intros H. destruct H. reflexivity.
Qed.
(** **** Exercise: 1 star, standard (zero_nbeq_plus_1) *)
Theorem zero_nbeq_plus_1 : forall n : nat,
0 =? (n + 1) = false.
Proof.
intros [].
- reflexivity.
- reflexivity.
Qed.
(* ================================================================= *)
(** ** 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.
Proof.
intros f x b.
rewrite -> x.
rewrite -> x.
reflexivity.
Qed.
(** **** 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.
Proof.
intros [] [].
- reflexivity.
- simpl. intros H. rewrite -> H. reflexivity.
- simpl. intros H. rewrite -> H. reflexivity.
- reflexivity.
Qed.
(** [] *)
(** **** 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 [S]s 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 :=
match m with
| Z => B Z
| A n => B n
| B n => A (incr n)
end.
Fixpoint bin_to_nat (m:bin) : nat :=
match m with
| Z => 0
| A n => 2 * (bin_to_nat n)
| B n => 1 + 2 * (bin_to_nat n)
end.
(** (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. *)
Example test_bin_incr1 : incr Z = B Z.
Proof. reflexivity. Qed.
Example test_bin_incr2 : incr (B Z) = A (B Z).
Proof. reflexivity. Qed.
Example test_bin_incr3 : incr (A (B Z)) = B (B Z).
Proof. reflexivity. Qed.
Example test_bin_incr4 : incr (B (B Z)) = A (A (B Z)).
Proof. reflexivity. Qed.
Example test_bin_incr5 : incr (A (A (B Z))) = B (A (B Z)).
Proof. reflexivity. Qed.
Example test_bin_to_nat0 : bin_to_nat Z = 0.
Proof. reflexivity. Qed.
Example test_bin_to_nat1 : bin_to_nat (B Z) = 1.
Proof. reflexivity. Qed.
Example test_bin_to_nat2 : bin_to_nat (A (B Z)) = 2.
Proof. reflexivity. Qed.
Example test_bin_to_nat3 : bin_to_nat (B (B Z)) = 3.
Proof. reflexivity. Qed.
Example test_bin_to_nat4 : bin_to_nat (A (A (B Z))) = 4.
Proof. reflexivity. Qed.
(* Do not modify the following line: *)
Definition manual_grade_for_binary : option (nat*string) := None.
(** [] *)
(* Wed Jan 9 12:02:44 EST 2019 *)
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