形式化方法作业

这里是必须完成的部分，到乘法交换律，可选部分未上传

(** * Basics: Functional Programming *)

(* $Date: 2012-01-27 20:02:32 -0500 (Fri, 27 Jan 2012) $ *)

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

(** * Enumerated Types *)

(** In Coq's programming language, almost nothing is built

in -- not even booleans or numbers! Instead, it provides powerful

tools for defining new types of data and functions that process

and transform them. *)

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

(** ** Days of the Week *)

(** Let's start with a very simple example. The following

declaration tells Coq that we are defining a new set of data

values -- a "type." The type is called [day], and its members are

[monday], [tuesday], etc. The lines of the definition can be read

"[monday] is a [day], [tuesday] is a [day], etc." *)

Inductive day : Type :=

| monday : day

| tuesday : day

| wednesday : day

| thursday : day

| friday : day

| saturday : day

| sunday : day.

(** Having defined [day], we can write functions that operate on

days. *)

Definition next_weekday (d:day) : day :=

match d with

| monday => tuesday

| tuesday => wednesday

| wednesday => thursday

| thursday => friday

| friday => monday

| saturday => monday

| sunday => monday

end.

(** One thing to note is that the argument and return types of

this function are explicitly declared. Like most functional

programming languages, Coq can often work out these types even if

they are not given explicitly -- i.e., it performs some _type

inference_ -- but we'll always include them to make reading

easier. *)

(** Having defined a function, we should check that it works on

some examples. There are actually three different ways to do this

in Coq. First, we can use the command [Eval simpl] to evaluate a

compound expression involving [next_weekday]. Uncomment the

following and see what they do. *)

Eval simpl in (next_weekday friday).

(* ==> monday : day *)

Eval simpl in (next_weekday (next_weekday saturday)).

(* ==> tuesday : day *)

(** If you have a computer handy, now would be an excellent

moment to fire up the Coq interpreter under your favorite IDE --

either CoqIde or Proof General -- and try this for yourself. Load

this file ([Basics.v]) from the book's accompanying Coq sources,

find the above example, submit it to Coq, and observe the

result. *)

(** The keyword [simpl] ("simplify") tells Coq precisely how to

evaluate the expression we give it. For the moment, [simpl] is

the only one we'll need; later on we'll see some alternatives that

are sometimes useful. *)

(** Second, we can record what we _expect_ the result to be in

the form of a Coq example: *)

Example test_next_weekday:

(next_weekday (next_weekday saturday)) = tuesday.

(** This declaration does two things: it makes an

assertion (that the second weekday after [saturday] is [tuesday]),

and it gives the assertion a name that can be used to refer to it

later. *)

(** Having made the assertion, we can also ask Coq to verify it,

like this: *)

Proof. simpl. reflexivity. Qed.

(** The details are not important for now (we'll come back to

them in a bit), but essentially this can be read as "The assertion

we've just made can be proved by observing that both sides of the

equality are the same after simplification." *)

(** Third, we can ask Coq to "extract," from a [Definition], a

program in some other, more conventional, programming

language (OCaml, Scheme, or Haskell) with a high-performance

compiler. This facility is very interesting, since it gives us a

way to construct _fully certified_ programs in mainstream

languages. Indeed, this is one of the main uses for which Coq was

developed. We won't have space to dig further into this topic,

but more information can be found in the Coq'Art book by Bertot

and Castéran, as well as the Coq reference manual. *)

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

(** ** Booleans *)

(** In a similar way, we can define the type [bool] of booleans,

with members [true] and [false]. *)

Inductive bool : Type :=

| true : bool

| false : bool.

(** Although we are rolling our own booleans here for the sake

of building up everything from scratch, Coq does, of course,

provide a default implementation of the booleans in its standard

library, together with a multitude of useful functions and

lemmas. (Take a look at [Coq.Init.Datatypes] in the Coq library

documentation if you're interested.) Whenever possible, we'll

name our own definitions and theorems so that they exactly

coincide with the ones in the standard library. *)

(** Functions over booleans can be defined in the same way as

above: *)

Definition negb (b:bool) : bool :=

match b with

| true => false

| false => true

end.

Definition andb (b1:bool) (b2:bool) : bool :=

match b1 with

| true => b2

| false => false

end.

Definition orb (b1:bool) (b2:bool) : bool :=

match b1 with

| true => true

| false => b2

end.

(** The last two illustrate the syntax for multi-argument

function definitions. *)

(** The following four "unit tests" constitute a complete

specification -- a truth table -- for the [orb] function: *)

Example test_orb1: (orb true false) = true.

Proof. simpl. reflexivity. Qed.

Example test_orb2: (orb false false) = false.

Proof. simpl. reflexivity. Qed.

Example test_orb3: (orb false true) = true.

Proof. simpl. reflexivity. Qed.

Example test_orb4: (orb true true) = true.

Proof. simpl. reflexivity. Qed.

(** _A note on notation_: We use square brackets to delimit

fragments of Coq code in comments in .v files; this convention,

also used by the [coqdoc] documentation tool, keeps them visually

separate from the surrounding text. In the html version of the

files, these pieces of text appear in a [different font]. *)

(** The following bit of Coq hackery defines a magic value

called [admit] that can fill a hole in an incomplete definition or

proof. We'll use it in the definition of [nandb] in the following

exercise. In general, your job in the exercises is to replace

[admit] or [Admitted] with real definitions or proofs. *)

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

(** **** Exercise: 1 star (nandb) *)

(** Complete the definition of the following function, then make

sure that the [Example] assertions below each can be verified by

Coq. *)

(** This function should return [true] if either or both of

its inputs are [false]. *)

Definition nandb (b1:bool) (b2:bool) : bool :=

match b1 with

| true => (negb b2)

| false => true

end.

(** Remove "[Admitted.]" and fill in each proof with

"[Proof. simpl. reflexivity. Qed.]" *)

Example test_nandb1: (nandb true false) = true.

Proof. simpl. reflexivity. Qed.

Example test_nandb2: (nandb false false) = true.

Proof. simpl. reflexivity. Qed.

Example test_nandb3: (nandb false true) = true.

Proof. simpl. reflexivity. Qed.

Example test_nandb4: (nandb true true) = false.

Proof. simpl. reflexivity. Qed.

(** [] *)

(** **** Exercise: 1 star (andb3) *)

Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool :=

andb (andb b1 b2) b3.

Example test_andb31: (andb3 true true true) = true.

Proof. simpl. reflexivity. Qed.

Example test_andb32: (andb3 false true true) = false.

Proof. simpl. reflexivity. Qed.

Example test_andb33: (andb3 true false true) = false.

Proof. simpl. reflexivity. Qed.

Example test_andb34: (andb3 true true false) = false.

Proof. simpl. reflexivity. Qed.

(** [] *)

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

(** ** Function Types *)

(** The [Check] command causes Coq to print the type of an

expression. For example, the type of [negb true] is [bool]. *)

Check true.

(* ===> true : bool *)

Check (negb true).

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

(** Functions like [negb] itself are also data values, just like

[true] and [false]. Their types are called _function types_, and

they are written with arrows. *)

Check negb.

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

(** The type of [negb], written [bool -> bool] and pronounced

"[bool] arrow [bool]," can be read, "Given an input of type

[bool], this function produces an output of type [bool]."

Similarly, the type of [andb], written [bool -> bool -> bool], can

be read, "Given two inputs, both of type [bool], this function

produces an output of type [bool]." *)

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

(** ** Numbers *)

(** _Technical digression_: Coq provides a fairly fancy module system,

to aid in organizing large developments. In this course, we won't

need most of its features, but one of them is useful: if we

enclose a collection of declarations between [Module X] and [End

X] markers, then, in the remainder of the file after the [End],

all these definitions will be referred to by names like [X.foo]

instead of just [foo]. This means that the new definition will

not clash with the unqualified name [foo] later, which would

otherwise be an error (a name can only be defined once in a given

scope). Here, we use this feature to introduce the definition of

the type [nat] in an inner module so that it does not shadow the

one from the standard library. *)

Module Playground1.

(** The types we have defined so far are examples of "enumerated

types": their definitions explicitly enumerate a finite set of

elements. A more interesting way of defining a type is to give a

collection of "inductive rules" describing its elements. For

example, we can define the natural numbers as follows: *)

Inductive nat : Type :=

| O : nat

| S : nat -> nat.

(** The clauses of this definition can be read:

- [O] is a natural number (note that this is the letter "[O]," not

the numeral "[0]").

- [S] is a "constructor" that takes a natural number and yields

another one -- that is, if [n] is a natural number, then [S n]

is too.

Let's look at this in a little more detail.

Every inductively defined set ([weekday], [nat], [bool], etc.) is

actually a set of _expressions_. The definition of [nat] says how

expressions in the set [nat] can be constructed:

- the expression [O] belongs to the set [nat];

- if [n] is an expression belonging to the set [nat], then [S n]

is also an expression belonging to the set [nat]; and

- expressions formed in these two ways are the only ones belonging

to the set [nat]. *)

(** These three conditions are the precise force of the

[Inductive] declaration. They imply that the expression [O], the

expression [S O], the expression [S (S O)], the expression

[S (S (S O))], and so on all belong to the set [nat], while other

expressions like [true], [andb true false], and [S (S false)] do

not.

We can write simple functions that pattern match on natural

numbers just as we did above -- for example, predecessor: *)

Definition pred (n : nat) : nat :=

match n with

| O => O

| S n' => n'

end.

(** The second branch can be read: "if [n] has the form [S n']

for some [n'], then return [n']." *)

End Playground1.

Definition minustwo (n : nat) : nat :=

match n with

| O => O

| S O => O

| S (S n') => n'

end.

(** Because natural numbers are such a pervasive form of data,

Coq provides a tiny bit of built-in magic for parsing and printing

them: ordinary arabic numerals can be used as an alternative to

the "unary" notation defined by the constructors [S] and [O]. Coq

prints numbers in arabic form by default: *)

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

Eval simpl in (minustwo 4).

(** The constructor [S] has the type [nat -> nat], just like the

functions [minustwo] and [pred]: *)

Check S.

Check pred.

Check minustwo.

(** These are all things that can be applied to a number to yield a

number. However, there is a fundamental difference: functions

like [pred] and [minustwo] come with _computation rules_

-- e.g., the definition of [pred] says that [pred n] can be

simplified to [match n with | O => O | S m' => m' end] -- while

the definition of [S] has no such behavior attached. Although it

is a function in the sense that it can be applied to an argument,

it does not _do_ anything at all! *)

(** For most function definitions over numbers, pure pattern

matching is not enough: we also need recursion. For example, to

check that a number [n] is even, we may need to recursively check

whether [n-2] is even. To write such functions, we use the

keyword [Fixpoint]. *)

Fixpoint evenb (n:nat) : bool :=

match n with

| O => true

| S O => false

| S (S n') => evenb n'

end.

(** When Coq checks this definition, it notes that [evenb] is

"decreasing on 1st argument." What this means is that we are

performing a _structural recursion_ (or _primitive recursion_)

over the argument [n] -- i.e., that we make recursive calls only

on strictly smaller values of [n]. This implies that all calls to

[evenb] will eventually terminate. Coq demands that some argument

of _every_ [Fixpoint] definition is decreasing. *)

(** We can define [oddb] by a similar [Fixpoint] declaration, but here

is a simpler definition that will be a bit easier to work with: *)

Definition oddb (n:nat) : bool := negb (evenb n).

Example test_oddb1: (oddb (S O)) = true.

Proof. simpl. reflexivity. Qed.

Example test_oddb2: (oddb (S (S (S (S O))))) = false.

Proof. simpl. reflexivity. Qed.

(** Naturally, we can also define multi-argument functions by

recursion. (Once again, we use a module to avoid polluting the

namespace.) *)

Module Playground2.

Fixpoint plus (n : nat) (m : nat) : nat :=

match n with

| O => m

| S n' => S (plus n' m)

end.

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

Eval simpl in (plus (S (S (S O))) (S (S O))).

(** The simplification that Coq performs to reach this conclusion can

be visualized as follows: *)

(* [plus (S (S (S O))) (S (S O))]

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

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

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

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

*)

(** As a notational convenience, if two or more arguments have

the same type, they can be written together. In the following

definition, [(n m : nat)] means just the same as if we had written

[(n : nat) (m : nat)]. *)

Fixpoint mult (n m : nat) : nat :=

match n with

| O => O

| S n' => plus m (mult n' m)

end.

(** You can match two expressions at once by putting a comma

between them: *)

Fixpoint minus (n m:nat) : nat :=

match n, m with

| O , _ => O

| S _ , O => n

| S n', S m' => minus n' m'

end.

(** The _ in the first line is a _wildcard pattern_. Writing _ in a

pattern is the same as writing some variable that doesn't get used

on the right-hand side. This avoids the need to invent a bogus

variable name. *)

End Playground2.

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

match power with

| O => S O

| S p => mult base (exp base p)

end.

Example test_mult1: (mult 3 3) = 9.

Proof. simpl. reflexivity. Qed.

(** **** Exercise: 1 star (factorial) *)

(** Recall the standard factorial function:

<<

factorial(0) = 1

factorial(n) = n * factorial(n-1) (if n>0)

>>

Translate this into Coq. *)

Fixpoint factorial (n:nat) : nat :=

match n with

|O=>1

|S n'=>n*(factorial n')

end.

Example test_factorial1: (factorial 3) = 6.

Proof. simpl. reflexivity. Qed.

Example test_factorial2: (factorial 5) = (mult 10 12).

Proof. simpl. reflexivity. Qed.

(** [] *)

(** We can make numerical expressions a little easier to read and

write by introducing "notations" for addition, multiplication, and

subtraction. *)

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

(at level 50, left associativity)

: nat_scope.

Notation "x - y" := (minus x y)

(at level 50, left associativity)

: nat_scope.

Notation "x * y" := (mult x y)

(at level 40, left associativity)

: nat_scope.

Check ((0 + 1) + 1).

(** Note that these do not change the definitions we've already

made: they are simply instructions to the Coq parser to accept [x

+ y] in place of [plus x y] and, conversely, to the Coq

pretty-printer to display [plus x y] as [x + y].

Each notation-symbol in Coq is active in a _notation scope_. Coq

tries to guess what scope you mean, so when you write [S(O*O)] it

guesses [nat_scope], but when you write the cartesian

product (tuple) type [bool*bool] it guesses [type_scope].

Occasionally you have to help it out with percent-notation by

writing [(x*y)%nat], and sometimes in Coq's feedback to you it

will use [%nat] to indicate what scope a notation is in.

Notation scopes also apply to numeral notation (3,4,5, etc.), so you

may sometimes see [0%nat] which means [O], or [0%Z] which means the

Integer zero. *)

(** When we say that Coq comes with nothing built-in, we really

mean it: even equality testing for numbers is a user-defined

operation! *)

(** The [beq_nat] function tests [nat]ural 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 beq_nat (n m : nat) : bool :=

match n with

| O => match m with

| O => true

| S m' => false

end

| S n' => match m with

| O => false

| S m' => beq_nat n' m'

end

end.

(** Similarly, the [ble_nat] function tests [nat]ural numbers for

[l]ess-or-[e]qual, yielding a [b]oolean. *)

Fixpoint ble_nat (n m : nat) : bool :=

match n with

| O => true

| S n' =>

match m with

| O => false

| S m' => ble_nat n' m'

end

end.

Example test_ble_nat1: (ble_nat 2 2) = true.

Proof. simpl. reflexivity. Qed.

Example test_ble_nat2: (ble_nat 2 4) = true.

Proof. simpl. reflexivity. Qed.

Example test_ble_nat3: (ble_nat 4 2) = false.

Proof. simpl. reflexivity. Qed.

(** **** Exercise: 2 stars (blt_nat) *)

(** The [blt_nat] function tests [nat]ural 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.

Note: If you have trouble with the [simpl] tactic, try using

[compute], which is like [simpl] on steroids. However, there is a

simple, elegant solution for which [simpl] suffices. *)

Fixpoint blt_nat (n m : nat) : bool :=

match n with

| O => match m with

| O => false

| S m' => true

end

| S n' => match m with

| O => false

| S m' => blt_nat n' m'

end

end.

Example test_blt_nat1: (blt_nat 2 2) = false.

Proof. simpl. reflexivity. Qed.

Example test_blt_nat2: (blt_nat 2 4) = true.

Proof. simpl. reflexivity. Qed.

Example test_blt_nat3: (blt_nat 4 2) = false.

Proof. simpl. reflexivity. Qed.

(** [] *)

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

(** * Proof By Simplification *)

(** Now that we've defined a few datatypes and functions, let's

turn to the question of how to state and prove properties of their

behavior. Actually, in a sense, we've already started doing this:

each [Example] in the previous sections makes a precise claim

about the behavior of some function on some particular inputs.

The proofs of these claims were always the same: use the

function's definition to simplify the expressions on both sides of

the [=] and notice that they become identical.

The same sort of "proof by simplification" can be used to prove

more interesting properties as well. For example, the fact that

[0] is a "neutral element" for [+] on the left can be proved

just by observing that [0 + n] reduces to [n] no matter what

[n] is, since the definition of [+] is recursive in its first

argument. *)

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

Proof.

simpl. reflexivity. Qed.

(** The [reflexivity] command implicitly simplifies both sides of the

equality before testing to see if they are the same, so we can

shorten the proof a little. *)

(** (It will be useful later to know that [reflexivity] actually

does somwhat more than [simpl] -- for example, it tries

"unfolding" defined terms, replacing them with their right-hand

sides. The reason for this difference is that, when reflexivity

succeeds, the whole goal is finished and we don't need to look at

whatever expanded expressions [reflexivity] has found; by

contrast, [simpl] is used in situations where we may have to read

and understand the new goal, so we would not want it blindly

expanding definitions.) *)

Theorem plus_O_n' : forall n:nat, 0 + n = n.

Proof.

reflexivity. Qed.

(** The form of this theorem and proof are almost exactly the

same as the examples above: the only differences are that we've

added the quantifier [forall n:nat] and that we've used the

keyword [Theorem] instead of [Example]. Indeed, the latter

difference is purely a matter of style; the keywords [Example] and

[Theorem] (and a few others, including [Lemma], [Fact], and

[Remark]) mean exactly the same thing to Coq.

The keywords [simpl] and [reflexivity] are examples of _tactics_.

A tactic is a command that is used between [Proof] and [Qed] to

tell Coq how it should check the correctness of some claim we are

making. We will see several more tactics in the rest of this

lecture, and yet more in future lectures. *)

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

(** What will Coq print in response to this query? *)

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

(** What about this one? *)

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

(** Explain the difference. [] *)

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

(** * The [intros] Tactic *)

(** Aside from unit tests, which apply functions to particular

arguments, most of the properties we will be interested in proving

about programs will begin with some quantifiers (e.g., "for all

numbers [n], ...") and/or hypothesis ("assuming [m=n], ..."). In

such situations, we will need to be able to reason by _assuming

the hypothesis_ -- i.e., we start by saying "OK, suppose [n] is

some arbitrary number," or "OK, suppose [m=n]."

The [intros] tactic permits us to do this by moving one or more

quantifiers or hypotheses from the goal to a "context" of current

assumptions.

For example, here is a slightly different proof of the same theorem. *)

Theorem plus_O_n'' : forall n:nat, 0 + n = n.

Proof.

intros n. reflexivity. Qed.

(** Step through this proof in Coq and notice how the goal and

context change. *)

Theorem plus_1_l : 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." *)

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

(** * Proof by Rewriting *)

(** Here is a slightly more interesting theorem: *)

Theorem plus_id_example : forall n m:nat,

n = m ->

n + n = m + m.

(** Instead of making a completely universal claim about all numbers

[n] and [m], this theorem talks about a more specialized property

that only holds when [n = m]. The arrow symbol is pronounced

"implies."

Since [n] and [m] are arbitrary numbers, we can't just use

simplification to prove this theorem. Instead, we prove it by

observing that, if we are assuming [n = m], then we can replace

[n] with [m] in the goal statement and obtain an equality with the

same expression on both sides. The tactic that tells Coq to

perform this replacement is called [rewrite]. *)

Proof.

intros n m. (* move both quantifiers into the context *)

intros H. (* move the hypothesis into the context *)

rewrite -> H. (* Rewrite the goal using the hypothesis *)

reflexivity. Qed.

(** The first line of the proof moves the universally quantified

variables [n] and [m] into the context. The second moves the

hypothesis [n = m] into the context and gives it the name [H].

The third tells Coq to rewrite the current goal ([n + n = m + m])

by replacing the left side of the equality hypothesis [H] with the

right side.

(The arrow symbol in the [rewrite] has nothing to do with

implication: it tells Coq to apply the rewrite from left to right.

To rewrite from right to left, you can use [rewrite <-]. Try

making this change in the above proof and see what difference it

makes in Coq's behavior.) *)

(** **** Exercise: 1 star (plus_id_exercise) *)

(** Remove "[Admitted.]" and fill in the proof. *)

Theorem plus_id_exercise : forall n m o : nat,

n = m -> m = o -> n + m = m + o.

Proof.

intros n m o.

intros H.

intros J.

rewrite->H.

rewrite->J.

reflexivity.

Qed.

(** [] *)

(** The [Admitted] command tells Coq that we want to give up

trying to prove this theorem and just accept it as a given. This

can be useful for developing longer proofs, since we can state

subsidiary facts that we believe will be useful for making some

larger argument, use [Admitted] to accept them on faith for the

moment, and continue thinking about the larger argument until we

are sure it makes sense; then we can go back and fill in the

proofs we skipped. Be careful, though: every time you say [admit]

or [Admitted] you are leaving a door open for total nonsense to

enter Coq's nice, rigorous, formally checked world! *)

(** We can also use the [rewrite] tactic with a previously proved

theorem instead of a hypothesis from the context. *)

Theorem mult_0_plus : forall n m : nat,

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

Proof.

intros n m.

rewrite -> plus_O_n.

reflexivity. Qed.

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

Theorem mult_1_plus : forall n m : nat,

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

Proof.

reflexivity.

Qed.

(** [] *)

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

(** * Case Analysis *)

(** Of course, not everything can be proved by simple

calculation: In general, unknown, hypothetical values (arbitrary

numbers, booleans, lists, etc.) can show up in the "head position"

of functions that we want to reason about, blocking

simplification. For example, if we try to prove the following

fact using the [simpl] tactic as above, we get stuck. *)

Theorem plus_1_neq_0_firsttry : forall n : nat,

beq_nat (n + 1) 0 = false.

Proof.

intros n. simpl. (* does nothing! *)

Admitted.

(** The reason for this is that the definitions of both

[beq_nat] and [+] begin by performing a [match] on their first

argument. But here, the first argument to [+] is the unknown

number [n] and the argument to [beq_nat] is the compound

expression [n + 1]; neither can be simplified.

What we need is to be able to consider the possible forms of [n]

separately. If [n] is [O], then we can calculate the final result

of [beq_nat (n + 1) 0] and check that it is, indeed, [false].

And if [n = S n'] for some [n'], then, although we don't know

exactly what number [n + 1] yields, we can calculate that, at

least, it will begin with one [S], and this is enough to calculate

that, again, [beq_nat (n + 1) 0] will yield [false].

The tactic that tells Coq to consider, separately, the cases where

[n = O] and where [n = S n'] is called [destruct]. *)

Theorem plus_1_neq_0 : forall n : nat,

beq_nat (n + 1) 0 = false.

Proof.

intros n. destruct n as [| n'].

reflexivity.

reflexivity. Qed.

(** The [destruct] generates _two_ subgoals, which we must then

prove, separately, in order to get Coq to accept the theorem as

proved. (No special command is needed for moving from one subgoal

to the other. When the first subgoal has been proved, it just

disappears and we are left with the other "in focus.") In this

proof, each of the subgoals is easily proved by a single use of

[reflexivity].

The annotation "[as [| n']]" is called an "intro pattern." It

tells Coq what variable names to introduce in each subgoal. In

general, what goes between the square brackets is a _list_ of

lists of names, separated by [|]. Here, the first component is

empty, since the [O] constructor is nullary (it doesn't carry any

data). The second component gives a single name, [n'], since [S]

is a unary constructor.

The [destruct] tactic can be used with any inductively defined

datatype. For example, we use it here to prove that boolean

negation is involutive -- i.e., that negation is its own

inverse. *)

Theorem negb_involutive : forall b : bool,

negb (negb b) = b.

Proof.

intros b. destruct b.

reflexivity.

reflexivity. Qed.

(** Note that the [destruct] here has no [as] clause because

none of the subcases of the [destruct] need to bind any variables,

so there is no need to specify any names. (We could also have

written "[as [|]]", or "[as []]".) In fact, we can omit the [as]

clause from _any_ [destruct] and Coq will fill in variable names

automatically. Although this is convenient, it is arguably bad

style, since Coq often makes confusing choices of names when left

to its own devices. *)

(** **** Exercise: 1 star (zero_nbeq_plus_1) *)

Theorem zero_nbeq_plus_1 : forall n : nat,

beq_nat 0 (n + 1) = false.

Proof.

intros n. destruct n.

reflexivity.

reflexivity. Qed.

(** [] *)

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

(** * Naming Cases *)

(** The fact that there is no explicit command for moving from

one branch of a case analysis to the next can make proof scripts

rather hard to read. In larger proofs, with nested case analyses,

it can even become hard to stay oriented when you're sitting with

Coq and stepping through the proof. (Imagine trying to remember

that the first five subgoals belong to the inner case analysis and

the remaining seven cases are what remains of the outer one...)

Disciplined use of indentation and comments can help, but a better

way is to use the [Case] tactic.

[Case] is not built into Coq: we need to define it ourselves.

There is no need to understand how it works -- just skip over the

definition to the example that follows. It uses some facilities

of Coq that we have not discussed -- the string library (just for

the concrete syntax of quoted strings) and the [Ltac] command,

which allows us to declare custom tactics. Kudos to Aaron

Bohannon for this nice hack! *)

Require String. Open Scope string_scope.

Ltac move_to_top x :=

match reverse goal with

| H : _ |- _ => try move x after H

end.

Tactic Notation "assert_eq" ident(x) constr(v) :=

let H := fresh in

assert (x = v) as H by reflexivity;

clear H.

Tactic Notation "Case_aux" ident(x) constr(name) :=

first [

set (x := name); move_to_top x

| assert_eq x name; move_to_top x

| fail 1 "because we are working on a different case" ].

Tactic Notation "Case" constr(name) := Case_aux Case name.

Tactic Notation "SCase" constr(name) := Case_aux SCase name.

Tactic Notation "SSCase" constr(name) := Case_aux SSCase name.

Tactic Notation "SSSCase" constr(name) := Case_aux SSSCase name.

Tactic Notation "SSSSCase" constr(name) := Case_aux SSSSCase name.

Tactic Notation "SSSSSCase" constr(name) := Case_aux SSSSSCase name.

Tactic Notation "SSSSSSCase" constr(name) := Case_aux SSSSSSCase name.

Tactic Notation "SSSSSSSCase" constr(name) := Case_aux SSSSSSSCase name.

(** Here's an example of how [Case] is used. Step through the

following proof and observe how the context changes. *)

Theorem andb_true_elim1 : forall b c : bool,

andb b c = true -> b = true.

Proof.

intros b c H.

destruct b.

Case "b = true".

reflexivity.

Case "b = false".

rewrite <- H. reflexivity. Qed.

(** [Case] does something very trivial: It simply adds a string

that we choose (tagged with the identifier "Case") to the context

for the current goal. When subgoals are generated, this string is

carried over into their contexts. When the last of these subgoals

is finally proved and the next top-level goal (a sibling of the

current one) becomes active, this string will no longer appear in

the context and we will be able to see that the case where we

introduced it is complete. Also, as a sanity check, if we try to

execute a new [Case] tactic while the string left by the previous

one is still in the context, we get a nice clear error message.

For nested case analyses (i.e., when we want to use a [destruct]

to solve a goal that has itself been generated by a [destruct]),

there is an [SCase] ("subcase") tactic. *)

(** **** Exercise: 2 stars (andb_true_elim2) *)

(** Prove [andb_true_elim2], marking cases (and subcases) when

you use [destruct]. *)

Theorem andb_true_elim2 : forall b c : bool,

andb b c = true -> c = true.

Proof.

intros b c H.

destruct c.

Case "c = true".

reflexivity.

Case "c = false".

rewrite <- H.

destruct b.

SCase "b = false".

reflexivity.

SCase "b = true".

reflexivity.

Qed.

(** [] *)

(** There are no hard and fast rules for how proofs should be

formatted in Coq -- in particular, where lines should be broken

and how sections of the proof should be indented to indicate their

nested structure. However, if the places where multiple subgoals

are generated are marked with explicit [Case] tactics placed at

the beginning of lines, then the proof will be readable almost no

matter what choices are made about other aspects of layout.

This is a good place to mention one other piece of (possibly

obvious) advice about line lengths. Beginning Coq users sometimes

tend to the extremes, either writing each tactic on its own line

or entire proofs on one line. Good style lies somewhere in the

middle. In particular, one reasonable convention is to limit

yourself to 80-character lines. Lines longer than this are hard

to read and can be inconvenient to display and print. Many

editors have features that help enforce this. *)

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

(** * Induction *)

(** We proved above that [0] is a neutral element for [+] on

the left using a simple partial evaluation argument. The fact

that it is also a neutral element on the _right_... *)

Theorem plus_0_r_firsttry : forall n:nat,

n + 0 = n.

(** ... cannot be proved in the same simple way. Just applying

[reflexivity] doesn't work: the [n] in [n + 0] is an arbitrary

unknown number, so the [match] in the definition of [+] can't be

simplified. And reasoning by cases using [destruct n] doesn't get

us much further: the branch of the case analysis where we assume [n

= 0] goes through, but in the branch where [n = S n'] for some [n']

we get stuck in exactly the same way. We could use [destruct n'] to

get one step further, but since [n] can be arbitrarily large, if we

try to keep on going this way we'll never be done. *)

Proof.

intros n.

simpl. (* Does nothing! *)

Admitted.

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

Theorem plus_0_r_secondtry : forall n:nat,

n + 0 = n.

Proof.

intros n. destruct n as [| n'].

Case "n = 0".

reflexivity. (* so far so good... *)

Case "n = S n'".

simpl. (* ...but here we are stuck again *)

Admitted.

(** To prove such facts -- indeed, to prove most interesting

facts about numbers, lists, and other inductively defined sets --

we need a more powerful reasoning principle: _induction_.

Recall (from high school) the principle of induction over natural

numbers: If [P(n)] is some proposition involving a natural number

[n] and we want to show that P holds for _all_ numbers [n], we can

reason like this:

- show that [P(O)] holds;

- show that, for any [n'], if [P(n')] holds, then so does

[P(S n')];

- conclude that [P(n)] holds for all [n].

In Coq, the steps are the same but the order is backwards: we

begin with the goal of proving [P(n)] for all [n] and break it

down (by applying the [induction] tactic) into two separate

subgoals: first showing [P(O)] and then showing [P(n') -> P(S

n')]. Here's how this works for the theorem we are trying to

prove at the moment: *)

Theorem plus_0_r : forall n:nat, n + 0 = n.

Proof.

intros n. induction n as [| n'].

Case "n = 0". reflexivity.

Case "n = S n'". simpl. rewrite -> IHn'. reflexivity. Qed.

(** Like [destruct], the [induction] tactic takes an [as...]

clause that specifies the names of the variables to be introduced

in the subgoals. In the first branch, [n] is replaced by [0] and

the goal becomes [0 + 0 = 0], which follows by simplification. In

the second, [n] is replaced by [S n'] and the assumption [n' + 0 =

n'] is added to the context (with the name [IHn'], i.e., the

Induction Hypothesis for [n']). The goal in this case becomes [(S

n') + 0 = S n'], which simplifies to [S (n' + 0) = S n'], which in

turn follows from the induction hypothesis. *)

Theorem minus_diag : forall n,

minus n n = 0.

Proof.

(* WORKED IN CLASS *)

intros n. induction n as [| n'].

Case "n = 0".

simpl. reflexivity.

Case "n = S n'".

simpl. rewrite -> IHn'. reflexivity. Qed.

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

Theorem mult_0_r : forall n:nat,

n * 0 = 0.

Proof.

intros n. induction n as [| n'].

Case "n = 0".

simpl. reflexivity.

Case "n = S n'".

simpl. rewrite -> IHn'. reflexivity. Qed.

Theorem plus_n_Sm : forall n m : nat,

S (n + m) = n + (S m).

Proof.

intros n m.

induction n as [| n'].

Case "n = 0".

simpl. reflexivity.

Case "n = S n'".

simpl. rewrite -> IHn'. reflexivity. Qed.

Theorem plus_comm : forall n m : nat,

n + m = m + n.

Proof.

intros n m.

induction n as [| n'].

Case "n=0".

rewrite->plus_0_r.

simpl.

reflexivity.

Case "n=S n'".

rewrite <- plus_n_Sm.

simpl.

rewrite -> IHn'.

reflexivity.

Qed.

(** [] *)

Fixpoint double (n:nat) :=

match n with

| O => O

| S n' => S (S (double n'))

end.

(** **** Exercise: 2 stars (double_plus) *)

Lemma double_plus : forall n, double n = n + n .

Proof.

intros n.

simpl.

induction n as [| n'].

Case "n=0".

simpl.

reflexivity.

Case "n=S n'".

simpl.

rewrite -> IHn'.

rewrite <- plus_n_Sm.

reflexivity.

Qed.

(** [] *)

(** **** Exercise: 1 star (destruct_induction) *)

(** Briefly explain the difference between the tactics

[destruct] and [induction].

(* FILL IN HERE *)

*)

(** [] *)

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

(** * Formal vs. Informal Proof *)

(** "Informal proofs are algorithms; formal proofs are code." *)

(** The question of what, exactly, constitutes a "proof" of a

mathematical claim has challenged philosophers for millenia. A

rough and ready definition, though, could be this: a proof of a

mathematical proposition [P] is a written (or spoken) text that

instills in the reader or hearer the certainty that [P] is true.

That is, a proof is an act of communication.

Now, acts of communication may involve different sorts of readers.

On one hand, the "reader" can be a program like Coq, in which case

the "belief" that is instilled is a simple mechanical check that

[P] can be derived from a certain set of formal logical rules, and

the proof is a recipe that guides the program in performing this

check. Such recipes are _formal_ proofs.

Alternatively, the reader can be a human being, in which case the

proof will be written in English or some other natural language,

thus necessarily _informal_. Here, the criteria for success are

less clearly specified. A "good" proof is one that makes the

reader believe [P]. But the same proof may be read by many

different readers, some of whom may be convinced by a particular

way of phrasing the argument, while others may not be. One reader

may be particularly pedantic, inexperienced, or just plain

thick-headed; the only way to convince them will be to make the

argument in painstaking detail. But another reader, more familiar

in the area, may find all this detail so overwhelming that they

lose the overall thread. All they want is to be told the main

ideas, because it is easier to fill in the details for themselves.

Ultimately, there is no universal standard, because there is no

single way of writing an informal proof that is guaranteed to

convince every conceivable reader. In practice, however,

mathematicians have developed a rich set of conventions and idioms

for writing about complex mathematical objects that, within a

certain community, make communication fairly reliable. The

conventions of this stylized form of communication give a fairly

clear standard for judging proofs good or bad.

Because we are using Coq in this course, we will be working

heavily with formal proofs. But this doesn't mean we can ignore

the informal ones! Formal proofs are useful in many ways, but

they are _not_ very efficient ways of communicating ideas between

human beings. *)

(** For example, here is a proof that addition is associative: *)

Theorem plus_assoc' : forall n m p : nat,

n + (m + p) = (n + m) + p.

Proof. intros n m p. induction n as [| n']. reflexivity.

simpl. rewrite -> IHn'. reflexivity. Qed.

(** Coq is perfectly happy with this as a proof. For a human,

however, it is difficult to make much sense of it. If you're used

to Coq you can probably step through the tactics one after the

other in your mind and imagine the state of the context and goal

stack at each point, but if the proof were even a little bit more

complicated this would be next to impossible. Instead, a

mathematician mighty write it like this: *)

(** - _Theorem_: For any [n], [m] and [p],

[[

n + (m + p) = (n + m) + p.

]]

_Proof_: By induction on [n].

- First, suppose [n = 0]. We must show

[[

0 + (m + p) = (0 + m) + p.

]]

This follows directly from the definition of [+].

- Next, suppose [n = S n'], where

[[

n' + (m + p) = (n' + m) + p.

]]

We must show

[[

(S n') + (m + p) = ((S n') + m) + p.

]]

By the definition of [+], this follows from

[[

S (n' + (m + p)) = S ((n' + m) + p),

]]

which is immediate from the induction hypothesis. [] *)

(** The overall form of the proof is basically similar. This is

no accident, of course: Coq has been designed so that its

[induction] tactic generates the same sub-goals, in the same

order, as the bullet points that a mathematician would write. But

there are significant differences of detail: the formal proof is

much more explicit in some ways (e.g., the use of [reflexivity])

but much less explicit in others; in particular, the "proof state"

at any given point in the Coq proof is completely implicit,

whereas the informal proof reminds the reader several times where

things stand. *)

(** Here is a formal proof that shows the structure more

clearly: *)

Theorem plus_assoc : forall n m p : nat,

n + (m + p) = (n + m) + p.

Proof.

intros n m p. induction n as [| n'].

Case "n = 0".

reflexivity.

Case "n = S n'".

simpl. rewrite -> IHn'. reflexivity. Qed.

(** **** Exercise: 2 stars (plus_comm_informal) *)

(** Translate your solution for [plus_comm] into an informal proof. *)

(** Theorem: Addition is commutative.

Proof: (* FILL IN HERE *)

[]

*)

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

(** Write an informal proof of the following theorem, using the

informal proof of [plus_assoc] as a model. Don't just

paraphrase the Coq tactics into English!

Theorem: [true = beq_nat n n] for any [n].

Proof: (* FILL IN HERE *)

[]

*)

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

Theorem beq_nat_refl : forall n : nat,

true = beq_nat n n.

Proof.

intros n. induction n as [| n'].

Case "n = 0".

reflexivity.

Case "n = S n'".

simpl. rewrite -> IHn'. reflexivity. Qed.

(** [] *)

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

(** * Proofs Within Proofs *)

(** In Coq, as in informal mathematics, large proofs are very

often broken into a sequence of theorems, with later proofs

referring to earlier theorems. Occasionally, however, a proof

will need some miscellaneous fact that is too trivial (and of too

little general interest) to bother giving it its own top-level

name. In such cases, it is convenient to be able to simply state

and prove the needed "sub-theorem" right at the point where it is

used. The [assert] tactic allows us to do this. For example, our

earlier proof of the [mult_0_plus] theorem referred to a previous

theorem named [plus_O_n]. We can also use [assert] to state and

prove [plus_O_n] in-line: *)

Theorem mult_0_plus' : forall n m : nat,

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

Proof.

intros n m.

assert (H: 0 + n = n).

Case "Proof of assertion". reflexivity.

rewrite -> H.

reflexivity. Qed.

(** The [assert] tactic introduces two sub-goals. The first is

the assertion itself; by prefixing it with [H:] we name the

assertion [H]. (Note that we could also name the assertion with

[as] just as we did above with [destruct] and [induction], i.e.,

[assert (0 + n = n) as H]. Also note that we mark the proof of

this assertion with a [Case], both for readability and so that,

when using Coq interactively, we can see when we're finished

proving the assertion by observing when the ["Proof of assertion"]

string disappears from the context.) The second goal is the same

as the one at the point where we invoke [assert], except that, in

the context, we have the assumption [H] that [0 + n = n]. That

is, [assert] generates one subgoal where we must prove the

asserted fact and a second subgoal where we can use the asserted

fact to make progress on whatever we were trying to prove in the

first place. *)

(** Actually, [assert] will turn out to be handy in many sorts of

situations. For example, suppose we want to prove that [(n + m)

+ (p + q) = (m + n) + (p + q)]. The only difference between the

two sides of the [=] is that the arguments [m] and [n] to the

first inner [+] are swapped, so it seems we should be able to

use the commutativity of addition ([plus_comm]) to rewrite one

into the other. However, the [rewrite] tactic is a little stupid

about _where_ it applies the rewrite. There are three uses of

[+] here, and it turns out that doing [rewrite -> plus_comm]

will affect only the _outer_ one. *)

Theorem plus_rearrange_firsttry : forall n m p q : nat,

(n + m) + (p + q) = (m + n) + (p + q).

Proof.

intros n m p q.

(* We just need to swap (n + m) for (m + n)...

it seems like plus_comm should do the trick! *)

assert (H: n + m = m + n).

Case "Proof of assertion".

rewrite -> plus_comm.

reflexivity.

rewrite -> H.

reflexivity.

Qed.

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

(** To get [plus_comm] to apply at the point where we want it, we can

introduce a local lemma stating that [n + m = m + n] (for

the particular [m] and [n] that we are talking about here), prove

this lemma using [plus_comm], and then use this lemma to do the

desired rewrite. *)

Theorem plus_rearrange : forall n m p q : nat,

(n + m) + (p + q) = (m + n) + (p + q).

Proof.

intros n m p q.

assert (H: n + m = m + n).

Case "Proof of assertion".

rewrite -> plus_comm. reflexivity.

rewrite -> H. reflexivity. Qed.

(** **** Exercise: 4 stars (mult_comm) *)

(** Use [assert] to help prove this theorem. You shouldn't need to

use induction. *)

Theorem plus_swap : forall n m p : nat,

n + (m + p) = m + (n + p).

Proof.

intros n m p.

rewrite -> plus_assoc'.

rewrite -> plus_assoc'.

assert (H: n + m = m + n).

Case "Proof of assertion".

rewrite -> plus_comm. reflexivity.

rewrite -> H. reflexivity. Qed.

(** Now prove commutativity of multiplication. (You will probably

need to define and prove a separate subsidiary theorem to be used

in the proof of this one.) You may find that [plus_swap] comes in

handy. *)

Lemma mult_xjt: forall n m : nat,

n + n * m = n * S m.

Proof.

intros n m.

induction n as [| n'].

Case "n=0".

reflexivity.

Case "n=S n'".

simpl.

rewrite <- IHn'.

rewrite -> plus_assoc.

rewrite -> plus_assoc.

assert (H: n' + m = m + n').

SCase "Proof of assertion".

rewrite -> plus_comm. reflexivity.

rewrite -> H.

reflexivity.

Qed.

Theorem mult_comm : forall m n : nat,

m * n = n * m.

Proof.

intros n m.

induction n as [| n'].

Case "n=0".

rewrite -> mult_0_r.

reflexivity.

Case "n=S n'".

simpl.

rewrite <- mult_xjt.

rewrite -> IHn'.

reflexivity.

Qed.

(** [] *)