(** * Basics: Functional Programming *)
(* $Date: 2012-07-21 13:38:33 -0400 (Sat, 21 Jul 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]. *)
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 CastÃ©ran, 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) *)
(* EXPECTED *)
(** 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 :=
(* FILL IN HERE *) admit.
(** Remove "[Admitted.]" and fill in each proof with
"[Proof. simpl. reflexivity. Qed.]" *)
Example test_nandb1: (nandb true false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb2: (nandb false false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb3: (nandb false true) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb4: (nandb true true) = false.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (andb3) *)
(* EXPECTED *)
Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool :=
(* FILL IN HERE *) admit.
Example test_andb31: (andb3 true true true) = true.
(* FILL IN HERE *) Admitted.
Example test_andb32: (andb3 false true true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb33: (andb3 true false true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb34: (andb3 true true false) = false.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** ** Function Types *)
(** 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_ 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 :=
(* FILL IN HERE *) admit.
Example test_factorial1: (factorial 3) = 6.
(* FILL IN HERE *) Admitted.
Example test_factorial2: (factorial 5) = (mult 10 12).
(* FILL IN HERE *) Admitted.
(** [] *)
(** We can make numerical expressions a little easier to read and
write by introducing "notations" for addition, multiplication, and
subtraction. *)
Notation "x + y" := (plus x y)
(at level 50, left associativity)
: nat_scope.
Notation "x - y" := (minus x y)
(at level 50, left associativity)
: nat_scope.
Notation "x * y" := (mult x y)
(at level 40, left associativity)
: nat_scope.
Check ((0 + 1) + 1).
(** 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) *)
(* EXPECTED *)
(** 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. *)
Definition blt_nat (n m : nat) : bool :=
(* FILL IN HERE *) admit.
Example test_blt_nat1: (blt_nat 2 2) = false.
(* FILL IN HERE *) Admitted.
Example test_blt_nat2: (blt_nat 2 4) = true.
(* FILL IN HERE *) Admitted.
Example test_blt_nat3: (blt_nat 4 2) = false.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** * Proof By Simplification *)
(** Now that we've defined a few datatypes and functions, let's
turn to 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). *)
(* FILL IN HERE *)
(** What about this one? *)
(* FILL IN HERE *)
(* Eval simpl in (forall n:nat, 0 + n = n). *)
(** Explain the difference.*)
(* FILL IN HERE *)
(** [] *)
(* ###################################################################### *)
(** * 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) *)
(* EXPECTED *)
(** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(** 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) *)
(* EXPECTED *)
Theorem mult_1_plus : forall n m : nat,
(1 + n) * m = m + (n * m).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** * 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) *)
(* EXPECTED *)
Theorem zero_nbeq_plus_1 : forall n : nat,
beq_nat 0 (n + 1) = false.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** * 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) *)
(* EXPECTED *)
(** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(** 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) *)
(* EXPECTED *)
Theorem mult_0_r : forall n:nat,
n * 0 = 0.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_n_Sm : forall n m : nat,
S (n + m) = n + (S m).
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_comm : forall n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
Fixpoint double (n:nat) :=
match n with
| O => O
| S n' => S (S (double n'))
end.
(** **** Exercise: 2 stars (double_plus) *)
(* EXPECTED *)
Lemma double_plus : forall n, double n = n + n .
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (destruct_induction) *)
(* EXPECTED *)
(** Briefly explain the difference between the tactics
[destruct] and [induction].
(* FILL IN HERE *) Admitted.
*)
(** [] *)
(* ###################################################################### *)
(** * 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) *)
(* EXPECTED *)
(** Translate your solution for [plus_comm] into an informal proof. *)
(** Theorem: Addition is commutative.
Proof:
(* FILL IN HERE *) Admitted.
[]
*)
(** **** Exercise: 2 stars, optional (beq_nat_refl_informal) *)
(* EXPECTED *)
(** 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 *) Admitted.
[]
*)
(** **** Exercise: 1 star, optional (beq_nat_refl) *)
(* EXPECTED *)
Theorem beq_nat_refl : forall n : nat,
true = beq_nat n n.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** * 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! *)
rewrite -> plus_comm.
(* Doesn't work...Coq rewrote the wrong plus! *)
Admitted.
(** 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, recommended (mult_comm) *)
(* EXPECTED *)
(** 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.
(* FILL IN HERE *) Admitted.
(** 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. *)
Theorem mult_comm : forall m n : nat,
m * n = n * m.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, optional (evenb_n__oddb_Sn) *)
(* EXPECTED *)
Theorem evenb_n__oddb_Sn : forall n : nat,
evenb n = negb (evenb (S n)).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** * More Exercises *)
(** **** Exercise: 3 stars, optional (more_exercises) *)
(** Take a piece of paper. For each of the following theorems, first
_think_ about whether (a) it can be proved using only
simplification and rewriting, (b) it also requires case
analysis ([destruct]), or (c) it also requires induction. Write
down your prediction. Then fill in the proof. (There is no need
to turn in your piece of paper; this is just to encourage you to
reflect before hacking!) *)
Theorem ble_nat_refl : forall n:nat,
true = ble_nat n n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_S : forall n:nat,
beq_nat 0 (S n) = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem andb_false_r : forall b : bool,
andb b false = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_ble_compat_l : forall n m p : nat,
ble_nat n m = true -> ble_nat (p + n) (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem S_nbeq_0 : forall n:nat,
beq_nat (S n) 0 = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_1_l : forall n:nat, 1 * n = n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem all3_spec : forall b c : bool,
orb
(andb b c)
(orb (negb b)
(negb c))
= true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_plus_distr_r : forall n m p : nat,
(n + m) * p = (n * p) + (m * p).
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_assoc : forall n m p : nat,
n * (m * p) = (n * m) * p.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, optional (plus_swap') *)
(** The [replace] tactic allows you to specify a particular subterm to
rewrite and what you want it rewritten to. More precisely,
[replace (t) with (u)] replaces (all copies of) expression [t] in
the goal by expression [u], and generates [t = u] as an additional
subgoal. This is often useful when a plain [rewrite] acts on the wrong
part of the goal.
Use the [replace] tactic to do a proof of [plus_swap'], just like
[plus_swap] but without needing [assert (n + m = m + n)].
*)
Theorem plus_swap' : forall n m p : nat,
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 4 stars, recommended (binary) *)
(** Consider a different, more efficient representation of natural
numbers using a binary rather than unary system. That is, instead
of saying that each natural number is either zero or the successor
of a natural number, we can say that each binary number is either
- zero,
- twice a binary number, or
- one more than twice a binary number.
(a) First, write an inductive definition of the type [bin]
corresponding to this description of binary numbers.
(Hint: recall that the definition of [nat] from class,
Inductive nat : Type :=
| O : nat
| S : nat -> nat.
says nothing about what [O] and [S] "mean". It just says "[O] is
a nat (whatever that is), and if [n] is a nat then so is [S n]".
The interpretation of [O] as zero and [S] as successor/plus one
comes from the way that we use nat values, by writing functions to
do things with them, proving things about them, and so on. Your
definition of [bin] should be correspondingly simple; it is the
functions you will write next that will give it mathematical
meaning.)
(b) Next, write an increment function for binary numbers, and a
function to convert binary numbers to unary numbers.
(c) Finally, prove that your increment and binary-to-unary
functions commute: that is, incrementing a binary number and
then converting it to unary yields the same result as first
converting it to unary and then incrementing.
*)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 5 stars (binary_inverse) *)
(** This exercise is a continuation of the previous exercise about
binary numbers. You will need your definitions and theorems from
the previous exercise to complete this one.
(a) First, write a function to convert natural numbers to binary
numbers. Then prove that starting with any natural number,
converting to binary, then converting back yields the same
natural number you started with.
(b) You might naturally think that we should also prove the
opposite direction: that starting with a binary number,
converting to a natural, and then back to binary yields the
same number we started with. However, it is not true!
Explain what the problem is.
(c) Define a function [normalize] from binary numbers to binary
numbers such that for any binary number b, converting to a
natural and then back to binary yields [(normalize b)]. Prove
it.
*)
(* FILL IN HERE *)
(** **** Exercise: 2 stars, optional (decreasing) *)
(** The requirement that some argument to each function be
"decreasing" 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.
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 _not_ accept
because of this restriction. *)
(* FILL IN HERE *)
(** [] *)