(** * Induction: Proof by Induction *)
(** The next line imports all of our definitions from the
previous chapter. *)
Require Export Basics.
(** For it to work, you need to use [coqc] to compile [Basics.v]
into [Basics.vo]. (This is like making a .class file from a .java
file, or a .o file from a .c file.)
Here are two ways to compile your code:
- CoqIDE:
Open [Basics.v].
In the "Compile" menu, click on "Compile Buffer".
- Command line:
Run [coqc Basics.v]
*)
(* ###################################################################### *)
(** * 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 -- you can 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". (* <----- here *)
reflexivity.
Case "b = false". (* <---- and here *)
rewrite <- H.
reflexivity.
Qed.
(** [Case] does something very straightforward: 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
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 (e.g., 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.
(* 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. *)
(* ###################################################################### *)
(** * Proof by Induction *)
(** We proved in the last chapter that [0] is a neutral element
for [+] on the left using a simple 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. *)
Proof.
intros n.
simpl. (* Does nothing! *)
Abort.
(** *** *)
(** 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 like this we'll never be done. *)
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 *)
Abort.
(** *** *)
(** 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 (basic_induction) *)
(** Prove the following lemmas using induction. You might need
previously proven results. *)
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.
Theorem plus_assoc : forall n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars (double_plus) *)
(** Consider the following function, which doubles its argument: *)
Fixpoint double (n:nat) :=
match n with
| O => O
| S n' => S (S (double n'))
end.
(** Use induction to prove this simple fact about [double]: *)
Lemma double_plus : forall n, double n = n + n .
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (destruct_induction) *)
(** Briefly explain the difference between the tactics
[destruct] and [induction].
(* FILL IN HERE *)
*)
(** [] *)
(* ###################################################################### *)
(** * 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! *)
Abort.
(** 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.
(* 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) *)
(** Prove the following simple fact: *)
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 (beq_nat_refl) *)
(** Prove the following theorem. Putting [true] on the left-hand side
of the equality may seem odd, but this is how the theorem is stated in
the standard library, so we follow suit. Since rewriting
works equally well in either direction, we will have no
problem using the theorem no matter which way we state it. *)
Theorem beq_nat_refl : forall n : nat,
true = beq_nat n n.
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: 3 stars (binary_commute) *)
(** Recall the [increment] and [binary-to-unary] functions that you
wrote for the [binary] exercise in the [Basics] chapter. Prove
that these 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.
(Before you start working on this exercise, please copy the
definitions from your solution to the [binary] exercise here so
that this file can be graded on its own. If you find yourself
wanting to change your original definitions to make the property
easier to prove, feel free to do so.) *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 5 stars, advanced (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.
Again, feel free to change your earlier definitions if this helps
here.
*)
(* FILL IN HERE *)
(** [] *)
(* ###################################################################### *)
(** * Advanced Material *)
(** ** 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 might write it something 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: 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, advanced (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 *)
[]
*)
(* $Date: 2013-09-26 14:40:26 -0400 (Thu, 26 Sep 2013) $ *)