(** * Induction: Proof by Induction *)
(** Before getting started, we need to import all of our
definitions from the previous chapter: *)
Require Export Basics.
(** For the [Require Export] to work, you first 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. There are two ways to do it:
- In CoqIDE:
Open [Basics.v]. In the "Compile" menu, click on "Compile
Buffer".
- From the command line:
[coqc Basics.v]
If you have trouble (e.g., if you get complaints about missing
identifiers later in the file), it may be because the "load path"
for Coq is not set up correctly. The [Print LoadPath.] command may
be helpful in sorting out such issues. *)
(* ################################################################# *)
(** * Proof by Induction *)
(** We proved in the last chapter that [0] is a neutral element
for [+] on the left, using an easy argument based on
simplification. We also observed that proving the fact that it is
also a neutral element on the _right_... *)
Theorem plus_n_O_firsttry : forall n:nat,
n = n + 0.
(** ... can't be done in the same simple way. Just applying
[reflexivity] doesn't work, since 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 fine, but in the branch where [n = S n'] for some [n'] we
get stuck in exactly the same way. *)
Theorem plus_n_O_secondtry : forall n:nat,
n = n + 0.
Proof.
intros n. destruct n as [| n'].
- (* n = 0 *)
reflexivity. (* so far so good... *)
- (* n = S n' *)
simpl. (* ...but here we are stuck again *)
Abort.
(** We could use [destruct n'] to get one step further, but,
since [n] can be arbitrarily large, if we just go on like this
we'll never finish. *)
(** To prove interesting facts about numbers, lists, and other
inductively defined sets, we usually need a more powerful
reasoning principle: _induction_.
Recall (from high school, a discrete math course, etc.) 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: 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: one where we must show [P(O)]
and another where we must show [P(n') -> P(S n')]. Here's how
this works for the theorem at hand: *)
Theorem plus_n_O : forall n:nat, n = n + 0.
Proof.
intros n. induction n as [| n' IHn'].
- (* n = 0 *) reflexivity.
- (* 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. Since there are two subgoals, the [as...] clause
has two parts, separated by [|]. (Strictly speaking, we can omit
the [as...] clause and Coq will choose names for us. In practice,
this is a bad idea, as Coq's automatic choices tend to be
confusing.)
In the first subgoal, [n] is replaced by [0]. No new variables
are introduced (so the first part of the [as...] is empty), and
the goal becomes [0 + 0 = 0], which follows by simplification.
In the second subgoal, [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']). These two names
are specified in the second part of the [as...] clause. The goal
in this case becomes [(S n') + 0 = S n'], which simplifies to
[S (n' + 0) = S n'], which in turn follows from [IHn']. *)
Theorem minus_diag : forall n,
minus n n = 0.
Proof.
(* WORKED IN CLASS *)
intros n. induction n as [| n' IHn'].
- (* n = 0 *)
simpl. reflexivity.
- (* n = S n' *)
simpl. rewrite -> IHn'. reflexivity. Qed.
(** (The use of the [intros] tactic in these proofs is actually
redundant. When applied to a goal that contains quantified
variables, the [induction] tactic will automatically move them
into the context as needed.) *)
(** **** Exercise: 2 stars, recommended (basic_induction) *)
(** Prove the following 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: 2 stars, optional (evenb_S) *)
(** One inconveninent aspect of our definition of [evenb n] is the
recursive call on [n - 2]. This makes proofs about [evenb n]
harder when done by induction on [n], since we may need an
induction hypothesis about [n - 2]. The following lemma gives an
alternative characterization of [evenb (S n)] that works better
with induction: *)
Theorem evenb_S : forall n : nat,
evenb (S n) = negb (evenb n).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 starM (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 often
broken into a sequence of theorems, with later proofs referring to
earlier theorems. But sometimes a proof will require 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 could instead 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). { 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]. (We can also name the assertion with [as] just as
we did above with [destruct] and [induction], i.e., [assert (0 + n
= n) as H].) Note that we surround the proof of this assertion
with curly braces [{ ... }], both for readability and so that,
when using Coq interactively, we can see more easily when we have
finished this sub-proof. The second goal is the same as the one
at the point where we invoke [assert] except that, in the context,
we now 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. *)
(** Another example of [assert]... *)
(** 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 not very smart 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)... seems
like plus_comm should do the trick! *)
rewrite -> plus_comm.
(* Doesn't work...Coq rewrote the wrong plus! *)
Abort.
(** To use [plus_comm] at the point where we need 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 it 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).
{ rewrite -> plus_comm. reflexivity. }
rewrite -> H. reflexivity. Qed.
(* ################################################################# *)
(** * Formal vs. Informal Proof *)
(** "_Informal proofs are algorithms; formal proofs are code_." *)
(** What constitutes a successful proof of a mathematical claim?
The question has challenged philosophers for millennia, but a
rough and ready definition 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 --
an unassailable argument for the truth of [P]. That is, a proof
is an act of communication.
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 that [P] can be mechanically
derived from a certain set of formal logical rules, and the proof
is a recipe that guides the program in checking this fact. 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,
and will thus necessarily be _informal_. Here, the criteria for
success are less clearly specified. A "valid" 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.
Some readers 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 other readers, 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, since it is easier for them to fill in the details for
themselves than to wade through a written presentation of them.
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 -- at least 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
completely forget about 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' IHn']. reflexivity.
simpl. rewrite -> IHn'. reflexivity. Qed.
(** Coq is perfectly happy with this. For a human, however, it
is difficult to make much sense of it. We can use comments and
bullets to show the structure a little 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' IHn'].
- (* n = 0 *)
reflexivity.
- (* n = S n' *)
simpl. rewrite -> IHn'. reflexivity. Qed.
(** ... and if you're used to Coq you may be able to 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.
A (pedantic) mathematician might write the proof 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. _Qed_. *)
(** The overall form of the proof is basically similar, and of
course 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). *)
(** **** Exercise: 2 stars, advanced, recommendedM (plus_comm_informal) *)
(** Translate your solution for [plus_comm] into an informal proof:
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)
*)
(** [] *)
(** **** Exercise: 2 stars, optionalM (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 *)
[] *)
(* ################################################################# *)
(** * More Exercises *)
(** **** Exercise: 3 stars, recommended (mult_comm) *)
(** Use [assert] to help prove this theorem. You shouldn't need to
use induction on [plus_swap]. *)
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: 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 you hack!) *)
Theorem leb_refl : forall n:nat,
true = leb 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,
leb n m = true -> leb (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 the [true] on the left-hand
side of the equality may look odd, but this is how the theorem is
stated in the Coq standard library, so we follow suit. Rewriting
works equally well in either direction, so 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: [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, recommendedM (binary_commute) *)
(** Recall the [incr] and [bin_to_nat] functions that you
wrote for the [binary] exercise in the [Basics] chapter. Prove
that the following diagram commutes:
incr
bin ----------------------> bin
| |
bin_to_nat | | bin_to_nat
| |
v v
nat ----------------------> nat
S
That is, incrementing a binary number and then converting it to
a (unary) natural number yields the same result as first converting
it to a natural number and then incrementing.
Name your theorem [bin_to_nat_pres_incr] ("pres" for "preserves").
Before you start working on this exercise, copy the definitions
from your solution to the [binary] exercise here so that this file
can be graded on its own. If you want to change your original
definitions to make the property easier to prove, feel free to
do so! *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 5 stars, advancedM (binary_inverse) *)
(** This exercise is a continuation of the previous exercise about
binary numbers. You will need your definitions and theorems from
there to complete this one; please copy them to this file to make
it self contained for grading.
(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, this is not true!
Explain what the problem is.
(c) Define a "direct" normalization function -- i.e., 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. (Warning: This
part is tricky!)
Again, feel free to change your earlier definitions if this helps
here. *)
(* FILL IN HERE *)
(** [] *)
(** $Date: 2016-10-07 14:01:19 -0400 (Fri, 07 Oct 2016) $ *)