(** * 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:
- In CoqIDE:
Open [Basics.v].
In the "Compile" menu, click on "Compile Buffer".
- From the command line:
Run [coqc Basics.v]
*)
(* ###################################################################### *)
(** * 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. The fact that it is also a neutral element on the
_right_... *)
Theorem plus_n_O_firsttry : forall n:nat,
n = n + 0.
(** ... cannot be proved 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. 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_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.
(** 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 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 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. 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'] -- notice that this name is
explicitly chosen in the [as...] clause of the call to [induction]
rather than letting Coq choose one arbitrarily). 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.
(** **** 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 that it
may need to perform a 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 a better characterization of [evenb (S n)]: *)
Theorem evenb_S : forall n : nat,
evenb (S n) = negb (evenb 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 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. *)
(** The [assert] tactic is 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 to, 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.
(* ###################################################################### *)
(** * More Exercises *)
(** **** Exercise: 3 stars, recommended (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: 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, recommended (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.
Name your theorem [bin_to_nat_pres_incr].
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
there 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, 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 *)
(** [] *)
(* ###################################################################### *)
(** * Formal vs. Informal Proof (Optional) *)
(** "Informal proofs are algorithms; formal proofs are code." *)
(** The question of what constitutes a proof of a mathematical
claim 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. 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, recommended (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: 2016-01-15 17:30:08 -0500 (Fri, 15 Jan 2016) $ *)