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h7x4 03d1b5e74a Complete most of Induction.v
2026-03-12 11:30:40 +09:00

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(** * Induction: Proof by Induction *)
(* ################################################################# *)
(** * Separate Compilation *)
(** Before getting started on this chapter, we need to import
all of our definitions from the previous chapter: *)
From LF Require Export Basics.
(** For this [Require] command to work, Rocq needs to be able to
find a compiled version of the previous chapter ([Basics.v]).
This compiled version, called [Basics.vo], is analogous to the
[.class] files compiled from [.java] source files and the [.o]
files compiled from [.c] files.
To compile [Basics.v] and obtain [Basics.vo], first make sure that
the files [Basics.v], [Induction.v], and [_CoqProject] are in
the current directory.
The [_CoqProject] file should contain just the following line:
-Q . LF
This maps the current directory ("[.]", which contains [Basics.v],
[Induction.v], etc.) to the prefix (or "logical directory")
"[LF]". Proof General, CoqIDE, and VSCoq read [_CoqProject]
automatically, to find out to where to look for the file
[Basics.vo] corresponding to the library [LF.Basics].
Once the files are in place, there are various ways to build
[Basics.vo] from an IDE, or you can build it from the command
line. From an IDE...
- In Proof General: The compilation can be made to happen
automatically when you submit the [Require] line above to PG, by
setting the emacs variable [coq-compile-before-require] to [t].
This can also be found in the menu: "Coq" > "Auto Compilation" >
"Compile Before Require".
- In CoqIDE: One thing you can do on all platforms is open
[Basics.v]; then, in the "Compile" menu, click on "Compile Buffer".
- For VSCode users, open the terminal pane at the bottom and then
follow the command line instructions below. (If you downloaded
the project setup .tgz file, just doing `make` should build all
the code.)
To compile [Basics.v] from the command line...
- First, generate a [Makefile] using the [rocq makefile] utility,
which comes installed with Rocq. (If you obtained the whole volume as
a single archive, a [Makefile] should already exist and you can
skip this step.)
rocq makefile -f _CoqProject *.v -o Makefile
You should rerun that command whenever you add or remove
Rocq files in this directory.
- Now you can compile [Basics.v] by running [make] with the
corresponding [.vo] file as a target:
make Basics.vo
All files in the directory can be compiled by giving no
arguments:
make
- Under the hood, [make] uses the Rocq compiler, [rocq compile]. You can
also run [rocq compile] directly:
rocq compile -Q . LF Basics.v
- Since [make] also calculates dependencies between source files
to compile them in the right order, [make] should generally be
preferred over running [rocq compile] explicitly. But as a last (but
not terrible) resort, you can simply compile each file manually
as you go. For example, before starting work on the present
chapter, you would need to run the following command:
rocq compile -Q . LF Basics.v
Then, once you've finished this chapter, you'd do
rocq compile -Q . LF Induction.v
to get ready to work on the next one. If you ever remove the
.vo files, you'd need to give both commands again (in that
order).
Troubleshooting:
- For many of the alternatives above you need to make sure that
the [rocq] executable is in your [PATH].
- If you get complaints about missing identifiers, it may be
because the "load path" for Rocq is not set up correctly. The
[Print LoadPath.] command may be helpful in sorting out such
issues.
- When trying to compile a later chapter, if you see a message like
Compiled library Induction makes inconsistent assumptions over
library Basics
a common reason is that the library [Basics] was modified and
recompiled without also recompiling [Induction] which depends
on it. Recompile [Induction], or everything if too many files
are affected (for instance by running [make] and if even this
doesn't work then [make clean; make]).
- If you get complaints about missing identifiers later in this
file it may be because the "load path" for Rocq is not set up
correctly. The [Print LoadPath.] command may be helpful in
sorting out such issues.
In particular, if you see a message like
Compiled library Foo makes inconsistent assumptions over
library Bar
check whether you have multiple installations of Rocq on your
machine. It may be that commands (like [rocq compile]) that you execute
in a terminal window are getting a different version of Rocq than
commands executed by Proof General or CoqIDE.
- One more tip for CoqIDE users: If you see messages like [Error:
Unable to locate library Basics], a likely reason is
inconsistencies between compiling things _within CoqIDE_ vs _using
[rocq] from the command line_. This typically happens when there
are two incompatible versions of Rocq installed on your
system (one associated with CoqIDE, and one associated with [rocq]
from the terminal). The workaround for this situation is
compiling using CoqIDE only (i.e. choosing "make" from the menu),
and avoiding using [rocq] directly at all. *)
(* ################################################################# *)
(** * Proof by Induction *)
(** We can prove that [0] is a neutral element for [+] on the _left_
using just [reflexivity]. But the proof that it is also a neutral
element on the _right_ ... *)
Theorem add_0_r_firsttry : forall n:nat,
n + 0 = n.
(** ... 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 just fine, but in the branch where [n = S n'] for
some [n'] we get stuck in exactly the same way. *)
Theorem add_0_r_secondtry : forall n:nat,
n + 0 = n.
Proof.
intros n. destruct n as [| n'] eqn:E.
- (* 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 a bit further, but,
since [n] can be arbitrarily large, we'll never get all the way
there if we just go on like this. *)
(** To prove interesting facts about numbers, lists, and other
inductively defined sets, we often need a more powerful reasoning
principle: _induction_.
Recall (from a discrete math course, probably) 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 Rocq, the steps are the same, except we typically encounter them
in reverse order: we begin with the goal of proving [P(n)] for all
[n] and apply the [induction] tactic to break it down 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 add_0_r : forall n:nat, n + 0 = n.
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 a vertical bar, [|]. (Strictly
speaking, we can omit the [as...] clause and Rocq will choose names
for us. In practice, this is a bad practice, as Rocq'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 easily 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' = (S n') + 0], which simplifies to
[S n' = S (n' + 0)], which in turn follows from [IHn']. *)
Theorem minus_n_n : 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, standard, especially useful (basic_induction)
Prove the following using induction. You might need previously
proven results. *)
Theorem mul_0_r : forall n:nat,
n * 0 = 0.
Proof.
intros n.
induction n as [| n' IHn'].
- simpl. reflexivity.
- simpl. rewrite -> IHn'. reflexivity.
Qed.
Theorem plus_n_Sm : forall n m : nat,
S (n + m) = n + (S m).
Proof.
intros n m.
induction n as [| n' IHn'].
- simpl. reflexivity.
- simpl. rewrite -> IHn'. reflexivity.
Qed.
Theorem add_comm : forall n m : nat,
n + m = m + n.
Proof.
intros n m.
induction n as [| n' IHn'].
- simpl. rewrite -> add_0_r. reflexivity.
- simpl. rewrite -> IHn', plus_n_Sm. reflexivity.
Qed.
Theorem add_assoc : forall n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
intros n m p.
induction n as [| n' IHn'].
- simpl. reflexivity.
- simpl. rewrite -> IHn'. reflexivity.
Qed.
(** **** Exercise: 2 stars, standard (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.
intros n.
induction n as [| n' IHn'].
- reflexivity.
- simpl. rewrite -> IHn', plus_n_Sm. reflexivity.
Qed.
(** **** Exercise: 2 stars, standard (eqb_refl)
The following theorem relates the computational equality [=?] on
[nat] with the definitional equality [=] on [bool]. *)
Theorem eqb_refl : forall n : nat,
(n =? n) = true.
Proof.
intros n.
induction n as [| n' IHn'].
- reflexivity.
- simpl. rewrite -> IHn'. reflexivity.
Qed.
(** **** Exercise: 2 stars, standard, optional (even_S)
One inconvenient aspect of our definition of [even n] is the
recursive call on [n - 2]. This makes proofs about [even 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 [even (S n)] that works better
with induction: *)
Theorem even_S : forall n : nat,
even (S n) = negb (even n).
Proof.
intros.
induction n as [| n' IHn'].
- reflexivity.
- rewrite -> IHn', negb_involutive. simpl. reflexivity.
Qed.
(* ################################################################# *)
(** * Proofs Within Proofs *)
(** In Rocq, 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 involve 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 use the required fact
"in place" and then prove it as a separate step. The [replace]
tactic allows us to do this. *)
Theorem mult_0_plus' : forall n m : nat,
(n + 0 + 0) * m = n * m.
Proof.
intros n m.
replace (n + 0 + 0) with n.
- reflexivity.
- rewrite add_comm. simpl. rewrite add_comm. reflexivity.
Qed.
(** The tactic [replace e1 with e2] tactic introduces two subgoals.
The first subgoal is the same as the one at the point where we
invoke [replace], except that [e1] is replaced by [e2]. The
second subgoal is the equality [e1 = e2] itself. *)
(** As another 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 ([add_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 -> add_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 add_comm should do the trick! *)
rewrite add_comm.
(* Doesn't work... Rocq rewrites the wrong plus! :-( *)
Abort.
(** To use [add_comm] at the point where we need it, we can rewrite
[n + m] to [m + n] using [replace] and then prove [n + m = m + n]
using [add_comm]. *)
Theorem plus_rearrange : forall n m p q : nat,
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
replace (n + m) with (m + n).
- reflexivity.
- rewrite add_comm. 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 answer 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 Rocq, 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 probably 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. 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 will 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 reasonably clear standard for judging
proofs good or bad.
Because we are using Rocq 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 add_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.
(** Rocq 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 add_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 Rocq you might 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 that
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 now show that
(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: Rocq 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 usually
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 Rocq proof is
completely implicit, whereas the informal proof reminds the reader
several times where things stand). *)
(** **** Exercise: 2 stars, advanced, optional (add_comm_informal)
Translate your solution for [add_comm] into an informal proof:
Theorem: Addition is commutative.
Theorem add_comm : forall n m : nat,
n + m = m + n.
Proof.
intros n m.
induction n as [| n' IHn'].
- simpl. rewrite -> add_0_r. reflexivity.
- simpl. rewrite -> IHn', plus_n_Sm. reflexivity.
Qed.
Theorem: For any natural number [n] and [m],
n + m = m + n
Proof: By induction on [n].
- First, suppose [n = 0]. We must show that
0 + m = m + 0
This follows directly from the definition of [+] and [add_0_r].
- Next, suppose [n = S n'], where
n' + m = m + n'
We must now show that
S(n') + m = m + S(n')
By the definition of [+], we have that
S(n' + m) = m + S(n')
By [plus_n_Sm], we further have that
S(n' + m) = S(m + n')
By the induction hypothesis, we can see that
S(m + n') = S(m + n')
_Qed_.
*)
(* Do not modify the following line: *)
Definition manual_grade_for_add_comm_informal : option (nat*string) := None.
(** [] *)
(** **** Exercise: 2 stars, standard, optional (eqb_refl_informal)
Write an informal proof of the following theorem, using the
informal proof of [add_assoc] as a model. Don't just
paraphrase the Rocq tactics into English!
Theorem: [(n =? n) = true] for any [n].
Proof: (* FILL IN HERE *)
*)
(* Do not modify the following line: *)
Definition manual_grade_for_eqb_refl_informal : option (nat*string) := None.
(** [] *)
(* ################################################################# *)
(** * More Exercises *)
(** **** Exercise: 3 stars, standard, especially useful (mul_comm)
Use [replace] to help prove [add_shuffle3]. You don't need to
use induction yet. *)
Theorem add_shuffle3 : forall n m p : nat,
n + (m + p) = m + (n + p).
Proof.
intros n m p.
rewrite -> add_assoc, add_assoc.
replace (m + n) with (n + m).
- reflexivity.
- rewrite -> add_comm. reflexivity.
Qed.
(** Now prove commutativity of multiplication. You will probably want
to look for (or define and prove) a "helper" theorem to be used in
the proof of this one. Hint: what is [n * (1 + k)]? *)
Theorem mul_comm : forall m n : nat,
m * n = n * m.
Proof.
intros m n.
induction n as [| n' IHn'].
- rewrite -> mul_0_r.
simpl.
reflexivity.
- simpl.
rewrite <- IHn'.
replace (m * S n') with (m + m * n').
+ reflexivity.
+ rewrite <- mult_n_Sm, add_comm. reflexivity.
Qed.
(** **** Exercise: 3 stars, standard, 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!) *)
(*
This seems like it would need to be proven for all numbers,
so it probably requires induction (c).
"Why is it not (b)?"
Because even if we destruct n into its variants and get rid of
the S() in the second variant, we're left n' + n' which requires our
original hypothesis.
*)
Theorem leb_refl : forall n:nat,
(n <=? n) = true.
Proof.
intros n.
induction n as [| n' IHn'].
- reflexivity.
- simpl. rewrite IHn'. reflexivity.
Qed.
(*
Here, it probably holds to just prove this for both variants of nat,
so maybe (b)
"Why is it not (a)?"
The terms are not just a syntactic reordering of each other.
*)
Theorem zero_neqb_S : forall n:nat,
0 =? (S n) = false.
Proof.
intros n.
destruct n.
- reflexivity.
- reflexivity.
Qed.
(*
This also seems like it would need to be proven for both variants of bool.
Maybe (b)
"Why is it not (a)?"
The terms here are not just a syntactic reordering of each other either.
*)
Theorem andb_false_r : forall b : bool,
andb b false = false.
Proof.
intros b.
destruct b.
- reflexivity.
- reflexivity.
Qed.
(*
This follows from the definition of nat, so maybe (a)
*)
Theorem S_neqb_0 : forall n:nat,
(S n) =? 0 = false.
Proof.
intros n.
reflexivity.
Qed.
(*
This needs to be proven for all n, so maybe (c)
"Why is it not (b)?"
Even if we destruct n, we're going to end up needing the very theorem itself
to get rid of "1 *". Usually, when you require the theorem you're trying to prove
while you're proving it, it's a sign that you should be using induction.
*)
Theorem mult_1_l : forall n:nat, 1 * n = n.
Proof.
intros n.
destruct n.
- reflexivity.
- simpl. rewrite add_comm, plus_O_n. reflexivity.
Qed.
(*
My prediction here was wrong, I noticed while I was solving with induction
and I suddenly didn't need the induction hypothesis. Not quite sure why simpl is
able to replace the "1 *" part.
*)
(* helper theorems for [all3_spec] *)
Theorem negb_or_negb : forall b c : bool, negb b || negb c = negb (b && c).
Proof.
intros b c.
destruct b.
- reflexivity.
- reflexivity.
Qed.
Theorem a_or_negb_a : forall b : bool, b || negb b = true.
Proof.
intros b.
destruct b.
- reflexivity.
- reflexivity.
Qed.
(*
This is probably (b), it does not follow from the definition of the bool structure
itself. We'll have to show that it works for all combinations. We _could_ possibly
rewrite the expression in terms of previously proven theorems about tautologies and
falsums of the logical operators, but that would just be outsourcing the
"check all combinations" work.
*)
Theorem all3_spec : forall b c : bool,
orb
(andb b c)
(orb (negb b)
(negb c))
= true.
Proof.
intros b c.
rewrite negb_or_negb, a_or_negb_a.
reflexivity.
Qed.
(*
I believe this one requires induction to move all the [S()] constructors
from one side to the other, for all possible [nat]s
*)
Theorem mult_plus_distr_r : forall n m p : nat,
(n + m) * p = (n * p) + (m * p).
Proof.
intros n m p.
induction n as [| n' IHn'].
- reflexivity.
- simpl. rewrite IHn'. rewrite add_assoc. reflexivity.
Qed.
(*
I believe this one also requires induction. For (b), we would have to destruct the number
all the way to its base in order to move around the constructors, which effectively
means we would be doing induction.
*)
Theorem mult_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', mult_plus_distr_r. reflexivity.
Qed.
(* ################################################################# *)
(** * Nat to Bin and Back to Nat *)
(** Recall the [bin] type we defined in [Basics]: *)
Inductive bin : Type :=
| Z
| B0 (n : bin)
| B1 (n : bin)
.
(** Before you start working on the next exercise, replace the stub
definitions of [incr] and [bin_to_nat], below, with your solution
from [Basics]. That will make it possible for this file to
be graded on its own. *)
Fixpoint incr (m:bin) : bin :=
match m with
| Z => B1 Z
| B0 b => B1 b
| B1 b => B0 (incr b)
end.
Fixpoint bin_to_nat (m:bin) : nat :=
match m with
| Z => O
| B0 m' => 2 * (bin_to_nat m')
| B1 m' => S (2 * (bin_to_nat m'))
end.
(** In [Basics], we did some unit testing of [bin_to_nat], but we
didn't prove its correctness. Now we'll do so. *)
(** **** Exercise: 3 stars, standard, especially useful (binary_commute)
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.
If you want to change your previous definitions of [incr] or [bin_to_nat]
to make the property easier to prove, feel free to do so! *)
Theorem bin_to_nat_pres_incr : forall b : bin,
bin_to_nat (incr b) = 1 + bin_to_nat b.
Proof.
intros b.
induction b as [| b' _| b'' IHb''].
- reflexivity.
- reflexivity.
- simpl.
rewrite IHb''.
simpl.
rewrite <- plus_n_Sm.
reflexivity.
Qed.
(** **** Exercise: 3 stars, standard (nat_bin_nat) *)
(** Write a function to convert natural numbers to binary numbers. *)
Fixpoint nat_to_bin (n:nat) : bin :=
match n with
| O => Z
| S n' => incr (nat_to_bin n')
end.
(** Prove that, if we start with any [nat], convert it to [bin], and
convert it back, we get the same [nat] which we started with.
Hint: This proof should go through smoothly using the previous
exercise about [incr] as a lemma. If not, revisit your definitions
of the functions involved and consider whether they are more
complicated than necessary: the shape of a proof by induction will
match the recursive structure of the program being verified, so
make the recursions as simple as possible. *)
Theorem nat_bin_nat : forall n, bin_to_nat (nat_to_bin n) = n.
Proof.
intros n.
induction n as [| n' IHn'].
- reflexivity.
- simpl. rewrite bin_to_nat_pres_incr, IHn'. reflexivity.
Qed.
(* ################################################################# *)
(** * Bin to Nat and Back to Bin (Advanced) *)
(** The opposite direction -- starting with a [bin], converting to [nat],
then converting back to [bin] -- turns out to be problematic. That
is, the following theorem does not hold. *)
Theorem bin_nat_bin_fails : forall b, nat_to_bin (bin_to_nat b) = b.
Abort.
(** Let's explore why that theorem fails, and how to prove a modified
version of it. We'll start with some lemmas that might seem
unrelated, but will turn out to be relevant. *)
(** **** Exercise: 2 stars, advanced (double_bin) *)
(** Prove this lemma about [double], which we defined earlier in the
chapter. *)
Lemma double_incr : forall n : nat, double (S n) = S (S (double n)).
Proof.
intros n.
destruct n.
- reflexivity.
- reflexivity.
Qed.
(** Now define a similar doubling function for [bin]. *)
(*
Fixpoint double_bin (b:bin) : bin :=
match b with
| Z => Z
| B0 Z => B1 Z
| B1 Z => B0 (B1 Z)
| B0 n' => B0 (double_bin n')
| B1 n' => B1 (double_bin n')
end.
*)
Definition double_bin (b:bin) : bin :=
nat_to_bin (double (bin_to_nat b)).
(** Check that your function correctly doubles zero. *)
Example double_bin_zero : double_bin Z = Z.
Proof. reflexivity. Qed.
(** Prove this lemma, which corresponds to [double_incr]. *)
Lemma double_incr_bin : forall b,
double_bin (incr b) = incr (incr (double_bin b)).
Proof.
intros b.
induction b as[| b0 IHb0 | b1 IHb1].
- reflexivity.
- simpl.
Abort.
(** [] *)
(** Let's return to our desired theorem: *)
Theorem bin_nat_bin_fails : forall b, nat_to_bin (bin_to_nat b) = b.
Abort.
(** The theorem fails because there are some [bin] such that we won't
necessarily get back to the _original_ [bin], but instead to an
"equivalent" [bin]. (We deliberately leave that notion undefined
here for you to think about.)
Explain in a comment, below, why this failure occurs. Your
explanation will not be graded, but it's important that you get it
clear in your mind before going on to the next part. If you're
stuck on this, think about alternative implementations of
[double_bin] that might have failed to satisfy [double_bin_zero]
yet otherwise seem correct. *)
(* FILL IN HERE *)
(** To solve that problem, we can introduce a _normalization_ function
that selects the simplest [bin] out of all the equivalent
[bin]. Then we can prove that the conversion from [bin] to [nat] and
back again produces that normalized, simplest [bin]. *)
(** **** Exercise: 4 stars, advanced (bin_nat_bin) *)
(** Define [normalize]. You will need to keep its definition as simple
as possible for later proofs to go smoothly. Do not use
[bin_to_nat] or [nat_to_bin], but do use [double_bin].
Hint: Structure the recursion such that it _always_ reaches the
end of the [bin] and _only_ processes each bit only once. Do not
try to "look ahead" at future bits. *)
Fixpoint normalize (b:bin) : bin
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(** It would be wise to do some [Example] proofs to check that your definition of
[normalize] works the way you intend before you proceed. They won't be graded,
but fill them in below. *)
(* FILL IN HERE *)
(** Finally, prove the main theorem. The inductive cases could be a
bit tricky.
Hint: Start by trying to prove the main statement, see where you
get stuck, and see if you can find a lemma -- perhaps requiring
its own inductive proof -- that will allow the main proof to make
progress. We have one lemma for the [B0] case (which also makes
use of [double_incr_bin]) and another for the [B1] case. *)
Theorem bin_nat_bin : forall b, nat_to_bin (bin_to_nat b) = normalize b.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* 2026-01-07 13:17 *)
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