(** * 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 Export] command to work, Coq needs to be
able to find a compiled version of [Basics.v], called [Basics.vo],
in a directory associated with the prefix [LF]. This file is
analogous to the [.class] files compiled from [.java] source files
and the [.o] files compiled from [.c] files.
First create a file named [_CoqProject] containing the following
line (if you obtained the whole volume "Logical Foundations" as a
single archive, a [_CoqProject] should already exist and you can
skip this step):
-Q . LF
This maps the current directory ("[.]", which contains [Basics.v],
[Induction.v], etc.) to the prefix (or "logical directory")
"[LF]". Proof General and CoqIDE read [_CoqProject]
automatically, so they know to where to look for the file
[Basics.vo] corresponding to the library [LF.Basics].
Once [_CoqProject] is thus created, there are various ways to
build [Basics.vo]:
- In Proof General or CoqIDE, the compilation should happen
automatically when you submit the [Require] line above to PG.
- For VSCode users, open the terminal pane at the bottom
and then use the command line instructions below.
(If you downloaded the project setup .tgz file, just doing `make`
should build all the code.)
- If you want to compile from the command line, generate a
[Makefile] using the [coq_makefile] utility, which comes
installed with Coq (if you obtained the whole volume as a
single archive, a [Makefile] should already exist and you can
skip this step):
coq_makefile -f _CoqProject *.v -o Makefile
Note: You should rerun that command whenever you add or remove
Coq files to the 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 Coq compiler, [coqc]. You can
also run [coqc] directly:
coqc -Q . LF Basics.v
But [make] also calculates dependencies between source files to
compile them in the right order, so [make] should generally be
preferred over explicit [coqc].
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.
In particular, if you see a message like
Compiled library Foo makes inconsistent assumptions over
library Bar
check whether you have multiple installations of Coq on your
machine. It may be that commands (like [coqc]) that you execute
in a terminal window are getting a different version of Coq than
commands executed by Proof General or CoqIDE.
- Another common reason is that the library [Bar] was modified and
recompiled without also recompiling [Foo] which depends on it.
Recompile [Foo], or everything if too many files are
affected. (Using the third solution above: [make clean; make].)
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
[coqc] from the command line_. This typically happens when there
are two incompatible versions of [coqc] installed on your
system (one associated with CoqIDE, and one associated with [coqc]
from the terminal). The workaround for this situation is
compiling using CoqIDE only (i.e. choosing "make" from the menu),
and avoiding using [coqc] 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 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 one step further, but,
since [n] can be arbitrarily large, we'll never get all the 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 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 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 [|]. (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' = (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.
(* FILL IN HERE *) Admitted.
Theorem plus_n_Sm : forall n m : nat,
S (n + m) = n + (S m).
Proof.
(* FILL IN HERE *) Admitted.
Theorem add_comm : forall n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem add_assoc : forall n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################################# *)
(** * 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 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 state and prove the
needed "sub-theorem" right at the point where it is used. The
[assert] tactic allows us to do this. *)
Theorem mult_0_plus' : forall n m : nat,
(n + 0 + 0) * m = n * m.
Proof.
intros n m.
assert (H: n + 0 + 0 = n).
{ rewrite add_comm. simpl. rewrite add_comm. 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 (n + 0
+ 0 = 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 [n + 0 + 0 = 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. *)
(** 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... Coq rewrites the wrong plus! :-( *)
Abort.
(** To use [add_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
[add_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 add_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 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.
(** 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 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 Coq 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: 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, especially useful (add_comm_informal)
Translate your solution for [add_comm] into an informal proof:
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)
*)
(* 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 Coq 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 [assert] 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.
(* FILL IN HERE *) Admitted.
(** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, standard, optional (plus_leb_compat_l)
If a hypothesis has the form [H: P -> a = b], then [rewrite H] will
rewrite [a] to [b] in the goal, and add [P] as a new subgoal. Use
that in the inductive step of this exercise. *)
Check leb.
Theorem plus_leb_compat_l : forall n m p : nat,
n <=? m = true -> (p + n) <=? (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** 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!) *)
Theorem leb_refl : forall n:nat,
(n <=? n) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_neqb_S : forall n: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 S_neqb_0 : forall n: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, standard, optional (add_shuffle3')
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 [add_shuffle3'], just like
[add_shuffle3] but without needing [assert]. *)
Theorem add_shuffle3' : forall n m p : nat,
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################################# *)
(** * 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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Fixpoint bin_to_nat (m:bin) : nat
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, standard (nat_bin_nat) *)
(** Write a function to convert natural numbers to binary numbers. *)
Fixpoint nat_to_bin (n:nat) : bin
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(** 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.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################################# *)
(** * 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.
(* FILL IN HERE *) Admitted.
(** Now define a similar doubling function for [bin]. *)
Definition double_bin (b:bin) : bin
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(** Check that your function correctly doubles zero. *)
Example double_bin_zero : double_bin Z = Z.
(* FILL IN HERE *) Admitted.
(** Prove this lemma, which corresponds to [double_incr]. *)
Lemma double_incr_bin : forall b,
double_bin (incr b) = incr (incr (double_bin b)).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** 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 process 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.
(** [] *)
(* 2023-12-29 17:12 *)