InductionProof by Induction
Separate Compilation
For this Require Export command to work, Coq needs to be
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and the .o files compiled from .c files.
First create a file named _CoqProject containing the following
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-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.
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Compiled library Foo makes inconsistent assumptions over
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Compiled library Foo makes inconsistent assumptions over
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- 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.)
Proof by Induction
... 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.
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 : ∀ 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.
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:
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:
- 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.
Theorem add_0_r : ∀ n:nat, n + 0 = n.
Proof.
intros n. induction n as [| n' IHn'].
- (* n = 0 *) reflexivity.
- (* n = S n' *) simpl. rewrite → IHn'. reflexivity. Qed.
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 : ∀ 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.
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 : ∀ n:nat,
n × 0 = 0.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_n_Sm : ∀ n m : nat,
S (n + m) = n + (S m).
Proof.
(* FILL IN HERE *) Admitted.
Theorem add_comm : ∀ n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem add_assoc : ∀ n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
(* FILL IN HERE *) Admitted.
☐
n × 0 = 0.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_n_Sm : ∀ n m : nat,
S (n + m) = n + (S m).
Proof.
(* FILL IN HERE *) Admitted.
Theorem add_comm : ∀ n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem add_assoc : ∀ 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:
Use induction to prove this simple fact about double:
Exercise: 2 stars, standard (eqb_refl)
The following theorem relates the computational equality =? on nat with the definitional equality = on bool.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:Proofs Within Proofs
Theorem mult_0_plus' : ∀ 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.
(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 : ∀ 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.
(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 : ∀ 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.
(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."
Theorem add_assoc' : ∀ 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.
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'' : ∀ 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.
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:
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).
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)
- 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.
- First, suppose n = 0. We must show that
Exercise: 2 stars, advanced, especially useful (add_comm_informal)
Translate your solution for add_comm into an informal proof:
(* Do not modify the following line: *)
Definition manual_grade_for_add_comm_informal : option (nat×string) := None.
☐
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!
(* Do not modify the following line: *)
Definition manual_grade_for_eqb_refl_informal : option (nat×string) := None.
☐
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 : ∀ n m p : nat,
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.
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)?
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 : ∀ n m p : nat,
n <=? m = true → (p + n) <=? (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
Theorem plus_leb_compat_l : ∀ 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 : ∀ n:nat,
(n <=? n) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_neqb_S : ∀ n:nat,
0 =? (S n) = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem andb_false_r : ∀ b : bool,
andb b false = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem S_neqb_0 : ∀ n:nat,
(S n) =? 0 = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_1_l : ∀ n:nat, 1 × n = n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem all3_spec : ∀ b c : bool,
orb
(andb b c)
(orb (negb b)
(negb c))
= true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_plus_distr_r : ∀ n m p : nat,
(n + m) × p = (n × p) + (m × p).
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_assoc : ∀ n m p : nat,
n × (m × p) = (n × m) × p.
Proof.
(* FILL IN HERE *) Admitted.
☐
(n <=? n) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_neqb_S : ∀ n:nat,
0 =? (S n) = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem andb_false_r : ∀ b : bool,
andb b false = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem S_neqb_0 : ∀ n:nat,
(S n) =? 0 = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_1_l : ∀ n:nat, 1 × n = n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem all3_spec : ∀ b c : bool,
orb
(andb b c)
(orb (negb b)
(negb c))
= true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_plus_distr_r : ∀ n m p : nat,
(n + m) × p = (n × p) + (m × p).
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_assoc : ∀ 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.
Theorem add_shuffle3' : ∀ n m p : nat,
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.
☐
n + (m + p) = m + (n + p).
Proof.
(* FILL IN HERE *) Admitted.
☐
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.
(* 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.
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!
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 SThat 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.
Theorem bin_to_nat_pres_incr : ∀ b : bin,
bin_to_nat (incr b) = 1 + bin_to_nat b.
Proof.
(* FILL IN HERE *) Admitted.
☐
bin_to_nat (incr b) = 1 + bin_to_nat b.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 3 stars, standard (nat_bin_nat)
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.
Bin to Nat and Back to Bin (Advanced)
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.
Prove this lemma about double, which we defined earlier in the
chapter.
Exercise: 2 stars, advanced (double_bin)
Now define a similar doubling function for bin.
Definition double_bin (b:bin) : bin
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Check that your function correctly doubles zero.
Prove this lemma, which corresponds to double_incr.
Lemma double_incr_bin : ∀ b,
double_bin (incr b) = incr (incr (double_bin b)).
Proof.
(* FILL IN HERE *) Admitted.
☐
double_bin (incr b) = incr (incr (double_bin b)).
Proof.
(* FILL IN HERE *) Admitted.
☐
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.
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.
Exercise: 4 stars, advanced (bin_nat_bin)
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 B_{0} case (which also makes
use of double_incr_bin) and another for the B_{1} case.
Theorem bin_nat_bin : ∀ b, nat_to_bin (bin_to_nat b) = normalize b.
Proof.
(* FILL IN HERE *) Admitted.
☐
Proof.
(* FILL IN HERE *) Admitted.
☐
(* 2023-12-29 17:12 *)