InductionProof by Induction
For the Require Export 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".
PG 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:
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.
- 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.
- In CoqIDE: Open Basics.v; then, in the "Compile" menu, click
on "Compile Buffer".
- From the command line: Generate a Makefile using the coq_makefile
utility, that 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):
- 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 plus_n_O_secondtry : ∀n:nat,
n = n + 0.
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 = n + 0.
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, if we just go on like this
we'll never finish.
To prove interesting facts about numbers, lists, and other
inductively defined sets, we usually need a more powerful
reasoning principle: induction.
Recall (from high school, a discrete math course, etc.) the
principle of induction over natural numbers: If P(n) is some
proposition involving a natural number n and we want to show
that P holds for all numbers n, we can reason like this:
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 plus_n_O : ∀n:nat, n = n + 0.
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_diag : ∀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, recommended (basic_induction)
Prove the following using induction. You might need previously proven results.
Theorem mult_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 plus_comm : ∀n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_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 plus_comm : ∀n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_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, optional (evenb_S)
One inconvenient aspect of our definition of evenb n is the recursive call on n - 2. This makes proofs about evenb n harder when done by induction on n, since we may need an induction hypothesis about n - 2. The following lemma gives an alternative characterization of evenb (S n) that works better with induction:Exercise: 1 star, standard, optional (destruct_induction)
Briefly explain the difference between the tactics destruct and induction.
(* Do not modify the following line: *)
Definition manual_grade_for_destruct_induction : option (nat*string) := None.
☐
Definition manual_grade_for_destruct_induction : option (nat*string) := None.
Proofs Within Proofs
Theorem mult_0_plus' : ∀n m : nat,
(0 + n) * m = n * m.
Proof.
intros n m.
assert (H: 0 + n = n). { reflexivity. }
rewrite → H.
reflexivity. Qed.
(0 + n) * m = n * m.
Proof.
intros n m.
assert (H: 0 + n = n). { reflexivity. }
rewrite → H.
reflexivity. Qed.
The assert tactic introduces two sub-goals. The first is
the assertion itself; by prefixing it with H: we name the
assertion H. (We can also name the assertion with as just as
we did above with destruct and induction, i.e., assert (0 + n
= n) as H.) Note that we surround the proof of this assertion
with curly braces { ... }, both for readability and so that,
when using Coq interactively, we can see more easily when we have
finished this sub-proof. The second goal is the same as the one
at the point where we invoke assert except that, in the context,
we now have the assumption H that 0 + n = n. That is,
assert generates one subgoal where we must prove the asserted
fact and a second subgoal where we can use the asserted fact to
make progress on whatever we were trying to prove in the first
place.
Another example of assert...
For example, suppose we want to prove that (n + m) + (p + q)
= (m + n) + (p + q). The only difference between the two sides of
the = is that the arguments m and n to the first inner +
are swapped, so it seems we should be able to use the
commutativity of addition (plus_comm) to rewrite one into the
other. However, the rewrite tactic is not very smart about
where it applies the rewrite. There are three uses of + here,
and it turns out that doing rewrite → plus_comm will affect
only the outer one...
Theorem plus_rearrange_firsttry : ∀n m p q : nat,
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
(* We just need to swap (n + m) for (m + n)... seems
like plus_comm should do the trick! *)
rewrite → plus_comm.
(* Doesn't work...Coq 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 plus_comm should do the trick! *)
rewrite → plus_comm.
(* Doesn't work...Coq rewrites the wrong plus! *)
Abort.
To use plus_comm at the point where we need it, we can introduce
a local lemma stating that n + m = m + n (for the particular m
and n that we are talking about here), prove this lemma using
plus_comm, and then use it to do the desired rewrite.
Theorem plus_rearrange : ∀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.
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
assert (H: n + m = m + n).
{ rewrite → plus_comm. reflexivity. }
rewrite → H. reflexivity. Qed.
Formal vs. Informal Proof
"Informal proofs are algorithms; formal proofs are code."
Theorem plus_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 plus_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 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:
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
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 fromS (n' + (m + p)) = S ((n' + m) + p),which is immediate from the induction hypothesis. Qed.
- First, suppose n = 0. We must show
Exercise: 2 stars, advanced, recommended (plus_comm_informal)
Translate your solution for plus_comm into an informal proof:
(* Do not modify the following line: *)
Definition manual_grade_for_plus_comm_informal : option (nat*string) := None.
☐
Definition manual_grade_for_plus_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 plus_assoc as a model. Don't just paraphrase the Coq tactics into English!☐
More Exercises
Exercise: 3 stars, standard, recommended (mult_comm)
Use assert to help prove this theorem. You shouldn't need to use induction on plus_swap.
Now prove commutativity of multiplication. You will probably
want to define and prove a "helper" theorem to be used
in the proof of this one. Hint: what is n * (1 + k)?
☐
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!)
Check leb.
Theorem leb_refl : ∀n:nat,
true = (n <=? n).
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_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 plus_ble_compat_l : ∀n m p : nat,
n <=? m = true → (p + n) <=? (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem S_nbeq_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.
☐
Theorem leb_refl : ∀n:nat,
true = (n <=? n).
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_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 plus_ble_compat_l : ∀n m p : nat,
n <=? m = true → (p + n) <=? (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem S_nbeq_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 (eqb_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.)Exercise: 2 stars, standard, 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.Exercise: 3 stars, standard, recommended (binary_commute)
Recall the incr and bin_to_nat functions that you wrote for the binary exercise in the Basics chapter. Prove that the following diagram commutes:incr bin ----------------------> bin | | bin_to_nat | | bin_to_nat | | v v nat ----------------------> nat 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. Name your theorem bin_to_nat_pres_incr ("pres" for "preserves").
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_binary_commute : option (nat*string) := None.
☐
(* Do not modify the following line: *)
Definition manual_grade_for_binary_commute : option (nat*string) := None.
Exercise: 5 stars, advanced (binary_inverse)
This is a further continuation of the previous exercises about binary numbers. You may find you need to go back and change your earlier definitions to get things to work here.
Prove that, if we start with any nat, convert it to binary, and
convert it back, we get the same nat we started with. (Hint: If
your definition of nat_to_bin involved any extra functions, you
may need to prove a subsidiary lemma showing how such functions
relate to nat_to_bin.)
Theorem nat_bin_nat : ∀n, bin_to_nat (nat_to_bin n) = n.
Proof.
(* FILL IN HERE *) Admitted.
(* Do not modify the following line: *)
Definition manual_grade_for_binary_inverse_a : option (nat*string) := None.
Proof.
(* FILL IN HERE *) Admitted.
(* Do not modify the following line: *)
Definition manual_grade_for_binary_inverse_a : option (nat*string) := None.
(b) One might naturally expect that we should also prove the
opposite direction — that starting with a binary number,
converting to a natural, and then back to binary should yield
the same number we started with. However, this is not the
case! Explain (in a comment) what the problem is.
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_binary_inverse_b : option (nat*string) := None.
(* Do not modify the following line: *)
Definition manual_grade_for_binary_inverse_b : option (nat*string) := None.
(c) Define a normalization function — i.e., a function
normalize going directly from bin to bin (i.e., not by
converting to nat and back) such that, for any binary number
b, converting b to a natural and then back to binary yields
(normalize b). Prove it. (Warning: This part is a bit
tricky — you may end up defining several auxiliary lemmas.
One good way to find out what you need is to 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.) Don't
define this using nat_to_bin and bin_to_nat!
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_binary_inverse_c : option (nat*string) := None.
☐
(* Do not modify the following line: *)
Definition manual_grade_for_binary_inverse_c : option (nat*string) := None.
(* Fri 30 Aug 2019 02:44:44 PM CEST *)