InductionProof by Induction
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coqc to compile Basics.v into Basics.vo. This is like
making a .class file from a .java file, or a .o file from a .c
file. There are two ways to do it:
- In CoqIDE:
- From the command line:
Proof by Induction
... cannot be proved in the same simple way. Just applying
reflexivity doesn't work, since the n in n + 0 is an arbitrary
unknown number, so the match in the definition of + can't be
simplified.
Proof.
intros n.
simpl. (* Does nothing! *)
Abort.
And reasoning by cases using destruct n doesn't get us much
further: the branch of the case analysis where we assume n = 0
goes through fine, but in the branch where n = S n' for some n' we
get stuck in exactly the same way. We could use destruct n' to
get one step further, but, since n can be arbitrarily large, if we
try to keep on like this we'll never be done.
Theorem plus_n_O_secondtry : ∀n:nat,
n = n + 0.
Proof.
intros n. destruct n as [| n'].
- (* n = 0 *)
reflexivity. (* so far so good... *)
- (* n = S n' *)
simpl. (* ...but here we are stuck again *)
Abort.
To prove interesting facts about numbers, lists, and other
inductively defined sets, we usually need a more powerful
reasoning principle: induction.
Recall (from high school, a discrete math course, etc.) the
principle of induction over natural numbers: If P(n) is some
proposition involving a natural number n and we want to show
that P holds for all numbers n, we can reason like this:
In Coq, the steps are the same but the order is backwards: we
begin with the goal of proving P(n) for all n and break it
down (by applying the induction tactic) into two separate
subgoals: first showing P(O) and then showing P(n') → P(S
n'). Here's how this works for the theorem at hand:
- 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.
Like destruct, the induction tactic takes an as...
clause that specifies the names of the variables to be introduced
in the subgoals. In the first branch, n is replaced by 0 and
the goal becomes 0 + 0 = 0, which follows by simplification. In
the second, n is replaced by S n' and the assumption n' + 0 =
n' is added to the context (with the name IHn', i.e., the
Induction Hypothesis for n' — notice that this name is
explicitly chosen in the as... clause of the call to induction
rather than letting Coq choose one arbitrarily). The goal in this
case becomes (S n') + 0 = S n', which simplifies to S (n' + 0)
= S n', which in turn follows from IHn'.
Theorem minus_diag : ∀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, 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.
Use induction to prove this simple fact about double:
☐
Exercise: 2 stars, optional (evenb_S)
One inconveninent aspect of our definition of evenb n is that it may need to perform a recursive call on n - 2. This makes proofs about evenb n harder when done by induction on n, since we may need an induction hypothesis about n - 2. The following lemma gives a better characterization of evenb (S n):
☐
(* FILL IN HERE *)
☐
Exercise: 1 star (destruct_induction)
Briefly explain the difference between the tactics destruct and induction.☐
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.
The assert tactic introduces two sub-goals. The first is
the assertion itself; by prefixing it with H: we name the
assertion H. (We can also name the assertion with as just as
we did above with destruct and induction, i.e., assert (0 + n
= n) as H.) Note that we surround the proof of this assertion
with curly braces { ... }, both for readability and so that,
when using Coq interactively, we can see more easily when we have
finished this sub-proof. The second goal is the same as the one
at the point where we invoke assert except that, in the context,
we now have the assumption H that 0 + n = n. That is,
assert generates one subgoal where we must prove the asserted
fact and a second subgoal where we can use the asserted fact to
make progress on whatever we were trying to prove in the first
place.
The assert tactic is handy in many sorts of situations. For
example, suppose we want to prove that (n + m) + (p + q) = (m +
n) + (p + q). The only difference between the two sides of the
= is that the arguments m and n to the first inner + are
swapped, so it seems we should be able to use the commutativity of
addition (plus_comm) to rewrite one into the other. However,
the rewrite tactic is a little stupid about where it applies
the rewrite. There are three uses of + here, and it turns out
that doing rewrite → plus_comm will affect only the outer
one...
Theorem plus_rearrange_firsttry : ∀n m p q : nat,
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
(* We just need to swap (n + m) for (m + n)...
it seems like plus_comm should do the trick! *)
rewrite → plus_comm.
(* Doesn't work...Coq rewrote the wrong plus! *)
Abort.
To get plus_comm to apply at the point where we want it to, we
can introduce a local lemma stating that n + m = m + n (for the
particular m and n that we are talking about here), prove this
lemma using plus_comm, and then use it to do the desired
rewrite.
Theorem plus_rearrange : ∀n m p q : nat,
(n + m) + (p + q) = (m + n) + (p + q).
Proof.
intros n m p q.
assert (H: n + m = m + n).
{ rewrite → plus_comm. reflexivity. }
rewrite → H. reflexivity. Qed.
More Exercises
Exercise: 3 stars, recommended (mult_comm)
Use assert to help prove this theorem. You shouldn't need to use induction on plus_swap.
Now prove commutativity of multiplication. (You will probably
need to define and prove a separate subsidiary theorem to be used
in the proof of this one. You may find that plus_swap comes in
handy.)
☐
Exercise: 3 stars, optional (more_exercises)
Take a piece of paper. For each of the following theorems, first think about whether (a) it can be proved using only simplification and rewriting, (b) it also requires case analysis (destruct), or (c) it also requires induction. Write down your prediction. Then fill in the proof. (There is no need to turn in your piece of paper; this is just to encourage you to reflect before you hack!)Theorem leb_refl : ∀n:nat,
true = leb n n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_S : ∀n:nat,
beq_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,
leb n m = true → leb (p + n) (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem S_nbeq_0 : ∀n:nat,
beq_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, optional (beq_nat_refl)
Prove the following theorem. (Putting the true on the left-hand side of the equality may look odd, but this is how the theorem is stated in the Coq standard library, so we follow suit. Rewriting works equally well in either direction, so we will have no problem using the theorem no matter which way we state it.)
☐
Use the replace tactic to do a proof of plus_swap', just like
plus_swap but without needing assert (n + m = m + n).
Exercise: 2 stars, optional (plus_swap')
The replace tactic allows you to specify a particular subterm to rewrite and what you want it rewritten to: replace (t) with (u) replaces (all copies of) expression t in the goal by expression u, and generates t = u as an additional subgoal. This is often useful when a plain rewrite acts on the wrong part of the goal.
☐
Before you start working on this exercise, please copy the
definitions from your solution to the binary exercise here so
that this file can be graded on its own. If you find yourself
wanting to change your original definitions to make the property
easier to prove, feel free to do so!
Exercise: 3 stars, 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:
bin --------- incr -------> bin
| |
bin_to_nat bin_to_nat
| |
v v
nat ---------- S ---------> nat
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.
Name your theorem bin_to_nat_pres_incr ("pres" for "preserves").
| |
bin_to_nat bin_to_nat
| |
v v
nat ---------- S ---------> nat
(* FILL IN HERE *)
☐
(a) First, write a function to convert natural numbers to binary
numbers. Then prove that starting with any natural number,
converting to binary, then converting back yields the same
natural number you started with.
(b) You might naturally think that we should also prove the
opposite direction: that starting with a binary number,
converting to a natural, and then back to binary yields the
same number we started with. However, this is not true!
Explain what the problem is.
(c) Define a "direct" normalization function — i.e., a function
normalize from binary numbers to binary numbers such that,
for any binary number b, converting to a natural and then back
to binary yields (normalize b). Prove it. (Warning: This
part is tricky!)
Again, feel free to change your earlier definitions if this helps
here.
Exercise: 5 stars, advanced (binary_inverse)
This exercise is a continuation of the previous exercise about binary numbers. You will need your definitions and theorems from there to complete this one.(* FILL IN HERE *)
☐
Formal vs. Informal Proof (Optional)
"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.
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.
... 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: true = beq_nat n n for any n.
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:☐
Exercise: 2 stars, optional (beq_nat_refl_informal)
Write an informal proof of the following theorem, using the informal proof of plus_assoc as a model. Don't just paraphrase the Coq tactics into English!☐