InductionProof by Induction
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We proved in the last chapter that 0 is a neutral element
for + on the left, using an easy argument based on
simplification. We also observed that proving the fact that it is
also a neutral element on the right...
- In CoqIDE:
- From the command line: Either
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'].
- (* 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'].
- (* 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.
Here is 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.
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'.
Here's the same proof, as written on paper:
- Theorem: For any n,
n = n + 0Proof: By induction on n.
- First, suppose n = 0. We must show
0 = 0 + 0.This follows directly from the definition of +.
- Next, suppose n = S n', where (as our IH)
n' = n' + 0.We must showS n' = (S n') + 0.By the definition of +, this follows fromS n' = S (n' + 0),which is immediate from the induction hypothesis. Qed.
- First, suppose n = 0. We must show
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, 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.
(* GRADE_THEOREM 0.5: mult_0_r *)
(* GRADE_THEOREM 0.5: plus_n_Sm *)
(* GRADE_THEOREM 0.5: plus_comm *)
(* GRADE_THEOREM 0.5: plus_assoc *)
☐
☐
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.
(* GRADE_THEOREM 0.5: mult_0_r *)
(* GRADE_THEOREM 0.5: plus_n_Sm *)
(* GRADE_THEOREM 0.5: plus_comm *)
(* GRADE_THEOREM 0.5: plus_assoc *)
Use induction to prove this simple fact about double:
☐
Exercise: 2 stars, 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 (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.
(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 rewrote 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 rewrote 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: 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, recommended (plus_comm_informal)
Translate your solution for plus_comm into an informal proof:☐
Exercise: 2 stars (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!☐
Exercise: 3 stars, recommended (binary_commute)
Recall the incr and bin_to_nat functions from the optional binary exercise in the Basics chapter. (If you haven't taken a look, take a look now.) Here's one way to solve it:
Inductive bin : Type :=
| BZ : bin (* binary zero *)
| T2 : bin → bin (* twice a binary number *)
| T2P1 : bin → bin. (* twice a binary number plus 1 *)
Fixpoint incr (m:bin) : bin :=
match m with
| BZ ⇒ T2P1 BZ
| T2 m' ⇒ T2P1 m'
| T2P1 m' ⇒ T2 (incr m')
end.
Fixpoint bin_to_nat (m:bin) : nat :=
match m with
| BZ ⇒ O
| T2 m' ⇒ 2 * bin_to_nat m'
| T2P1 m' ⇒ 1 + 2 * bin_to_nat m'
end.
| BZ : bin (* binary zero *)
| T2 : bin → bin (* twice a binary number *)
| T2P1 : bin → bin. (* twice a binary number plus 1 *)
Fixpoint incr (m:bin) : bin :=
match m with
| BZ ⇒ T2P1 BZ
| T2 m' ⇒ T2P1 m'
| T2P1 m' ⇒ T2 (incr m')
end.
Fixpoint bin_to_nat (m:bin) : nat :=
match m with
| BZ ⇒ O
| T2 m' ⇒ 2 * bin_to_nat m'
| T2P1 m' ⇒ 1 + 2 * bin_to_nat m'
end.
Prove (in Coq) 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.
Name your theorem bin_to_nat_pres_incr ("pres" for "preserves").
bin ----------------------> bin
| |
bin_to_nat | | bin_to_nat
| |
v v
nat ----------------------> nat
S
(* FILL IN HERE *)
☐