Induction
Overview
Theorem plus_n_O : ∀ n:nat, n = n + 0.
Proof.
intros n. induction n as [| n' IHn'].
- (* n = 0 *) reflexivity.
- (* n = S n' *) simpl plus. rewrite <- IHn'. reflexivity. Qed.
Proof.
intros n. induction n as [| n' IHn'].
- (* n = 0 *) reflexivity.
- (* n = S n' *) simpl plus. 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 + 0 Proof: 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 show
S n' = (S n') + 0. By the definition of +, this follows from
S 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 minus. reflexivity.
- (* n = S n' *)
simpl minus. rewrite → IHn'. reflexivity. Qed.
minus n n = 0.
Proof.
(* WORKED IN CLASS *)
intros n. induction n as [| n' IHn'].
- (* n = 0 *)
simpl minus. reflexivity.
- (* n = S n' *)
simpl minus. 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 (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.
☐
Fixpoint double (n:nat) :=
match n with
| O ⇒ O
| S n' ⇒ S (S (double n'))
end.
Lemma double_plus : ∀ n, double n = n + n.
Proof.
(* FILL IN HERE *) Admitted.
☐
match n with
| O ⇒ O
| S n' ⇒ S (S (double n'))
end.
Lemma double_plus : ∀ n, double n = n + n.
Proof.
(* FILL IN HERE *) Admitted.
☐
(* As we've done on previous days, write an informal proof of
double_plus in a comment. You can follow the basic outline of
the informal proof of plus_n_O above. Be sure to state your goal
and IH in each case! *)
(* Do not modify the following line: *)
Definition manual_grade_for_double_plus_informal : option (nat×string) := None.
☐
double_plus in a comment. You can follow the basic outline of
the informal proof of plus_n_O above. Be sure to state your goal
and IH in each case! *)
(* Do not modify the following line: *)
Definition manual_grade_for_double_plus_informal : option (nat×string) := None.
☐
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 = leb n n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_S : ∀ n:nat,
eqb 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,
eqb (S n) 0 = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_1_l : ∀ n:nat, 1 × n = n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem leb_refl : ∀ n:nat,
true = leb n n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_S : ∀ n:nat,
eqb 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,
eqb (S n) 0 = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_1_l : ∀ n:nat, 1 × n = n.
Proof.
(* FILL IN HERE *) Admitted.
A good additional exercise is to draw out the truth table for the
following combination of boolean operations, for all b and
c.
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.
☐
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.)Applying Theorems to Arguments
Coq prints the statement of the plus_comm theorem in the same
way that it prints the type of any term that we ask it to
Check. Why?
The reason is that the identifier plus_comm actually refers to a
proof object -- a data structure that represents a logical
derivation establishing of the truth of the statement ∀ n m
: nat, n + m = m + n. The type of this object is the statement
of the theorem that it is a proof of.
Intuitively, this makes sense because the statement of a theorem
tells us what we can use that theorem for, just as the type of a
computational object tells us what we can do with that object --
e.g., if we have a term of type nat → nat → nat, we can give
it two nats as arguments and get a nat back. Similarly, if we
have an object of type n = m → n + n = m + m and we provide it
an "argument" of type n = m, we can derive n + n = m + m.
Operationally, this analogy goes even further: by applying a
theorem, as if it were a function, to hypotheses with matching
types, we can specialize its result without having to resort to
intermediate assertions. For example, suppose we wanted to prove
the following result:
It appears at first sight that we ought to be able to prove this
by rewriting with plus_comm twice to make the two sides match.
The problem, however, is that the second rewrite will undo the
effect of the first.
Proof.
intros x y z.
rewrite plus_comm.
rewrite plus_comm.
(* We are back where we started... *)
Abort.
intros x y z.
rewrite plus_comm.
rewrite plus_comm.
(* We are back where we started... *)
Abort.
One simple way of fixing this problem, using only tools that we
already know, is to use assert to derive a specialized version
of plus_comm that can be used to rewrite exactly where we
want. Note that the H we get out is specialized to y and z,
unlike the universally quantified plus_comm.
Lemma plus_comm3_take2 :
∀ x y z, x + (y + z) = (z + y) + x.
Proof.
intros x y z.
rewrite plus_comm.
assert (H : y + z = z + y).
{ rewrite plus_comm. reflexivity. }
rewrite H.
reflexivity.
Qed.
∀ x y z, x + (y + z) = (z + y) + x.
Proof.
intros x y z.
rewrite plus_comm.
assert (H : y + z = z + y).
{ rewrite plus_comm. reflexivity. }
rewrite H.
reflexivity.
Qed.
A more elegant alternative is to apply plus_comm directly to the
arguments we want to instantiate it with, in much the same way as
we apply a polymorphic function to a type argument.
Lemma plus_comm3_take3 :
∀ x y z, x + (y + z) = (z + y) + x.
Proof.
intros x y z.
rewrite plus_comm.
rewrite (plus_comm y z).
reflexivity.
Qed.
∀ x y z, x + (y + z) = (z + y) + x.
Proof.
intros x y z.
rewrite plus_comm.
rewrite (plus_comm y z).
reflexivity.
Qed.
You can "use theorems as functions" in this way with almost all
tactics that take a theorem name as an argument. Recall that we
did this with apply:
Example lemma_application_ex1 :
∀ (n m : nat),
47 = n →
n = m →
47 = m.
Proof.
intros n m Hn Hnm.
apply (trans_eq _ _ _ _ Hn Hnm).
Qed.
∀ (n m : nat),
47 = n →
n = m →
47 = m.
Proof.
intros n m Hn Hnm.
apply (trans_eq _ _ _ _ Hn Hnm).
Qed.
Our second example uses this helper lemma from day 12 to use
destruct on an applied theorem.
Lemma orb_true :
∀ a b : bool, a || b = true → a = true ∨ b = true.
Proof.
intros a b. apply orb_true_iff.
Qed.
Example lemma_application_ex2 :
∀ b : bool, false || b = true → b = true.
Proof.
intros b H. destruct (orb_true _ _ H) as [Hcontra | Hb].
- discriminate Hcontra.
- apply Hb.
Qed.
∀ a b : bool, a || b = true → a = true ∨ b = true.
Proof.
intros a b. apply orb_true_iff.
Qed.
Example lemma_application_ex2 :
∀ b : bool, false || b = true → b = true.
Proof.
intros b H. destruct (orb_true _ _ H) as [Hcontra | Hb].
- discriminate Hcontra.
- apply Hb.
Qed.
We will see many more examples of the idioms from this section in
later chapters.
Structural induction
- First, show that P is true of l when l is nil.
- Then show that P is true of l when l is cons n l' for some number n and some smaller list l', assuming that P is true for l'.
Theorem app_assoc : ∀ A (l m n:list A),
l ++ m ++ n = (l ++ m) ++ n.
Proof.
intros A l m n.
induction l as [|a l' IH].
- reflexivity.
- simpl app. rewrite IH. reflexivity.
Qed.
l ++ m ++ n = (l ++ m) ++ n.
Proof.
intros A l m n.
induction l as [|a l' IH].
- reflexivity.
- simpl app. rewrite IH. reflexivity.
Qed.
Notice that, as when doing induction on natural numbers, the
as... clause provided to the induction tactic gives a name to
the induction hypothesis corresponding to the smaller list l1'
in the cons case. Once again, this Coq proof is not especially
illuminating as a static written document -- it is easy to see
what's going on if you are reading the proof in an interactive Coq
session and you can see the current goal and context at each
point, but this state is not visible in the written-down parts of
the Coq proof. So a natural-language proof -- one written for
human readers -- will need to include more explicit signposts; in
particular, it will help the reader stay oriented if we remind
them exactly what the induction hypothesis is in the second
case.
Theorem: For all A-lists l1, l2, and l3,
(l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof: Given a type A, let lists l1, l2, and l3 be given.
By induction on l1.
- First, suppose l1 = []. We must show
([] ++ l2) ++ l3 = [] ++ (l2 ++ l3), which follows directly from the definition of ++. - Next, suppose l1 = n::l1', with
(l1' ++ l2) ++ l3 = l1' ++ (l2 ++ l3) (the induction hypothesis). We must show
((a :: l1') ++ l2) ++ l3 = (a :: l1') ++ (l2 ++ l3). By the definition of ++, this follows from
a :: ((l1' ++ l2) ++ l3) = a :: (l1' ++ (l2 ++ l3)), which is immediate from the induction hypothesis. ☐
Practice Exercises
Exercise: 2 stars, standard, optional (poly_exercises)
Here are two simple exercises; they're optional, just for practice.
Theorem app_nil_r : ∀ (X:Type), ∀ l:list X,
l ++ [] = l.
Proof.
(* FILL IN HERE *) Admitted.
Lemma app_length : ∀ (X:Type) (l1 l2 : list X),
length (l1 ++ l2) = length l1 + length l2.
Proof.
(* FILL IN HERE *) Admitted.
☐
l ++ [] = l.
Proof.
(* FILL IN HERE *) Admitted.
Lemma app_length : ∀ (X:Type) (l1 l2 : list X),
length (l1 ++ l2) = length l1 + length l2.
Proof.
(* FILL IN HERE *) Admitted.
☐
(* You can do structural induction on anything defined using
`Inductive`. Here's a lemma about binary trees. *)
Fixpoint size {X : Type} (t : bt X) : nat :=
match t with
| empty ⇒ 0
| node l _ r ⇒ 1 + size l + size r
end.
Lemma size_length_inorder : ∀ (X : Type) (t : bt X),
size t = length (inorder t).
Proof.
induction t as [| l IHl v r IHr].
- reflexivity.
- simpl size.
simpl inorder.
rewrite app_length.
simpl length.
rewrite <- plus_n_Sm.
rewrite IHl.
rewrite IHr.
reflexivity.
Qed.
`Inductive`. Here's a lemma about binary trees. *)
Fixpoint size {X : Type} (t : bt X) : nat :=
match t with
| empty ⇒ 0
| node l _ r ⇒ 1 + size l + size r
end.
Lemma size_length_inorder : ∀ (X : Type) (t : bt X),
size t = length (inorder t).
Proof.
induction t as [| l IHl v r IHr].
- reflexivity.
- simpl size.
simpl inorder.
rewrite app_length.
simpl length.
rewrite <- plus_n_Sm.
rewrite IHl.
rewrite IHr.
reflexivity.
Qed.
Exercise: 2 stars, standard, optional (more_poly_exercises)
Here are some slightly more interesting ones...
Theorem rev_app_distr: ∀ X (l1 l2 : list X),
rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof.
(* FILL IN HERE *) Admitted.
Theorem rev_involutive : ∀ X : Type, ∀ l : list X,
rev (rev l) = l.
Proof.
(* FILL IN HERE *) Admitted.
☐
rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof.
(* FILL IN HERE *) Admitted.
Theorem rev_involutive : ∀ X : Type, ∀ l : list X,
rev (rev l) = l.
Proof.
(* FILL IN HERE *) Admitted.
☐
Lemma map_length :
∀ A B (f:A→B) (l:list A),
length (map f l) = length l.
Proof.
(* FILL IN HERE *) Admitted.
☐
∀ A B (f:A→B) (l:list A),
length (map f l) = length l.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 3 stars, standard (map_rev)
Show that map and rev commute. You may need to define an auxiliary lemma.
Theorem map_rev : ∀ (X Y : Type) (f : X → Y) (l : list X),
map f (rev l) = rev (map f l).
Proof.
(* FILL IN HERE *) Admitted.
☐
map f (rev l) = rev (map f l).
Proof.
(* FILL IN HERE *) Admitted.
☐
Fixpoint eq_strand (dna1 dna2 : strand) : bool :=
match (dna1, dna2) with
| ([], []) ⇒ true
| ([], _) ⇒ false
| (_, []) ⇒ false
| (b1 :: dna1', b2 :: dna2') ⇒
eq_base b1 b2 && eq_strand dna1' dna2'
end.
Lemma eq_strand_refl : ∀ (dna : strand),
eq_strand dna dna = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
match (dna1, dna2) with
| ([], []) ⇒ true
| ([], _) ⇒ false
| (_, []) ⇒ false
| (b1 :: dna1', b2 :: dna2') ⇒
eq_base b1 b2 && eq_strand dna1' dna2'
end.
Lemma eq_strand_refl : ∀ (dna : strand),
eq_strand dna dna = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
More Exercises
Exercise: 3 stars, standard, optional (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.)
A hint on helper lemmas: rather than just trying to prove some
other helper lemma, try stating it and using Admitted to treat
it as proven. Then check and see if your helper lemma actually,
well, helps! If it does, great---it's worth trying to prove
it. If not, well... find something else.
Relatedly, the admit tactic is a nice way to ignore a case you
don't want to bother proving for now. If you've used admit, then
you won't be able to finish the proof with a Qed (only
Admitted), but at least you can look ahead!
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.
Theorem plus_swap' : ∀ 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.
☐
One last hard one
Exercise: 4 stars, standard (bin_nat_bin)
Recall the incr and bin_to_nat functions from our early work on recursion. Here's one way to solve it:
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.
Fixpoint nat_to_bin (n:nat) : bin :=
match n with
| O ⇒ BZ
| S n' ⇒ incr (nat_to_bin n')
end.
Definition double_bin (b:bin) : bin :=
match b with
| BZ ⇒ BZ
| _ ⇒ T2 b
end.
Fixpoint normalize (b:bin) : bin :=
match b with
| BZ ⇒ BZ
| T2 b' ⇒ double_bin (normalize b')
| T2P1 b' ⇒ T2P1 (normalize b')
end.
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.
Fixpoint nat_to_bin (n:nat) : bin :=
match n with
| O ⇒ BZ
| S n' ⇒ incr (nat_to_bin n')
end.
Definition double_bin (b:bin) : bin :=
match b with
| BZ ⇒ BZ
| _ ⇒ T2 b
end.
Fixpoint normalize (b:bin) : bin :=
match b with
| BZ ⇒ BZ
| T2 b' ⇒ double_bin (normalize b')
| T2P1 b' ⇒ T2P1 (normalize b')
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").
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").
(* FILL IN HERE *)
Prove that bin_to_nat undoes the work of nat_to_bin.
Now prove that nat_to_bin undoes the work of bin_to_nat... up
to normalization.
Before trying to prove this lemma---which will need some helper
lemmas about incr and double_bin---ask yourself... why is
normalize there, and what is it doing?
Theorem bin_nat_bin : ∀ b, nat_to_bin (bin_to_nat b) = normalize b.
Proof.
(* FILL IN HERE *) Admitted.
☐
Proof.
(* FILL IN HERE *) Admitted.
☐
(* 2022-01-12 10:20 *)