Meta Topics

Software Technology: Proof by Induction, Structured Data

Acknowledgements

This material is taken verbatim from Software Foundations V4.0

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Assumed Knowledge: * Some familiarity with the basics of Coq
Learning Outcomes: * Write your first proofs by induction, rewriting, and case analysis in Coq.
Grab the Coq source file Induction.v
First, we import all of our definitions from the previous chapter.

Require Export Basics.

For the Require Export to work, you first need to use 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:
    Open Basics.v. In the "Compile" menu, click on "Compile Buffer".
  • From the command line:
    Run coqc Basics.v

Proof by Induction

We proved in the last chapter that 0 is a neutral element for + on the left using an easy argument based on simplification. The fact that it is also a neutral element on the right...

Theorem plus_n_O_firsttry : n:nat,
  n = n + 0.

... 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:
  • 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.
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:

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.

Exercise: 2 stars (double_plus)

Consider the following function, which doubles its argument:

Fixpoint double (n:nat) :=
  match n with
  | OO
  | S n'S (S (double n'))
  end.

Use induction to prove this simple fact about double:

Lemma double_plus : n, double n = n + n .
Proof.
  (* FILL IN HERE *) Admitted.

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):

Theorem evenb_S : n : nat,
  evenb (S n) = negb (evenb n).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star (destruct_induction)

Briefly explain the difference between the tactics destruct and induction.
(* FILL IN HERE *)

Proofs Within Proofs

In Coq, as in informal mathematics, large proofs are often broken into a sequence of theorems, with later proofs referring to earlier theorems. But sometimes a proof will require some miscellaneous fact that is too trivial and of too little general interest to bother giving it its own top-level name. In such cases, it is convenient to be able to simply state and prove the needed "sub-theorem" right at the point where it is used. The assert tactic allows us to do this. For example, our earlier proof of the mult_0_plus theorem referred to a previous theorem named plus_O_n. We could instead use assert to state and prove plus_O_n in-line:

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.

Theorem plus_swap : n m p : nat,
  n + (m + p) = m + (n + p).
Proof.
  (* FILL IN HERE *) Admitted.

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.)

Theorem mult_comm : m n : nat,
  m * n = n * m.
Proof.
  (* FILL IN HERE *) Admitted.

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.)

Theorem beq_nat_refl : n : nat,
  true = beq_nat n n.
Proof.
  (* FILL IN HERE *) Admitted.

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.
Use the replace tactic to do a proof of plus_swap', just like plus_swap but without needing assert (n + m = m + n).

Theorem plus_swap' : n m p : nat,
  n + (m + p) = m + (n + p).
Proof.
  (* FILL IN HERE *) Admitted.

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").
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!

(* FILL IN 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.
(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.

(* FILL IN HERE *)

Formal vs. Informal Proof (Optional)

"Informal proofs are algorithms; formal proofs are code."
The question of what constitutes a proof of a mathematical claim has challenged philosophers for millennia, but a rough and ready definition could be this: A proof of a mathematical proposition P is a written (or spoken) text that instills in the reader or hearer the certainty that P is true. That is, a proof is an act of communication.
Acts of communication may involve different sorts of readers. On one hand, the "reader" can be a program like Coq, in which case the "belief" that is instilled is that P can be mechanically derived from a certain set of formal logical rules, and the proof is a recipe that guides the program in checking this fact. Such recipes are formal proofs.
Alternatively, the reader can be a human being, in which case the proof will be written in English or some other natural language, and will thus necessarily be informal. Here, the criteria for success are less clearly specified. A "valid" proof is one that makes the reader believe P. But the same proof may be read by many different readers, some of whom may be convinced by a particular way of phrasing the argument, while others may not be. Some readers may be particularly pedantic, inexperienced, or just plain thick-headed; the only way to convince them will be to make the argument in painstaking detail. But other readers, more familiar in the area, may find all this detail so overwhelming that they lose the overall thread; all they want is to be told the main ideas, since it is easier for them to fill in the details for themselves than to wade through a written presentation of them. Ultimately, there is no universal standard, because there is no single way of writing an informal proof that is guaranteed to convince every conceivable reader.
In practice, however, mathematicians have developed a rich set of conventions and idioms for writing about complex mathematical objects that — at least within a certain community — make communication fairly reliable. The conventions of this stylized form of communication give a fairly clear standard for judging proofs good or bad.
Because we are using Coq in this course, we will be working heavily with formal proofs. But this doesn't mean we can completely forget about informal ones! Formal proofs are useful in many ways, but they are not very efficient ways of communicating ideas between human beings.
For example, here is a proof that addition is associative:

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:
  • 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 from
        S (n' + (m + p)) = S ((n' + m) + p),
      which is immediate from the induction hypothesis. Qed.
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).

Exercise: 2 stars, advanced, recommended (plus_comm_informal)

Translate your solution for plus_comm into an informal proof:
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)

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!
Theorem: true = beq_nat n n for any n.
Proof: (* FILL IN HERE *)
Grab the Coq source file Lists.v

Require Export Induction.
Module NatList.

Pairs of Numbers

In an Inductive type definition, each constructor can take any number of arguments — none (as with true and O), one (as with S), or more than one, as here:

Inductive natprod : Type :=
| pair : nat nat natprod.

This declaration can be read: "There is one way to construct a pair of numbers: by applying the constructor pair to two arguments of type nat."

Check (pair 3 5).

Here are two simple functions for extracting the first and second components of a pair. The definitions also illustrate how to do pattern matching on two-argument constructors.

Definition fst (p : natprod) : nat :=
  match p with
  | pair x yx
  end.

Definition snd (p : natprod) : nat :=
  match p with
  | pair x yy
  end.

Compute (fst (pair 3 5)).
(* ===> 3 *)

Since pairs are used quite a bit, it is nice to be able to write them with the standard mathematical notation (x,y) instead of pair x y. We can tell Coq to allow this with a Notation declaration.

Notation "( x , y )" := (pair x y).

The new notation can be used both in expressions and in pattern matches (indeed, we've seen it already in the previous chapter — this works because the pair notation is actually provided as part of the standard library):

Compute (fst (3,5)).

Definition fst' (p : natprod) : nat :=
  match p with
  | (x,y) ⇒ x
  end.

Definition snd' (p : natprod) : nat :=
  match p with
  | (x,y) ⇒ y
  end.

Definition swap_pair (p : natprod) : natprod :=
  match p with
  | (x,y) ⇒ (y,x)
  end.

Let's try to prove a few simple facts about pairs.
If we state things in a particular (and slightly peculiar) way, we can complete proofs with just reflexivity (and its built-in simplification):

Theorem surjective_pairing' : (n m : nat),
  (n,m) = (fst (n,m), snd (n,m)).
Proof.
  reflexivity. Qed.

But reflexivity is not enough if we state the lemma in a more natural way:

Theorem surjective_pairing_stuck : (p : natprod),
  p = (fst p, snd p).
Proof.
  simpl. (* Doesn't reduce anything! *)
Abort.

We have to expose the structure of p so that simpl can perform the pattern match in fst and snd. We can do this with destruct.

Theorem surjective_pairing : (p : natprod),
  p = (fst p, snd p).
Proof.
  intros p. destruct p as [n m]. simpl. reflexivity. Qed.

Notice that, unlike its behavior with nats, destruct doesn't generate an extra subgoal here. That's because natprods can only be constructed in one way.

Exercise: 1 star (snd_fst_is_swap)

Theorem snd_fst_is_swap : (p : natprod),
  (snd p, fst p) = swap_pair p.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, optional (fst_swap_is_snd)

Theorem fst_swap_is_snd : (p : natprod),
  fst (swap_pair p) = snd p.
Proof.
  (* FILL IN HERE *) Admitted.

Lists of Numbers

Generalizing the definition of pairs, we can describe the type of lists of numbers like this: "A list is either the empty list or else a pair of a number and another list."

Inductive natlist : Type :=
  | nil : natlist
  | cons : nat natlist natlist.

For example, here is a three-element list:

Definition mylist := cons 1 (cons 2 (cons 3 nil)).

As with pairs, it is more convenient to write lists in familiar programming notation. The following declarations allow us to use :: as an infix cons operator and square brackets as an "outfix" notation for constructing lists.

Notation "x :: l" := (cons x l)
                     (at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x ; .. ; y ]" := (cons x .. (cons y nil) ..).

It is not necessary to understand the details of these declarations, but in case you are interested, here is roughly what's going on. The right associativity annotation tells Coq how to parenthesize expressions involving several uses of :: so that, for example, the next three declarations mean exactly the same thing:

Definition mylist1 := 1 :: (2 :: (3 :: nil)).
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [1;2;3].

The at level 60 part tells Coq how to parenthesize expressions that involve both :: and some other infix operator. For example, since we defined + as infix notation for the plus function at level 50,
  Notation "x + y" := (plus x y)
                      (at level 50, left associativity).
the + operator will bind tighter than ::, so 1 + 2 :: [3] will be parsed, as we'd expect, as (1 + 2) :: [3] rather than 1 + (2 :: [3]).
(Expressions like "1 + 2 :: [3]" can be a little confusing when you read them in a .v file. The inner brackets, around 3, indicate a list, but the outer brackets, which are invisible in the HTML rendering, are there to instruct the "coqdoc" tool that the bracketed part should be displayed as Coq code rather than running text.)
The second and third Notation declarations above introduce the standard square-bracket notation for lists; the right-hand side of the third one illustrates Coq's syntax for declaring n-ary notations and translating them to nested sequences of binary constructors.

Repeat

A number of functions are useful for manipulating lists. For example, the repeat function takes a number n and a count and returns a list of length count where every element is n.

Fixpoint repeat (n count : nat) : natlist :=
  match count with
  | Onil
  | S count'n :: (repeat n count')
  end.

Length

The length function calculates the length of a list.

Fixpoint length (l:natlist) : nat :=
  match l with
  | nilO
  | h :: tS (length t)
  end.

Append

The app function concatenates (appends) two lists.

Fixpoint app (l1 l2 : natlist) : natlist :=
  match l1 with
  | nill2
  | h :: th :: (app t l2)
  end.

Actually, app will be used a lot in some parts of what follows, so it is convenient to have an infix operator for it.

Notation "x ++ y" := (app x y)
                     (right associativity, at level 60).

Example test_app1: [1;2;3] ++ [4;5] = [1;2;3;4;5].
Proof. reflexivity. Qed.
Example test_app2: nil ++ [4;5] = [4;5].
Proof. reflexivity. Qed.
Example test_app3: [1;2;3] ++ nil = [1;2;3].
Proof. reflexivity. Qed.

Head (with default) and Tail

Here are two smaller examples of programming with lists. The hd function returns the first element (the "head") of the list, while tl returns everything but the first element (the "tail"). Of course, the empty list has no first element, so we must pass a default value to be returned in that case.

Definition hd (default:nat) (l:natlist) : nat :=
  match l with
  | nildefault
  | h :: th
  end.

Definition tl (l:natlist) : natlist :=
  match l with
  | nilnil
  | h :: tt
  end.

Example test_hd1: hd 0 [1;2;3] = 1.
Proof. reflexivity. Qed.
Example test_hd2: hd 0 [] = 0.
Proof. reflexivity. Qed.
Example test_tl: tl [1;2;3] = [2;3].
Proof. reflexivity. Qed.

Exercises

Exercise: 2 stars, recommended (list_funs)

Complete the definitions of nonzeros, oddmembers and countoddmembers below. Have a look at the tests to understand what these functions should do.

Fixpoint nonzeros (l:natlist) : natlist
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_nonzeros:
  nonzeros [0;1;0;2;3;0;0] = [1;2;3].
  (* FILL IN HERE *) Admitted.

Fixpoint oddmembers (l:natlist) : natlist
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_oddmembers:
  oddmembers [0;1;0;2;3;0;0] = [1;3].
  (* FILL IN HERE *) Admitted.

Fixpoint countoddmembers (l:natlist) : nat
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_countoddmembers1:
  countoddmembers [1;0;3;1;4;5] = 4.
  (* FILL IN HERE *) Admitted.

Example test_countoddmembers2:
  countoddmembers [0;2;4] = 0.
  (* FILL IN HERE *) Admitted.

Example test_countoddmembers3:
  countoddmembers nil = 0.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, advanced (alternate)

Complete the definition of alternate, which "zips up" two lists into one, alternating between elements taken from the first list and elements from the second. See the tests below for more specific examples.
Note: one natural and elegant way of writing alternate will fail to satisfy Coq's requirement that all Fixpoint definitions be "obviously terminating." If you find yourself in this rut, look for a slightly more verbose solution that considers elements of both lists at the same time. (One possible solution requires defining a new kind of pairs, but this is not the only way.)

Fixpoint alternate (l1 l2 : natlist) : natlist
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_alternate1:
  alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6].
  (* FILL IN HERE *) Admitted.

Example test_alternate2:
  alternate [1] [4;5;6] = [1;4;5;6].
  (* FILL IN HERE *) Admitted.

Example test_alternate3:
  alternate [1;2;3] [4] = [1;4;2;3].
  (* FILL IN HERE *) Admitted.

Example test_alternate4:
  alternate [] [20;30] = [20;30].
  (* FILL IN HERE *) Admitted.

Bags via Lists

A bag (or multiset) is like a set, except that each element can appear multiple times rather than just once. One possible implementation is to represent a bag of numbers as a list.

Definition bag := natlist.

Exercise: 3 stars, recommended (bag_functions)

Complete the following definitions for the functions count, sum, add, and member for bags.

Fixpoint count (v:nat) (s:bag) : nat
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

All these proofs can be done just by reflexivity.

Example test_count1: count 1 [1;2;3;1;4;1] = 3.
 (* FILL IN HERE *) Admitted.
Example test_count2: count 6 [1;2;3;1;4;1] = 0.
 (* FILL IN HERE *) Admitted.

Multiset sum is similar to set union: sum a b contains all the elements of a and of b. (Mathematicians usually define union on multisets a little bit differently, which is why we don't use that name for this operation.) For sum we're giving you a header that does not give explicit names to the arguments. Moreover, it uses the keyword Definition instead of Fixpoint, so even if you had names for the arguments, you wouldn't be able to process them recursively. The point of stating the question this way is to encourage you to think about whether sum can be implemented in another way — perhaps by using functions that have already been defined.

Definition sum : bag bag bag
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_sum1: count 1 (sum [1;2;3] [1;4;1]) = 3.
 (* FILL IN HERE *) Admitted.

Definition add (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_add1: count 1 (add 1 [1;4;1]) = 3.
 (* FILL IN HERE *) Admitted.
Example test_add2: count 5 (add 1 [1;4;1]) = 0.
 (* FILL IN HERE *) Admitted.

Definition member (v:nat) (s:bag) : bool
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_member1: member 1 [1;4;1] = true.
 (* FILL IN HERE *) Admitted.

Example test_member2: member 2 [1;4;1] = false.
 (* FILL IN HERE *) Admitted.

Exercise: 3 stars, optional (bag_more_functions)

Here are some more bag functions for you to practice with.
When remove_one is applied to a bag without the number to remove, it should return the same bag unchanged.

Fixpoint remove_one (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_remove_one1:
  count 5 (remove_one 5 [2;1;5;4;1]) = 0.
  (* FILL IN HERE *) Admitted.

Example test_remove_one2:
  count 5 (remove_one 5 [2;1;4;1]) = 0.
  (* FILL IN HERE *) Admitted.

Example test_remove_one3:
  count 4 (remove_one 5 [2;1;4;5;1;4]) = 2.
  (* FILL IN HERE *) Admitted.

Example test_remove_one4:
  count 5 (remove_one 5 [2;1;5;4;5;1;4]) = 1.
  (* FILL IN HERE *) Admitted.

Fixpoint remove_all (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_remove_all1: count 5 (remove_all 5 [2;1;5;4;1]) = 0.
 (* FILL IN HERE *) Admitted.
Example test_remove_all2: count 5 (remove_all 5 [2;1;4;1]) = 0.
 (* FILL IN HERE *) Admitted.
Example test_remove_all3: count 4 (remove_all 5 [2;1;4;5;1;4]) = 2.
 (* FILL IN HERE *) Admitted.
Example test_remove_all4: count 5 (remove_all 5 [2;1;5;4;5;1;4;5;1;4]) = 0.
 (* FILL IN HERE *) Admitted.

Fixpoint subset (s1:bag) (s2:bag) : bool
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_subset1: subset [1;2] [2;1;4;1] = true.
 (* FILL IN HERE *) Admitted.
Example test_subset2: subset [1;2;2] [2;1;4;1] = false.
 (* FILL IN HERE *) Admitted.

Exercise: 3 stars, recommended (bag_theorem)

Write down an interesting theorem bag_theorem about bags involving the functions count and add, and prove it. Note that, since this problem is somewhat open-ended, it's possible that you may come up with a theorem which is true, but whose proof requires techniques you haven't learned yet. Feel free to ask for help if you get stuck!

(*
Theorem bag_theorem : ...
Proof.
  ...
Qed.
*)



Reasoning About Lists

As with numbers, simple facts about list-processing functions can sometimes be proved entirely by simplification. For example, the simplification performed by reflexivity is enough for this theorem...

Theorem nil_app : l:natlist,
  [] ++ l = l.
Proof. reflexivity. Qed.

... because the [] is substituted into the match "scrutinee" in the definition of app, allowing the match itself to be simplified.
Also, as with numbers, it is sometimes helpful to perform case analysis on the possible shapes (empty or non-empty) of an unknown list.

Theorem tl_length_pred : l:natlist,
  pred (length l) = length (tl l).
Proof.
  intros l. destruct l as [| n l'].
  - (* l = nil *)
    reflexivity.
  - (* l = cons n l' *)
    reflexivity. Qed.

Here, the nil case works because we've chosen to define tl nil = nil. Notice that the as annotation on the destruct tactic here introduces two names, n and l', corresponding to the fact that the cons constructor for lists takes two arguments (the head and tail of the list it is constructing).
Usually, though, interesting theorems about lists require induction for their proofs.

Micro-Sermon

Simply reading example proof scripts will not get you very far! It is important to work through the details of each one, using Coq and thinking about what each step achieves. Otherwise it is more or less guaranteed that the exercises will make no sense when you get to them. 'Nuff said.

Induction on Lists

Proofs by induction over datatypes like natlist are a little less familiar than standard natural number induction, but the idea is equally simple. Each Inductive declaration defines a set of data values that can be built up using the declared constructors: a boolean can be either true or false; a number can be either O or S applied to another number; a list can be either nil or cons applied to a number and a list.
Moreover, applications of the declared constructors to one another are the only possible shapes that elements of an inductively defined set can have, and this fact directly gives rise to a way of reasoning about inductively defined sets: a number is either O or else it is S applied to some smaller number; a list is either nil or else it is cons applied to some number and some smaller list; etc. So, if we have in mind some proposition P that mentions a list l and we want to argue that P holds for all lists, we can reason as follows:
  • 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'.
Since larger lists can only be built up from smaller ones, eventually reaching nil, these two arguments together establish the truth of P for all lists l. Here's a concrete example:

Theorem app_assoc : l1 l2 l3 : natlist,
  (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof.
  intros l1 l2 l3. induction l1 as [| n l1' IHl1'].
  - (* l1 = nil *)
    reflexivity.
  - (* l1 = cons n l1' *)
    simpl. rewrite IHl1'. 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.
For comparison, here is an informal proof of the same theorem.
Theorem: For all lists l1, l2, and l3, (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof: 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
      ((n :: l1') ++ l2) ++ l3 = (n :: l1') ++ (l2 ++ l3).
    By the definition of ++, this follows from
      n :: ((l1' ++ l2) ++ l3) = n :: (l1' ++ (l2 ++ l3)),
    which is immediate from the induction hypothesis.

Reversing a list

For a slightly more involved example of inductive proof over lists, suppose we use app to define a list-reversing function rev:

Fixpoint rev (l:natlist) : natlist :=
  match l with
  | nilnil
  | h :: trev t ++ [h]
  end.

Example test_rev1: rev [1;2;3] = [3;2;1].
Proof. reflexivity. Qed.
Example test_rev2: rev nil = nil.
Proof. reflexivity. Qed.

Proofs about reverse

Now let's prove some theorems about our newly defined rev. For something a bit more challenging than what we've seen, let's prove that reversing a list does not change its length. Our first attempt gets stuck in the successor case...

Theorem rev_length_firsttry : l : natlist,
  length (rev l) = length l.
Proof.
  intros l. induction l as [| n l' IHl'].
  - (* l =  *)
    reflexivity.
  - (* l = n :: l' *)
    (* This is the tricky case.  Let's begin as usual
       by simplifying. *)

    simpl.
    (* Now we seem to be stuck: the goal is an equality
       involving ++, but we don't have any useful equations
       in either the immediate context or in the global
       environment!  We can make a little progress by using
       the IH to rewrite the goal... *)

    rewrite IHl'.
    (* ... but now we can't go any further. *)
Abort.

So let's take the equation relating ++ and length that would have enabled us to make progress and prove it as a separate lemma.

Theorem app_length : l1 l2 : natlist,
  length (l1 ++ l2) = (length l1) + (length l2).
Proof.
  (* WORKED IN CLASS *)
  intros l1 l2. induction l1 as [| n l1' IHl1'].
  - (* l1 = nil *)
    reflexivity.
  - (* l1 = cons *)
    simpl. rewrite IHl1'. reflexivity. Qed.

Note that, to make the lemma as general as possible, we quantify over all natlists, not just those that result from an application of rev. This should seem natural, because the truth of the goal clearly doesn't depend on the list having been reversed. Moreover, it is easier to prove the more general property.
Now we can complete the original proof.

Theorem rev_length : l : natlist,
  length (rev l) = length l.
Proof.
  intros l. induction l as [| n l' IHl'].
  - (* l = nil *)
    reflexivity.
  - (* l = cons *)
    simpl. rewrite app_length, plus_comm.
    rewrite IHl'. reflexivity. Qed.

For comparison, here are informal proofs of these two theorems:
Theorem: For all lists l1 and l2, length (l1 ++ l2) = length l1 + length l2.
Proof: By induction on l1.
  • First, suppose l1 = []. We must show
      length ([] ++ l2) = length [] + length l2,
    which follows directly from the definitions of length and ++.
  • Next, suppose l1 = n::l1', with
      length (l1' ++ l2) = length l1' + length l2.
    We must show
      length ((n::l1') ++ l2) = length (n::l1') + length l2).
    This follows directly from the definitions of length and ++ together with the induction hypothesis.
Theorem: For all lists l, length (rev l) = length l.
Proof: By induction on l.
  • First, suppose l = []. We must show
      length (rev []) = length [],
    which follows directly from the definitions of length and rev.
  • Next, suppose l = n::l', with
      length (rev l') = length l'.
    We must show
      length (rev (n :: l')) = length (n :: l').
    By the definition of rev, this follows from
      length ((rev l') ++ [n]) = S (length l')
    which, by the previous lemma, is the same as
      length (rev l') + length [n] = S (length l').
    This follows directly from the induction hypothesis and the definition of length.
The style of these proofs is rather longwinded and pedantic. After the first few, we might find it easier to follow proofs that give fewer details (which can easily work out in our own minds or on scratch paper if necessary) and just highlight the non-obvious steps. In this more compressed style, the above proof might look like this:
Theorem: For all lists l, length (rev l) = length l.
Proof: First, observe that length (l ++ [n]) = S (length l) for any l (this follows by a straightforward induction on l). The main property again follows by induction on l, using the observation together with the induction hypothesis in the case where l = n'::l'.
Which style is preferable in a given situation depends on the sophistication of the expected audience and how similar the proof at hand is to ones that the audience will already be familiar with. The more pedantic style is a good default for our present purposes.

SearchAbout

We've seen that proofs can make use of other theorems we've already proved, e.g., using rewrite. But in order to refer to a theorem, we need to know its name! Indeed, it is often hard even to remember what theorems have been proven, much less what they are called.
Coq's SearchAbout command is quite helpful with this. Typing SearchAbout foo will cause Coq to display a list of all theorems involving foo. For example, try uncommenting the following line to see a list of theorems that we have proved about rev:

(*  SearchAbout rev. *)

Keep SearchAbout in mind as you do the following exercises and throughout the rest of the book; it can save you a lot of time!
If you are using ProofGeneral, you can run SearchAbout with C-c C-a C-a. Pasting its response into your buffer can be accomplished with C-c C-;.

List Exercises, Part 1

Exercise: 3 stars (list_exercises)

More practice with lists:

Theorem app_nil_r : l : natlist,
  l ++ [] = l.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem rev_involutive : l : natlist,
  rev (rev l) = l.
Proof.
  (* FILL IN HERE *) Admitted.

There is a short solution to the next one. If you find yourself getting tangled up, step back and try to look for a simpler way.

Theorem app_assoc4 : l1 l2 l3 l4 : natlist,
  l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
  (* FILL IN HERE *) Admitted.

An exercise about your implementation of nonzeros:

Lemma nonzeros_app : l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars (beq_natlist)

Fill in the definition of beq_natlist, which compares lists of numbers for equality. Prove that beq_natlist l l yields true for every list l.

Fixpoint beq_natlist (l1 l2 : natlist) : bool
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_beq_natlist1 :
  (beq_natlist nil nil = true).
 (* FILL IN HERE *) Admitted.

Example test_beq_natlist2 :
  beq_natlist [1;2;3] [1;2;3] = true.
(* FILL IN HERE *) Admitted.

Example test_beq_natlist3 :
  beq_natlist [1;2;3] [1;2;4] = false.
 (* FILL IN HERE *) Admitted.

Theorem beq_natlist_refl : l:natlist,
  true = beq_natlist l l.
Proof.
  (* FILL IN HERE *) Admitted.

List Exercises, Part 2

Exercise: 3 stars, advanced (bag_proofs)

Here are a couple of little theorems to prove about your definitions about bags earlier in the file.

Theorem count_member_nonzero : (s : bag),
  leb 1 (count 1 (1 :: s)) = true.
Proof.
  (* FILL IN HERE *) Admitted.

The following lemma about leb might help you in the next proof.

Theorem ble_n_Sn : n,
  leb n (S n) = true.
Proof.
  intros n. induction n as [| n' IHn'].
  - (* 0 *)
    simpl. reflexivity.
  - (* S n' *)
    simpl. rewrite IHn'. reflexivity. Qed.

Theorem remove_decreases_count: (s : bag),
  leb (count 0 (remove_one 0 s)) (count 0 s) = true.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, optional (bag_count_sum)

Write down an interesting theorem bag_count_sum about bags involving the functions count and sum, and prove it.

(* FILL IN HERE *)

Exercise: 4 stars, advanced (rev_injective)

Prove that the rev function is injective — that is,
    (l1 l2 : natlist), rev l1 = rev l2  l1 = l2.
(There is a hard way and an easy way to do this.)

(* FILL IN HERE *)

Options

Suppose we want to write a function that returns the nth element of some list. If we give it type nat natlist nat, then we'll have to choose some number to return when the list is too short...

Fixpoint nth_bad (l:natlist) (n:nat) : nat :=
  match l with
  | nil ⇒ 42 (* arbitrary! *)
  | a :: l'match beq_nat n O with
               | truea
               | falsenth_bad l' (pred n)
               end
  end.

This solution is not so good: If nth_bad returns 42, we can't tell whether that value actually appears on the input without further processing. A better alternative is to change the return type of nth_bad to include an error value as a possible outcome. We call this type natoption.

Inductive natoption : Type :=
  | Some : nat natoption
  | None : natoption.

We can then change the above definition of nth_bad to return None when the list is too short and Some a when the list has enough members and a appears at position n. We call this new function nth_error to indicate that it may result in an error.

Fixpoint nth_error (l:natlist) (n:nat) : natoption :=
  match l with
  | nilNone
  | a :: l'match beq_nat n O with
               | trueSome a
               | falsenth_error l' (pred n)
               end
  end.

Example test_nth_error1 : nth_error [4;5;6;7] 0 = Some 4.
Proof. reflexivity. Qed.
Example test_nth_error2 : nth_error [4;5;6;7] 3 = Some 7.
Proof. reflexivity. Qed.
Example test_nth_error3 : nth_error [4;5;6;7] 9 = None.
Proof. reflexivity. Qed.

(In the HTML version, the boilerplate proofs of these examples are elided. Click on a box if you want to see one.)
This example is also an opportunity to introduce one more small feature of Coq's programming language: conditional expressions...

Fixpoint nth_error' (l:natlist) (n:nat) : natoption :=
  match l with
  | nilNone
  | a :: l'if beq_nat n O then Some a
               else nth_error' l' (pred n)
  end.

Coq's conditionals are exactly like those found in any other language, with one small generalization. Since the boolean type is not built in, Coq actually allows conditional expressions over any inductively defined type with exactly two constructors. The guard is considered true if it evaluates to the first constructor in the Inductive definition and false if it evaluates to the second.
The function below pulls the nat out of a natoption, returning a supplied default in the None case.

Definition option_elim (d : nat) (o : natoption) : nat :=
  match o with
  | Some n'n'
  | Noned
  end.

Exercise: 2 stars (hd_error)

Using the same idea, fix the hd function from earlier so we don't have to pass a default element for the nil case.

Definition hd_error (l : natlist) : natoption
  (* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example test_hd_error1 : hd_error [] = None.
 (* FILL IN HERE *) Admitted.

Example test_hd_error2 : hd_error [1] = Some 1.
 (* FILL IN HERE *) Admitted.

Example test_hd_error3 : hd_error [5;6] = Some 5.
 (* FILL IN HERE *) Admitted.

Exercise: 1 star, optional (option_elim_hd)

This exercise relates your new hd_error to the old hd.

Theorem option_elim_hd : (l:natlist) (default:nat),
  hd default l = option_elim default (hd_error l).
Proof.
  (* FILL IN HERE *) Admitted.

End NatList.

Partial Maps

As a final illustration of how data structures can be defined in Coq, here is a simple partial map data type, analogous to the map or dictionary data structures found in most programming languages.
First, we define a new inductive datatype id to serve as the "keys" of our partial maps.

Inductive id : Type :=
  | Id : nat id.

Internally, an id is just a number. Introducing a separate type by wrapping each nat with the tag Id makes definitions more readable and gives us the flexibility to change representations later if we wish.
We'll also need an equality test for ids:

Definition beq_id x1 x2 :=
  match x1, x2 with
  | Id n1, Id n2beq_nat n1 n2
  end.

Exercise: 1 star (beq_id_refl)

Theorem beq_id_refl : x, true = beq_id x x.
Proof.
  (* FILL IN HERE *) Admitted.
Now we define the type of partial maps:

Module PartialMap.
Import NatList.

Inductive partial_map : Type :=
  | empty : partial_map
  | record : id nat partial_map partial_map.

This declaration can be read: "There are two ways to construct a partial_map: either using the constructor empty to represent an empty partial map, or by applying the constructor record to a key, a value, and an existing partial_map to construct a partial_map with an additional key-to-value mapping."
The update function overrides the entry for a given key in a partial map (or adds a new entry if the given key is not already present).

Definition update (d : partial_map)
                  (key : id) (value : nat)
                  : partial_map :=
  record key value d.

Last, the find function searches a partial_map for a given key. It returns None if the key was not found and Some val if the key was associated with val. If the same key is mapped to multiple values, find will return the first one it encounters.

Fixpoint find (key : id) (d : partial_map) : natoption :=
  match d with
  | emptyNone
  | record k v d'if beq_id key k
                     then Some v
                     else find key d'
  end.

Exercise: 1 star (update_eq)

Theorem update_eq :
  (d : partial_map) (k : id) (v: nat),
    find k (update d k v) = Some v.
Proof.
 (* FILL IN HERE *) Admitted.

Exercise: 1 star (update_neq)

Theorem update_neq :
  (d : partial_map) (m n : id) (o: nat),
    beq_id m n = false find m (update d n o) = find m d.
Proof.
 (* FILL IN HERE *) Admitted.

End PartialMap.

Exercise: 2 stars (baz_num_elts)

Consider the following inductive definition:

Inductive baz : Type :=
  | Baz1 : baz baz
  | Baz2 : baz bool baz.

How many elements does the type baz have? (Answer in English or the natural language of your choice.)
(* FILL IN HERE *)