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 with nybble, and here):

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

Functions for extracting the first and second components of a pair can then be defined by pattern matching.

Since pairs will be used heavily in what follows, 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.

The new notation can be used both in expressions and in pattern matches.

Note that pattern-matching on a pair (with parentheses: (x, y)) is not to be confused with the "multiple pattern" syntax (with no parentheses: x, y) that we have seen previously. The above examples illustrate pattern matching on a pair with elements x and y, whereas, for example, the definition of minus in Basics performs pattern matching on the values n and m:
       Fixpoint minus (n m : nat) : nat :=
         match n, m with
         | O , _ ⇒ O
         | S _ , O ⇒ n
         | S n', S m' ⇒ minus n' m'
         end. The distinction is minor, but it is worth knowing that they are not the same. For instance, the following definitions are ill-formed:
        (* Can't match on a pair with multiple patterns: *)
        Definition bad_fst (p : natprod) : nat :=
          match p with
          | x, y ⇒ x
          end.

        (* Can't match on multiple values with pair patterns: *)
        Definition bad_minus (n m : nat) : nat :=
          match n, m with
          | (O , _ ) ⇒ O
          | (S _ , O ) ⇒ n
          | (S n', S m') ⇒ bad_minus n' m'
          end. Now let's try to prove a few simple facts about pairs. If we state properties of pairs in a slightly peculiar way, we can sometimes complete their proofs with just reflexivity (and its built-in simplification):

But just reflexivity is not enough if we state the lemma in the most natural way:

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

Notice that, unlike its behavior with nats, where it generates two subgoals, destruct generates just one subgoal here. That's because natprods can only be constructed in one way.

Exercise: 1 star, standard (snd_fst_is_swap)

Exercise: 1 star, standard, optional (fst_swap_is_snd)

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."

For example, here is a three-element list:

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.

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

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

Next let's look at several functions for constructing and manipulating lists. First, the repeat function takes a number n and a count and returns a list of length count in which every element is n.

Length

The length function calculates the length of a list.

Append

The app function concatenates (appends) two lists.

Since app will be used extensively, it is again convenient to have an infix operator for it.

Head 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"). Since the empty list has no first element, we pass a default value to be returned in that case.

Exercises

Exercise: 2 stars, standard, especially useful (list_funs)

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

For countoddmembers, we're giving you a header that uses keyword Definition instead of Fixpoint. The point of stating the question this way is to encourage you to implement the function by using already-defined functions, rather than writing your own recursive definition.

Exercise: 3 stars, advanced (alternate)

Complete the following definition of alternate, which interleaves 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. Hint: there is an elegant way of writing alternate that fails to satisfy Coq's requirement that all Fixpoint definitions be structurally recursing, as mentioned in Basics. If you encounter that difficulty, consider pattern matching against both lists at the same time with the "multiple pattern" syntax we've seen before.

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 representation for a bag of numbers is as a list.

Exercise: 3 stars, standard, especially useful (bag_functions)

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

All these proofs can be done just by reflexivity.

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 -- using max instead of sum -- which is why we don't call this operation union.) We've deliberately given you a header that does not give explicit names to the arguments. Implement sum with an already-defined function without changing the header.

Exercise: 3 stars, standard, 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. (This exercise is optional, but students following the advanced track will need to fill in the definition of remove_one for a later exercise.)

Exercise: 2 stars, standard, especially useful (add_inc_count)

Adding a value to a bag should increase the value's count by one. State this as a theorem and prove it in Coq.

Reasoning About Lists

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

...because the [] is substituted into the "scrutinee" (the expression whose value is being "scrutinized" by the match) 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.

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. We'll see how to do this next. (Micro-Sermon: As we get deeper into this material, simply reading proof scripts will not get you very far! It is important to step through the details of each one using Coq and think 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. For example, a boolean can be either true or false; a number can be either O or S applied to another number; and 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. This last 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 always be broken down into smaller ones, eventually reaching nil, these two arguments together establish the truth of P for all lists l. Here's a concrete example:

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 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. ☐

Generalizing Statements

In some situations, it can be necessary to generalize a statement for being able to prove it by induction. Intuitively, the reason for this is that a more general statement also yields a more general (and hence stronger) inductive hypothesis. For this reason, when being stuck in a proof, it can make sense to step back and see whether it is possible to prove a stronger statement.

To get a more general inductive hypothesis, we can generalize the statement as follows: For a slightly more involved example of inductive proof over lists, suppose we use app to define a list-reversing function rev:

For something a bit more challenging than the proofs we've seen so far, let's prove that reversing a list does not change its length. Our first attempt gets stuck in the successor case...

A first attempt may be to prove exactly the statement that we are missing at this point. However, we will fail to to so since the inductive hypothesis resulting from it is not general enough to conclude the proof.

We can slightly strengthen the lemma to not only work on reversed lists but general lists.

This generalized lemma would be sufficient to conclude our original proof. Still, we could prove an even more general lemma about the length of appended lists. So let's take the equation relating ++ and length that would have enabled us to make progress at the point where we got stuck and state it as a separate lemma.

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.

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 plus.
• 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 reading a couple like this, we might find it easier to follow proofs that give fewer details (which we 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, 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 they will already be familiar with. The more pedantic style is a good default for our present purposes.

Search

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 Search command is quite helpful with this. Let's say you've forgotten the name of a theorem about rev. The command Search rev will cause Coq to display a list of all theorems involving rev.

Or say you've forgotten the name of the theorem showing that plus is commutative. You can use a pattern to search for all theorems involving the equality of two additions.

You'll see a lot of results there, nearly all of them from the standard library. To restrict the results, you can search inside a particular module:

You can also make the search more precise by using variables in the search pattern instead of wildcards:

The question mark in front of the variable is needed to indicate that it is a variable in the search pattern, rather than a variable that is expected to be in scope currently. Keep Search in mind as you do the following exercises and throughout the rest of the book; it can save you a lot of time! Your IDE likely has its own functionality to help with searching. For example, in ProofGeneral, you can run Search with C-c C-a C-a, and paste its response into your buffer with C-c C-;.

List Exercises, Part 1

Exercise: 3 stars, standard (list_exercises)

More practice with lists:

An involution is a function that is its own inverse. That is, applying the function twice yield the original input.

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.

An exercise about your implementation of nonzeros:

Exercise: 2 stars, standard (eqblist)

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

List Exercises, Part 2

Here are a couple of little theorems to prove about your definitions about bags above.

Exercise: 1 star, standard (count_member_nonzero)

The following lemma about leb might help you in the next exercise (it will also be useful in later chapters).

Before doing the next exercise, make sure you've filled in the definition of remove_one above.

Exercise: 3 stars, advanced (remove_does_not_increase_count)

Exercise: 3 stars, standard, optional (bag_count_sum)

Write down an interesting theorem bag_count_sum about bags involving the functions count and sum, and prove it using Coq. (You may find that the difficulty of the proof depends on how you defined count! Hint: If you defined count using =? you may find it useful to know that destruct works on arbitrary expressions, not just simple identifiers.)

Exercise: 3 stars, advanced (involution_injective)

Prove that every involution is injective. Involutions were defined above in rev_involutive. An injective function is one-to-one: it maps distinct inputs to distinct outputs, without any collisions.

Exercise: 2 stars, advanced (rev_injective)

Prove that rev is injective. Do not prove this by induction -- that would be hard. Instead, re-use the same proof technique that you used for involution_injective. Do not try to use that exercise directly as a lemma: the types are not the same.

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...

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.

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. As we see here, constructors of inductive definitions can be capitalized.

(In the HTML version, the boilerplate proofs of these examples are elided. Click on a box if you want to see one.) The function below pulls the nat out of a natoption, returning a supplied default in the None case.

Exercise: 2 stars, standard (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.

Exercise: 1 star, standard, optional (option_elim_hd)

This exercise relates your new hd_error to the old hd.

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.

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 flexibility to change representations later if we want to. We'll also need an equality test for ids:

Exercise: 1 star, standard (eqb_id_refl)

Now we define the type of partial maps:

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 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 by shadowing it with a new one (or simply adds a new entry if the given key is not already present).

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.

Exercise: 1 star, standard (update_eq)

Exercise: 1 star, standard (update_neq)

Exercise: 2 stars, standard, optional (baz_num_elts)

Consider the following inductive definition:

How many elements does the type baz have? (Explain in words, in a comment.)

Labeled Trees

To illustrate how induction on more complex datatypes works, we will introduce the datatype btree of labeled trees that are at most binary: This means that a each node in tree is labeled with a natural number and has at most two children. We can describe such a datatype as follows:

We define a function size that counts the number of nodes in the tree.

Function sumLabels sums up the labels at all nodes.

Function incrementLabels increments all tree labels by one.

We introduce a helper lemma for doing a specific arithmetic transformation that we will use in the Theorem that we want to prove next. It is only requires a combination of associativity and commutativity of addition but since it involves several of such steps it makes the main proof to have it as a seperate lemma.

We can state the following theorem on the sum of labels of a tree with incremented labels. The proof proceeds by induction on the tree t. Consequently, we will need to to prove three cases:

• The base case t = leaf n
• The case t = unary n t' for a unary tree t with inductive hypothesis IHt' for the child tree t'
• The case t = binary n t₁' t₂' for a binary tree t with inductive hyptheses IHt1' for child tree t₁'

and inductive hypothesis IHt2' for child tree t₂'