This line imports all of our definitions from Basics.v. For it to work, you need 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.)

Here are two ways to compile your code:

• CoqIDE
  
  1. Open Basics.v. 2. In the "Compile" menu, click on "Compile Buffer".
  
  
• Command line
  
  1. Run: coqc Basics.v

Technical note: In this file, we again use the Module feature to wrap all of the definitions for pairs and lists of numbers in a module so that, later, we can use the same names for the generic ("polymorphic") versions of the same operations.

Pairs of numbers

Each constructor of an inductive type can take any number of parameters---none (as with true and O), one (as with S), or more than one:

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

Here are some simple function definitions illustrating pattern matching on two-argument constructors:

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 instruct Coq to allow this with a Notation declaration.

The new notation is supported both in expressions and in pattern matches:

Let's try and prove a few simple facts about pairs. If we state the lemmas in a particular way, we can prove them with just partial evaluation

But that's not enough if we state the lemma in a more natural way:

We have to expose the structure of p so that simple can perform the pattern match in fst and snd.

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

Notice that Coq allows us to use the notation we introduced for pairs in the "as..." pattern telling it what variables to bind.

Exercise: 2 stars

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Exercise: 2 stars, optional

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Lists of numbers

Generalizing the definition of pairs a little, we can describe the type of lists of numbers like this: "A list can be 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 mathematical notation. The following two 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 fully understand the second line of the first declaration (or any of the right-hand side of the second), 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:

The at level 60 part tells Coq how to parenthesize expressions that involve both :: and some other infix operator. For example, if we define + as infix notation for the plus function at level 50,

Notation "x + y" := (plus x y)
                    (at level 50, left associativity).

then + will bind tighter than ::, and 1 + 2 :: [3] will be parsed correctly as (1 + 2) :: [3] rather than 1 + (2 :: [3]).

The second declaration introduces the standard square-brackets notation for lists; its right-hand side illustrates Coq's syntax for declaring n-ary notations and translating them to nested sequences of binary constructors.

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.

The length function calculates the length of a list.

The app function concatenates two lists.

In fact, app will be used so pervasively in some parts of what follows that it is convenient to have an infix operator for it.

Here are two more small examples. The hd function returns the first element (the "head") of the list, while tl ("tail") returns everything but the first element.

Bags via lists

A bag (or multiset) is a set where each element can appear any finite number of times. One reasonable implementation of bags is to represent a bag of numbers as a list.

Exercise: 2 stars, optional (bags)

As an exercise, complete the following definitions for functions like count and union on bags

All these proofs can be done just by reflexivity.

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Reasoning about lists

Just as with numbers, simple facts about list-processing functions can some-times be proved entirely by simplification. For example, simplification is enough for this theorem...

... because the [] is substituted into the match position in the definition of app, allowing the match itself to be simplified.

Also like 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 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.

Induction on lists

Proofs by induction over non-numeric data types are perhaps a little less familiar than natural number induction, but the basic idea is equally simple. Each Inductive declaration defines a set of data values that can be built up from the declared constructors: a number can be either O or S applied to a number; a boolean can be either true or false; 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', asssuming that P is true for l'.

Since larger lists can only be built up from smaller ones, stopping eventually with nil, these two things together establish the truth of P for all lists l.

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. A human-readable (informal) proof needs to include more explicit -- in particular, it helps the reader a lot to be reminded exactly what the induction hypothesis is in the second case.

Theorem: For all l1, l2, and l3, l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3.

Proof: By induction on l1.

• First, suppose l = []. We must show:
  
         [] ++ (l2 ++ l3) = ([] ++ l2) ++ l3
  
  which follows directly from the definition of ++.
  
  
• Next, suppose l = 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. ☐

An exercise to be worked together:

Exercise: 1 star (list_funs)

As an exercise, complete the definitions of nonzeros, oddmembers and countoddmembers below

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Exercise: 2 stars (alternate)

Complete the definition of alternate.

This exercise illustrates the fact that it sometimes requires a little extra thought to satisfy Coq's requirement that all Fixpoint definitions be "obviously terminating." There is an easy way to write the alternate function using just a single match...end, but Coq will not accept it as obviously terminating. Look for a slightly more verbose solution with two nested match...end constructs. Note that each match must be terminated by an end.

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For a slightly more involved example of an inductive proof over lists, suppose we define a "cons on the right" function snoc like this...

... and use it to define a list-reversing function rev line this:

Now we prove some more list theorems using our newly defined snoc and rev. Let's try something a little more intricate: proving that reversing a list does not change its length. Our first attempt at this proof gets stuck in the successor case...

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

Now we can complete the original proof.

Let's take a look at informal proofs of these two theorems:

Theorem: For all numbers n and lists l, length (snoc l n) = S (length l).

Proof: By induction on l.

• First, suppose l = []. We must show
  
          length (snoc [] n) = S (length []),
  
  which follows directly from the definitions of length and snoc.
  
  
• Next, suppose l = n'::l', with
  
          length (snoc l' n) = S (length l').
  
  We must show
  
          length (snoc (n' :: l') n) = S (length (n' :: l'))
  
  By the definitions of length and snoc, this follows from
  
          S (length (snoc l' n)) = S (S (length l')),
  
  which is immediate from 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 (snoc (rev l') n) = S (length l')
  
  which, by the previous lemma, is the same as
  
          S (length (rev l')) = S (length l').
  
  This is immediate from the induction hypothesis. ☐

Obviously, the style of these proofs is rather longwinded and pedantic. After we've seen a few of them, we might begin to find it easier to follow proofs that give a little less detail overall (since we can easily work them 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 more like this:

Theorem: For all lists l, length (rev l) = length l.

Proof: First, observe that

       length (snoc l n) = S (length l)

for any l. This follows by a straightforward induction on l.

The main property now follows by another straightforward 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 on how similar the proof at hand is to ones that the audience will already be familiar with. The more pedantic style is usually a safe fallback.

List exercises, Part 1

Exercise: 2 stars (list_exercises)

More practice with lists

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

An exercise about your implementation of nonzeros

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List exercises, part 2

Exercise: 2 stars (list_design)

Design exercise:

• Write down a non-trivial theorem involving cons (::), snoc, and append (++).
• Prove it.

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Exercise: 2 stars, optional (bag_proofs)

If you did the optional exercise about bags above, here are a couple of little theorems to prove about your definitions

The following lemma about ble_nat might help you below

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Options

Here is another type definition that is quite useful in day-to-day programming:

We can use natoption as a way of returning "error codes" from functions. For example, 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 return some number when the list is too short!

On the other hand, if we give it type nat -> natlist -> natoption, then we can return None when the list is too short and Some a when the list has enough members and a appears at position n.

This example is also an opportunity to introduce one more small feature of Coq's programming language: conditional expressions.

Coq's conditionals are exactly like those in every 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.

This function pulls the nat out of a natoption, returning a supplied default in the None case.

Exercise: 2 stars

Using the same idea, fix the hd function from earlier so we don't have to return an arbitrary element

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Exercise: 2 stars

This exercise relates your new hd_opt to the old hd

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Exercise: 2 stars (beq_natlist)

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

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The apply tactic

We often encounter situations where the goal to be proved is exactly the same as some hypothesis in the context or some previously proved lemma.

The apply tactic also works with CONDITIONAL hypotheses and lemmas: if the statement being applied is an implication, then the premises of this implication will be added to the list of subgoals needing to be proved.

You may find it instructive to experiment with this proof and see if there is a way to complete it using just rewrite instead of apply.

Typically, when we use apply H, the statement H will begin with a forall binding some "universal variables." When Coq matches the current goal against the conclusion of H, it will try to find appropriate values for these variables. For example, when we do apply eq2 in the following proof, the universal variable q in eq2 gets instantiated with n and r gets instantiated with m.

Exercise: 2 stars

Complete the following proof without using simpl.

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To use the apply tactic, the (conclusion of the) fact being applied must match the goal EXACTLY -- for example, apply will not work if the left and right sides of the equality are swapped.

When you find yourself in such a situation, one thing to do is to go back and try reorganizing the statement of whatever you are trying to prove so that things appear the right way around when they are needed. But if this is not possible or convenient, then the following lemma can be used to swap the sides of an equality statement in the goal so that some other lemma can be applied.

Exercise: 3 stars (apply_exercises)

Here are two exercises for you

This one is a little tricky. The first line of the proof is provided, because it uses an idea we haven't seen before. Notice that we don't introduce m before performing induction. This leave it general, so that the IH doesn't specify a particular m, but lets us pick. We'll talk more about this later in the course.

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Exercise: 3 stars (beq_nat_sym_informal)

Provide an informal proof of this lemma:

Theorem: For any nats n m and bool b, if beq_nat n m = b then beq_nat m n = b.

Proof: (* FILL IN HERE *)
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Exercise: 1 star (apply_rewrite)

Briefly explain the difference between the tactics apply and rewrite. Are there situations where either one can be applied?

(* FILL IN HERE *)

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Varying the Induction Hypothesis

One subtlety in these inductive proofs is worth noticing here. For example, look back at the proof of the ass_app theorem. The induction hypothesis (in the second subgoal generated by the induction tactic) is

l1' ++ l2 ++ l3 = (l1' ++ l2) ++ l3 .

(Note that, because we've defined ++ to be right associative, the expression on the left of the = is the same as writing l1' ++ (l2 ++ l3).)

This hypothesis makes a statement about l1' together with the particular lists l2 and l3. The lists l2 and l3, which were introduced into the context by the intros at the top of the proof, are "held constant" in the induction hypothesis. If we set up the proof slightly differently by introducing just n into the context at the top, then we get an induction hypothesis that makes a stronger claim:

forall l2 l3, l1' ++ l2 ++ l3 = (l1' ++ l2) ++ l3

(Use Coq to see the difference for yourself.)

In the present case, the difference between the two proofs is minor, since the definition of the ++ function just examines its first argument and doesn't do anything interesting with its second argument. But we'll soon come to situations where setting up the induction hypothesis one way or the other can make the difference between a proof working and failing.

Exercise: 2 stars

Give an alternate proof of the associativity of ++ with a more general induction hypothesis. Complete the following (leaving the first line unchanged).

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Additional Exercises

Exercise: 1 star (destruct_induction)

Briefly explain the difference between the tactics destruct and induction.

(* FILL IN HERE *)

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Put beq_nat_sym into the top-level namespace, so that we can use it later without having to write NatList.beq_nat_sym.