Inductively Defined Propositions

In the Logic chapter, we looked at several ways of writing propositions, including conjunction, disjunction, and existential quantification. In this chapter, we bring yet another new tool into the mix: inductively defined propositions. To begin, some examples...

Example: The Collatz Conjecture

The Collatz Conjecture is a famous open problem in number theory. Its statement is quite simple. First, we define a function f on numbers, as follows:

Next, we look at what happens when we repeatedly apply f to some given starting number. For example, f 12 is 6, and f 6 is 3, so by repeatedly applying f we get the sequence 12, 6, 3, 10, 5, 16, 8, 4, 2, 1. Similarly, if we start with 19, we get the longer sequence 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Both of these sequences eventually reach 1. The question posed by Collatz was: Is the sequence starting from any natural number guaranteed to reach 1 eventually? To formalize this question in Coq, we might try to define a recursive function that calculates the total number of steps that it takes for such a sequence to reach 1.

This definition is rejected by Coq's termination checker, since the argument to the recursive call, f n, is not "obviously smaller" than n. Indeed, this isn't just a pointless limitation: functions in Coq are required to be total, to ensure logical consistency. Moreover, we can't fix it by devising a more clever termination checker: deciding whether this particular function is total would be equivalent to settling the Collatz conjecture! Fortunately, there is another way to do it: We can express the concept "reaches 1 eventually" as an inductively defined property of numbers:

What we've done here is to use Coq's Inductive definition mechanism to characterize the property "Collatz holds for..." by stating two different ways in which it can hold: (1) Collatz holds for 1 and (2) if Collatz holds for f n then it holds for n. For particular numbers, we can now argue that the Collatz sequence reaches 1 like this (again, we'll go through the details of how it works a bit later in the chapter):

The Collatz conjecture then states that the sequence beginning from any number reaches 1:

If you succeed in proving this conjecture, you've got a bright future as a number theorist! But don't spend too long on it -- it's been open since 1937.

Example: Ordering

A binary relation on a set X is a family of propositions parameterized by two elements of X -- i.e., a proposition about pairs of elements of X. For example, one familiar binary relation on nat is le, the less-than-or-equal-to relation. We've already seen how to define it as a boolean computation. Here is a "direct" propositional definition.

The following definition says that there are two ways to show that one number is less than or equal to another: either observe that they are the same number, or, if the second has the form S m, give evidence that the first is less than or equal to m.

(By "reserving" the notation before defining the Inductive, we can use it in the definition.)

Example: Transitive Closure

As another example, the transitive closure of a relation R is the smallest relation that contains R and that is transitive.

For example, suppose we define a "parent of" relation on a group of people...

Then we can define "ancestor of" as its transitive closure:

Exercise: 1 star, standard, optional (close_refl_trans)

How would you modify this definition so that it defines reflexive and transitive closure? How about reflexive, symmetric, and transitive closure?

Example: Permutations

The familiar mathematical concept of permutation also has an elegant formulation as an inductive relation. For simplicity, let's focus on permutations of lists with exactly three elements.

This definition says:

• If l₂ can be obtained from l₁ by swapping the first and second elements, then l₂ is a permutation of l₁.
• If l₂ can be obtained from l₁ by swapping the second and third elements, then l₂ is a permutation of l₁.
• If l₂ is a permutation of l₁ and l₃ is a permutation of l₂, then l₃ is a permutation of l₁.

Exercise: 1 star, standard, optional (perm)

According to this definition, is [1;2;3] a permutation of [3;2;1]? Is [1;2;3] a permutation of itself?

Example: Evenness (yet again)

We've already seen two ways of stating a proposition that a number n is even: We can say (1) even n = true, or (2) ∃ k, n = double k. A third possibility, which we'll use as a running example for the rest of this chapter, is to say that n is even if we can establish its evenness from the following rules:

• The number 0 is even.
• If n is even, then S (S n) is even.

(Defining evenness in this way may seem a bit confusing, since we have already seen another perfectly good way of doing it -- "n is even if it is equal to the result of doubling some number". It makes a convenient running example because it is simple and compact, but we will see more compelling examples in future chapters.) To illustrate how this new definition of evenness works, let's imagine using it to show that 4 is even. First, we give the rules names for easy reference:

• Rule ev_0: The number 0 is even.
• Rule ev_SS: If n is even, then S (S n) is even.

Now, by rule ev_SS, it suffices to show that 2 is even. This, in turn, is again guaranteed by rule ev_SS, as long as we can show that 0 is even. But this last fact follows directly from the ev_0 rule. We can translate the informal definition of evenness from above into a formal Inductive declaration, where each "way that a number can be even" corresponds to a separate constructor:

This definition is interestingly different from previous uses of Inductive. For one thing, we are defining not a Type (like nat) or a function yielding a Type (like list), but rather a function from nat to Prop -- that is, a property of numbers. But what is really new is that, because the nat argument of ev appears to the right of the colon on the first line, it is allowed to take different values in the types of different constructors: 0 in the type of ev_0 and S (S n) in the type of ev_SS. Accordingly, the type of each constructor must be specified explicitly (after a colon), and each constructor's type must have the form ev n for some natural number n. In contrast, recall the definition of list:
    Inductive list (X:Type) : Type :=
      | nil
      | cons (x : X) (l : list X). or equivalently:
    Inductive list (X:Type) : Type :=
      | nil : list X
      | cons (x : X) (l : list X) : list X. This definition introduces the X parameter globally, to the left of the colon, forcing the result of nil and cons to be the same type (i.e., list X). But if we had tried to bring nat to the left of the colon in defining ev, we would have seen an error:

In an Inductive definition, an argument to the type constructor on the left of the colon is called a "parameter", whereas an argument on the right is called an "index" or "annotation." For example, in Inductive list (X : Type) := ..., the X is a parameter, while in Inductive ev : nat → Prop := ..., the unnamed nat argument is an index. We can think of this as defining a Coq property ev : nat → Prop, together with "evidence constructors" ev_0 : ev 0 and ev_SS : ∀ n, ev n → ev (S (S n)). These evidence constructors can be thought of as "primitive evidence of evenness", and they can be used just like proven theorems. In particular, we can use Coq's apply tactic with the constructor names to obtain evidence for ev of particular numbers...

... or we can use function application syntax to combine several constructors:

In this way, we can also prove theorems that have hypotheses involving ev.

Exercise: 1 star, standard (ev_double)

Using Evidence in Proofs

Besides constructing evidence that numbers are even, we can also destruct such evidence, reasoning about how it could have been built. Defining ev with an Inductive declaration tells Coq not only that the constructors ev_0 and ev_SS are valid ways to build evidence that some number is ev, but also that these two constructors are the only ways to build evidence that numbers are ev. In other words, if someone gives us evidence E for the assertion ev n, then we know that E must be one of two things:

• E is ev_0 (and n is O), or
• E is ev_SS n' E' (and n is S (S n'), where E' is evidence for ev n').

This suggests that it should be possible to analyze a hypothesis of the form ev n much as we do inductively defined data structures; in particular, it should be possible to argue by case analysis or by induction on such evidence. Let's look at a few examples to see what this means in practice.

Inversion on Evidence

Suppose we are proving some fact involving a number n, and we are given ev n as a hypothesis. We already know how to perform case analysis on n using destruct or induction, generating separate subgoals for the case where n = O and the case where n = S n' for some n'. But for some proofs we may instead want to analyze the evidence for ev n directly. As a tool for such proofs, we can formalize the intuitive characterization that we gave above for evidence of ev n, using destruct.

Facts like this are often called "inversion lemmas" because they allow us to "invert" some given information to reason about all the different ways it could have been derived. Here, there are two ways to prove ev n, and the inversion lemma makes this explicit. We can use the inversion lemma that we proved above to help structure proofs:

Note how the inversion lemma produces two subgoals, which correspond to the two ways of proving ev. The first subgoal is a contradiction that is discharged with discriminate. The second subgoal makes use of injection and rewrite. Coq provides a handy tactic called inversion that factors out this common pattern, saving us the trouble of explicitly stating and proving an inversion lemma for every Inductive definition we make. Here, the inversion tactic can detect (1) that the first case, where n = 0, does not apply and (2) that the n' that appears in the ev_SS case must be the same as n. It includes an "as" annotation similar to destruct, allowing us to assign names rather than have Coq choose them.

The inversion tactic can apply the principle of explosion to "obviously contradictory" hypotheses involving inductively defined properties, something that takes a bit more work using our inversion lemma. Compare:

Exercise: 1 star, standard (inversion_practice)

Prove the following result using inversion. (For extra practice, you can also prove it using the inversion lemma.)

Exercise: 1 star, standard (ev5_nonsense)

Prove the following result using inversion.

The inversion tactic does quite a bit of work. For example, when applied to an equality assumption, it does the work of both discriminate and injection. In addition, it carries out the intros and rewrites that are typically necessary in the case of injection. It can also be applied to analyze evidence for arbitrary inductively defined propositions, not just equality. As examples, we'll use it to re-prove some theorems from chapter Tactics. (Here we are being a bit lazy by omitting the as clause from inversion, thereby asking Coq to choose names for the variables and hypotheses that it introduces.)

Here's how inversion works in general.

• Suppose the name H refers to an assumption P in the current context, where P has been defined by an Inductive declaration.
• Then, for each of the constructors of P, inversion H generates a subgoal in which H has been replaced by the specific conditions under which this constructor could have been used to prove P.
• Some of these subgoals will be self-contradictory; inversion throws these away.
• The ones that are left represent the cases that must be proved to establish the original goal. For those, inversion adds to the proof context all equations that must hold of the arguments given to P -- e.g., n' = n in the proof of evSS_ev).

The ev_double exercise above shows that our new notion of evenness is implied by the two earlier ones (since, by even_bool_prop in chapter Logic, we already know that those are equivalent to each other). To show that all three coincide, we just need the following lemma.

We could try to proceed by case analysis or induction on n. But since ev is mentioned in a premise, this strategy seems unpromising, because (as we've noted before) the induction hypothesis will talk about n-1 (which is not even!). Thus, it seems better to first try inversion on the evidence for ev. Indeed, the first case can be solved trivially. And we can seemingly make progress on the second case with a helper lemma.

Unfortunately, the second case is harder. We need to show ∃ n₀, S (S n') = double n₀, but the only available assumption is E', which states that ev n' holds. Since this isn't directly useful, it seems that we are stuck and that performing case analysis on E was a waste of time. If we look more closely at our second goal, however, we can see that something interesting happened: By performing case analysis on E, we were able to reduce the original result to a similar one that involves a different piece of evidence for ev: namely E'. More formally, we could finish our proof if we could show that
        ∃ k', n' = double k', which is the same as the original statement, but with n' instead of n. Indeed, it is not difficult to convince Coq that this intermediate result would suffice.

Unfortunately, now we are stuck. To see this clearly, let's move E' back into the goal from the hypotheses.

Now it is obvious that we are trying to prove another instance of the same theorem we set out to prove -- only here we are talking about n' instead of n.

Induction on Evidence

If this story feels familiar, it is no coincidence: We encountered similar problems in the Induction chapter, when trying to use case analysis to prove results that required induction. And once again the solution is... induction! The behavior of induction on evidence is the same as its behavior on data: It causes Coq to generate one subgoal for each constructor that could have used to build that evidence, while providing an induction hypothesis for each recursive occurrence of the property in question. To prove that a property of n holds for all even numbers (i.e., those for which ev n holds), we can use induction on ev n. This requires us to prove two things, corresponding to the two ways in which ev n could have been constructed. If it was constructed by ev_0, then n=0 and the property must hold of 0. If it was constructed by ev_SS, then the evidence of ev n is of the form ev_SS n' E', where n = S (S n') and E' is evidence for ev n'. In this case, the inductive hypothesis says that the property we are trying to prove holds for n'. Let's try proving that lemma again:

Here, we can see that Coq produced an IH that corresponds to E', the single recursive occurrence of ev in its own definition. Since E' mentions n', the induction hypothesis talks about n', as opposed to n or some other number. The equivalence between the second and third definitions of evenness now follows.

As we will see in later chapters, induction on evidence is a recurring technique across many areas -- in particular for formalizing the semantics of programming languages. The following exercises provide simple examples of this technique, to help you familiarize yourself with it.

Exercise: 2 stars, standard (ev_sum)

Exercise: 4 stars, advanced, optional (ev'_ev)

In general, there may be multiple ways of defining a property inductively. For example, here's a (slightly contrived) alternative definition for ev:

Prove that this definition is logically equivalent to the old one. To streamline the proof, use the technique (from the Logic chapter) of applying theorems to arguments, and note that the same technique works with constructors of inductively defined propositions.

Exercise: 3 stars, advanced, especially useful (ev_ev__ev)

Exercise: 3 stars, standard, optional (ev_plus_plus)

This exercise can be completed without induction or case analysis. But, you will need a clever assertion and some tedious rewriting. Hint: Is (n+m) + (n+p) even?

Inductive Relations

A proposition parameterized by a number (such as ev) can be thought of as a property -- i.e., it defines a subset of nat, namely those numbers for which the proposition is provable. In the same way, a two-argument proposition can be thought of as a relation -- i.e., it defines a set of pairs for which the proposition is provable.

Just like properties, relations can be defined inductively. One useful example is the "less than or equal to" relation on numbers that we briefly saw above.

(We've written the definition a bit differently this time, giving explicit names to the arguments to the constructors and moving them to the left of the colons.) Proofs of facts about ≤ using the constructors le_n and le_S follow the same patterns as proofs about properties, like ev above. We can apply the constructors to prove ≤ goals (e.g., to show that 3<=3 or 3<=6), and we can use tactics like inversion to extract information from ≤ hypotheses in the context (e.g., to prove that (2 ≤ 1) → 2+2=5.) Here are some sanity checks on the definition. (Notice that, although these are the same kind of simple "unit tests" as we gave for the testing functions we wrote in the first few lectures, we must construct their proofs explicitly -- simpl and reflexivity don't do the job, because the proofs aren't just a matter of simplifying computations.)

The "strictly less than" relation n < m can now be defined in terms of le.

Exercise: 2 stars, standard, optional (total_relation)

Define an inductive binary relation total_relation that holds between every pair of natural numbers.

Exercise: 2 stars, standard, optional (empty_relation)

Define an inductive binary relation empty_relation (on numbers) that never holds.

From the definition of le, we can sketch the behaviors of destruct, inversion, and induction on a hypothesis H providing evidence of the form le e₁ e₂. Doing destruct H will generate two cases. In the first case, e₁ = e₂, and it will replace instances of e₂ with e₁ in the goal and context. In the second case, e₂ = S n' for some n' for which le e₁ n' holds, and it will replace instances of e₂ with S n'. Doing inversion H will remove impossible cases and add generated equalities to the context for further use. Doing induction H will, in the second case, add the induction hypothesis that the goal holds when e₂ is replaced with n'. Here are a number of facts about the ≤ and < relations that we are going to need later in the course. The proofs make good practice exercises.

Exercise: 5 stars, standard, optional (le_and_lt_facts)

Hint: May be easiest to prove by induction on n.

Exercise: 4 stars, standard, optional (more_le_exercises)

Hint: May be easiest to prove by induction on m.

Hint: The next two can easily be proved without using induction.

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

We can define three-place relations, four-place relations, etc., in just the same way as binary relations. For example, consider the following three-place relation on numbers:

• Which of the following propositions are provable?
  • R 1 1 2
  • R 2 2 6
• If we dropped constructor c₅ from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer.
• If we dropped constructor c₄ from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer.

Exercise: 3 stars, standard, optional (R_fact)

The relation R above actually encodes a familiar function. Figure out which function; then state and prove this equivalence in Coq.

Exercise: 3 stars, advanced (subsequence)

A list is a subsequence of another list if all of the elements in the first list occur in the same order in the second list, possibly with some extra elements in between. For example,
      [1;2;3] is a subsequence of each of the lists
      [1;2;3]
      [1;1;1;2;2;3]
      [1;2;7;3]
      [5;6;1;9;9;2;7;3;8] but it is not a subsequence of any of the lists
      [1;2]
      [1;3]
      [5;6;2;1;7;3;8].

• Define an inductive proposition subseq on list nat that captures what it means to be a subsequence. (Hint: You'll need three cases.)
• Prove subseq_refl that subsequence is reflexive, that is, any list is a subsequence of itself.
• Prove subseq_app that for any lists l₁, l₂, and l₃, if l₁ is a subsequence of l₂, then l₁ is also a subsequence of l₂ ++ l₃.
• (Harder) Prove subseq_trans that subsequence is transitive -- that is, if l₁ is a subsequence of l₂ and l₂ is a subsequence of l₃, then l₁ is a subsequence of l₃.

Exercise: 2 stars, standard, optional (R_provability2)

Suppose we give Coq the following definition:
    Inductive R : nat → list nat → Prop :=
      | c₁ : R 0 []
      | c₂ n l (H: R n l) : R (S n) (n :: l)
      | c₃ n l (H: R (S n) l) : R n l. Which of the following propositions are provable?

• R 2 [1;0]
• R 1 [1;2;1;0]
• R 6 [3;2;1;0]

A Digression on Notation

There are several equivalent ways of writing inductive types.  We've mostly seen this style...

... which omits the result types because they are all the same (i.e., bin). It is completely equivalent to this...

... where we fill them in, and this...

... where we put everything on the right of the colon. For inductively defined propositions, we need to explicitly give the result type for each constructor (because they are not all the same), so the first style doesn't make sense, but we can use either the second or the third interchangeably.

Case Study: Regular Expressions

The ev property provides a simple example for illustrating inductive definitions and the basic techniques for reasoning about them, but it is not terribly exciting -- after all, it is equivalent to the two non-inductive definitions of evenness that we had already seen, and does not seem to offer any concrete benefit over them. To give a better sense of the power of inductive definitions, we now show how to use them to model a classic concept in computer science: regular expressions. Regular expressions are a simple language for describing sets of strings. Their syntax is defined as follows:

Note that this definition is polymorphic: Regular expressions in reg_exp T describe strings with characters drawn from T -- that is, lists of elements of T. (Technical aside: We depart slightly from standard practice in that we do not require the type T to be finite. This results in a somewhat different theory of regular expressions, but the difference is not significant for present purposes.) We connect regular expressions and strings via the following rules, which define when a regular expression matches some string:

• The expression EmptySet does not match any string.
• The expression EmptyStr matches the empty string [].
• The expression Char x matches the one-character string [x].
• If re₁ matches s₁, and re₂ matches s₂, then App re₁ re₂ matches s₁ ++ s₂.
• If at least one of re₁ and re₂ matches s, then Union re₁ re₂ matches s.
• Finally, if we can write some string s as the concatenation of a sequence of strings s = s_1 ++ ... ++ s_k, and the expression re matches each one of the strings s_i, then Star re matches s.
  In particular, the sequence of strings may be empty, so Star re always matches the empty string [] no matter what re is.

We can easily translate this informal definition into an Inductive one as follows. We use the notation s =~ re in place of exp_match s re.

Notice that these rules are not quite the same as the informal ones that we gave at the beginning of the section. First, we don't need to include a rule explicitly stating that no string matches EmptySet; we just don't happen to include any rule that would have the effect of some string matching EmptySet. (Indeed, the syntax of inductive definitions doesn't even allow us to give such a "negative rule.") Second, the informal rules for Union and Star correspond to two constructors each: MUnionL / MUnionR, and MStar0 / MStarApp. The result is logically equivalent to the original rules but more convenient to use in Coq, since the recursive occurrences of exp_match are given as direct arguments to the constructors, making it easier to perform induction on evidence. (The exp_match_ex₁ and exp_match_ex₂ exercises below ask you to prove that the constructors given in the inductive declaration and the ones that would arise from a more literal transcription of the informal rules are indeed equivalent.) Let's illustrate these rules with a few examples.

(Notice how the last example applies MApp to the string [1] directly. Since the goal mentions [1; 2] instead of [1] ++ [2], Coq wouldn't be able to figure out how to split the string on its own.) Using inversion, we can also show that certain strings do not match a regular expression:

We can define helper functions for writing down regular expressions. The reg_exp_of_list function constructs a regular expression that matches exactly the list that it receives as an argument:

We can also prove general facts about exp_match. For instance, the following lemma shows that every string s that matches re also matches Star re.

(Note the use of app_nil_r to change the goal of the theorem to exactly the shape expected by MStarApp.)

Exercise: 3 stars, standard (exp_match_ex₁)

The following lemmas show that the informal matching rules given at the beginning of the chapter can be obtained from the formal inductive definition.

The next lemma is stated in terms of the fold function from the Poly chapter: If ss : list (list T) represents a sequence of strings s₁, ..., sn, then fold app ss [] is the result of concatenating them all together.

Since the definition of exp_match has a recursive structure, we might expect that proofs involving regular expressions will often require induction on evidence. For example, suppose we want to prove the following intuitive result: If a regular expression re matches some string s, then all elements of s must occur as character literals somewhere in re. To state this as a theorem, we first define a function re_chars that lists all characters that occur in a regular expression:

The main theorem:

Something interesting happens in the MApp case. We obtain two induction hypotheses: One that applies when x occurs in s₁ (which matches re₁), and a second one that applies when x occurs in s₂ (which matches re₂).

Here again we get two induction hypotheses, and they illustrate why we need induction on evidence for exp_match, rather than induction on the regular expression re: The latter would only provide an induction hypothesis for strings that match re, which would not allow us to reason about the case In x s₂.

Exercise: 4 stars, standard (re_not_empty)

Write a recursive function re_not_empty that tests whether a regular expression matches some string. Prove that your function is correct.

The remember Tactic

One potentially confusing feature of the induction tactic is that it will let you try to perform an induction over a term that isn't sufficiently general. The effect of this is to lose information (much as destruct without an eqn: clause can do), and leave you unable to complete the proof. Here's an example:

Now, just doing an inversion on H₁ won't get us very far in the recursive cases. (Try it!). So we need induction (on evidence!). Here is a naive first attempt.

But now, although we get seven cases (as we would expect from the definition of exp_match), we have lost a very important bit of information from H₁: the fact that s₁ matched something of the form Star re. This means that we have to give proofs for all seven constructors of this definition, even though all but two of them (MStar0 and MStarApp) are contradictory. We can still get the proof to go through for a few constructors, such as MEmpty...

... but most cases get stuck. For MChar, for instance, we must show
      s₂ =~ Char x' →
      x'::s₂ =~ Char x' which is clearly impossible.

The problem here is that induction over a Prop hypothesis only works properly with hypotheses that are "completely general," i.e., ones in which all the arguments are variables, as opposed to more complex expressions like Star re. (In this respect, induction on evidence behaves more like destruct-without-eqn: than like inversion.) A possible, but awkward, way to solve this problem is "manually generalizing" over the problematic expressions by adding explicit equality hypotheses to the lemma:

We can now proceed by performing induction over evidence directly, because the argument to the first hypothesis is sufficiently general, which means that we can discharge most cases by inverting the re' = Star re equality in the context. This works, but it makes the statement of the lemma a bit ugly. Fortunately, there is a better way...

The tactic remember e as x causes Coq to (1) replace all occurrences of the expression e by the variable x, and (2) add an equation x = e to the context. Here's how we can use it to show the above result:

We now have Heqre' : re' = Star re.

The Heqre' is contradictory in most cases, allowing us to conclude immediately.

The interesting cases are those that correspond to Star. Note that the induction hypothesis IH₂ on the MStarApp case mentions an additional premise Star re'' = Star re, which results from the equality generated by remember.

Exercise: 4 stars, standard, optional (exp_match_ex₂)

The MStar'' lemma below (combined with its converse, the MStar' exercise above), shows that our definition of exp_match for Star is equivalent to the informal one given previously.

Exercise: 5 stars, advanced (weak_pumping)

One of the first really interesting theorems in the theory of regular expressions is the so-called pumping lemma, which states, informally, that any sufficiently long string s matching a regular expression re can be "pumped" by repeating some middle section of s an arbitrary number of times to produce a new string also matching re. (For the sake of simplicity in this exercise, we consider a slightly weaker theorem than is usually stated in courses on automata theory -- hence the name weak_pumping.) To get started, we need to define "sufficiently long." Since we are working in a constructive logic, we actually need to be able to calculate, for each regular expression re, the minimum length for strings s to guarantee "pumpability."

You may find these lemmas about the pumping constant useful when proving the pumping lemma below.

Next, it is useful to define an auxiliary function that repeats a string (appends it to itself) some number of times.

This auxiliary lemma might also be useful in your proof of the pumping lemma.

The (weak) pumping lemma itself says that, if s =~ re and if the length of s is at least the pumping constant of re, then s can be split into three substrings s₁ ++ s₂ ++ s₃ in such a way that s₂ can be repeated any number of times and the result, when combined with s₁ and s₃, will still match re. Since s₂ is also guaranteed not to be the empty string, this gives us a (constructive!) way to generate strings matching re that are as long as we like.

Complete the proof below. Several of the lemmas about le that were in an optional exercise earlier in this chapter may also be useful.

Exercise: 5 stars, advanced, optional (pumping)

Now here is the usual version of the pumping lemma. In addition to requiring that s₂ ≠ [], it also requires that length s₁ + length s₂ ≤ pumping_constant re.

You may want to copy your proof of weak_pumping below.

Case Study: Improving Reflection

We've seen in the Logic chapter that we sometimes need to relate boolean computations to statements in Prop. But performing this conversion as we did there can result in tedious proof scripts. Consider the proof of the following theorem:

In the first branch after destruct, we explicitly apply the eqb_eq lemma to the equation generated by destructing n =? m, to convert the assumption n =? m = true into the assumption n = m; then we had to rewrite using this assumption to complete the case. We can streamline this sort of reasoning by defining an inductive proposition that yields a better case-analysis principle for n =? m. Instead of generating the assumption (n =? m) = true, which usually requires some massaging before we can use it, this principle gives us right away the assumption we really need: n = m. Following the terminology introduced in Logic, we call this the "reflection principle for equality on numbers," and we say that the boolean n =? m is reflected in the proposition n = m.

The reflect property takes two arguments: a proposition P and a boolean b. It states that the property P reflects (intuitively, is equivalent to) the boolean b: that is, P holds if and only if b = true. To see this, notice that, by definition, the only way we can produce evidence for reflect P true is by showing P and then using the ReflectT constructor. If we invert this statement, this means that we can extract evidence for P from a proof of reflect P true. Similarly, the only way to show reflect P false is by tagging evidence for ¬ P with the ReflectF constructor. To put this observation to work, we first prove that the statements P ↔ b = true and reflect P b are indeed equivalent. First, the left-to-right implication:

Now you prove the right-to-left implication:

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

We can think of reflect as a variant of the usual "if and only if" connective; the advantage of reflect is that, by destructing a hypothesis or lemma of the form reflect P b, we can perform case analysis on b while at the same time generating appropriate hypothesis in the two branches (P in the first subgoal and ¬ P in the second). Let's use reflect to produce a smoother proof of filter_not_empty_In. We begin by recasting the eqb_eq lemma in terms of reflect:

The proof of filter_not_empty_In now goes as follows. Notice how the calls to destruct and rewrite in the earlier proof of this theorem are combined here into a single call to destruct. (To see this clearly, execute the two proofs of filter_not_empty_In with Coq and observe the differences in proof state at the beginning of the first case of the destruct.)

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

Use eqbP as above to prove the following: