IndPropInductively Defined Propositions


Set Warnings "-notation-overridden,-parsing".
From LF Require Export Logic.
From Coq Require Export Lia.

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: inductive definitions.
Note: For the sake of simplicity, most of this chapter uses an inductive definition of "evenness" as a running example. You may find this confusing, since we already have a perfectly good way of defining evenness as a proposition (n is even if it is equal to the result of doubling some number); if so, rest assured that we will see many more compelling examples of inductively defined propositions toward the end of this chapter and in future chapters.
In past chapters, we have seen two ways of stating that a number n is even: We can say
(1) evenb n = true, or
(2) k, n = double k.
Yet another possibility is to say that n is even if we can establish its evenness from the following rules:
  • Rule ev_0: The number 0 is even.
  • Rule ev_SS: If n is even, then S (S n) is even.
To illustrate how this new definition of evenness works, let's imagine using it to show that 4 is even. 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 will see many definitions like this one during the rest of the course. For purposes of informal discussions, it is helpful to have a lightweight notation that makes them easy to read and write. Inference rules are one such notation. (We'll use ev for the name of this property, since even is already used.)
   (ev_0)  

ev 0
ev n (ev_SS)  

ev (S (S n))
Each of the textual rules that we started with is reformatted here as an inference rule; the intended reading is that, if the premises above the line all hold, then the conclusion below the line follows. For example, the rule ev_SS says that, if n satisfies ev, then S (S n) also does. If a rule has no premises above the line, then its conclusion holds unconditionally.
We can represent a proof using these rules by combining rule applications into a proof tree. Here's how we might transcribe the above proof that 4 is even:

                             -------- (ev_0)
                              ev 0
                             -------- (ev_SS)
                              ev 2
                             -------- (ev_SS)
                              ev 4
(Why call this a "tree", rather than a "stack", for example? Because, in general, inference rules can have multiple premises. We will see examples of this shortly.)

Inductive Definition of Evenness

Putting all of this together, we can translate the definition of evenness into a formal Coq definition using an Inductive declaration, where each constructor corresponds to an inference rule:

Inductive ev : nat Prop :=
| ev_0 : ev 0
| ev_SS (n : nat) (H : ev n) : ev (S (S n)).
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).
This definition introduces the X parameter globally, to the left of the colon, forcing the result of nil and cons to be the same (i.e., list X). Had we tried to bring nat to the left of the colon in defining ev, we would have seen an error:

Fail Inductive wrong_ev (n : nat) : Prop :=
| wrong_ev_0 : wrong_ev 0
| wrong_ev_SS (H: wrong_ev n) : wrong_ev (S (S n)).
(* ===> Error: Last occurrence of "wrong_ev" must have "n"
        as 1st argument in "wrong_ev 0". *)

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; in Inductive ev : nat Prop := ..., the unnamed nat argument is an index.
We can think of the definition of ev 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)).
Such "evidence constructors" have the same status as proven theorems. In particular, we can use Coq's apply tactic with the rule names to prove ev for particular numbers...

Theorem ev_4 : ev 4.
Proof. apply ev_SS. apply ev_SS. apply ev_0. Qed.
... or we can use function application syntax:

Theorem ev_4' : ev 4.
Proof. apply (ev_SS 2 (ev_SS 0 ev_0)). Qed.
We can also prove theorems that have hypotheses involving ev.

Theorem ev_plus4 : n, ev n ev (4 + n).
Proof.
  intros n. simpl. intros Hn.
  apply ev_SS. apply ev_SS. apply Hn.
Qed.

Exercise: 1 star, standard (ev_double)

Theorem ev_double : n,
  ev (double n).
Proof.
  (* FILL IN HERE *) Admitted.

Using Evidence in Proofs

Besides constructing evidence that numbers are even, we can also destruct such evidence, which amounts to reasoning about how it could have been built.
Introducing 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 induction and case analysis 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 that ev n directly. As a tool, we can prove our characterization of evidence for ev n, using destruct.

Theorem ev_inversion :
   (n : nat), ev n
    (n = 0) ( n', n = S (S n') ev n').
Proof.
  intros n E.
  destruct E as [ | n' E'] eqn:EE.
  - (* E = ev_0 : ev 0 *)
    left. reflexivity.
  - (* E = ev_SS n' E' : ev (S (S n')) *)
    right. n'. split. reflexivity. apply E'.
Qed.
The following theorem can easily be proved using destruct on evidence.

Theorem ev_minus2 : n,
  ev n ev (pred (pred n)).
Proof.
  intros n E.
  destruct E as [| n' E'] eqn:EE.
  - (* E = ev_0 *) simpl. apply ev_0.
  - (* E = ev_SS n' E' *) simpl. apply E'.
Qed.
However, this variation cannot easily be handled with just destruct.

Theorem evSS_ev : n,
  ev (S (S n)) ev n.
Intuitively, we know that evidence for the hypothesis cannot consist just of the ev_0 constructor, since O and S are different constructors of the type nat; hence, ev_SS is the only case that applies. Unfortunately, destruct is not smart enough to realize this, and it still generates two subgoals. Even worse, in doing so, it keeps the final goal unchanged, failing to provide any useful information for completing the proof.
Proof.
  intros n E.
  destruct E as [| n' E'] eqn:EE.
  - (* E = ev_0. *)
    (* We must prove that n is even from no assumptions! *)
Abort.
What happened, exactly? Calling destruct has the effect of replacing all occurrences of the property argument by the values that correspond to each constructor. This is enough in the case of ev_minus2 because that argument n is mentioned directly in the final goal. However, it doesn't help in the case of evSS_ev since the term that gets replaced (S (S n)) is not mentioned anywhere.
If we remember that term S (S n), the proof goes through. (We'll discuss remember in more detail below.)

Theorem evSS_ev_remember : n,
  ev (S (S n)) ev n.
Proof.
  intros n E. remember (S (S n)) as k eqn:Hk. destruct E as [|n' E'] eqn:EE.
  - (* E = ev_0 *)
    (* Now we do have an assumption, in which k = S (S n) has been
       rewritten as 0 = S (S n) by destruct. That assumption
       gives us a contradiction. *)

    discriminate Hk.
  - (* E = ev_S n' E' *)
    (* This time k = S (S n) has been rewritten as S (S n') = S (S n). *)
    injection Hk as Heq. rewrite <- Heq. apply E'.
Qed.
Alternatively, the proof is straightforward using our inversion lemma.

Theorem evSS_ev : n, ev (S (S n)) ev n.
Proof.
 intros n H. apply ev_inversion in H.
 destruct H as [H0|H1].
 - discriminate H0.
 - destruct H1 as [n' [Hnm Hev]]. injection Hnm as Heq.
   rewrite Heq. apply Hev.
Qed.
Note how both proofs produce 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 that common pattern.
The inversion tactic can detect (1) that the first case (n = 0) does not apply and (2) that the n' that appears in the ev_SS case must be the same as n. It has an "as" variant similar to destruct, allowing us to assign names rather than have Coq choose them.

Theorem evSS_ev' : n,
  ev (S (S n)) ev n.
Proof.
  intros n E.
  inversion E as [| n' E' Heq].
  (* We are in the E = ev_SS n' E' case now. *)
  apply E'.
Qed.
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. For example:

Theorem one_not_even : ¬ ev 1.
Proof.
  intros H. apply ev_inversion in H.
  destruct H as [ | [m [Hm _]]].
  - discriminate H.
  - discriminate Hm.
Qed.

Theorem one_not_even' : ¬ ev 1.
  intros H. inversion H. Qed.

Exercise: 1 star, standard (inversion_practice)

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

Theorem SSSSev__even : n,
  ev (S (S (S (S n)))) ev n.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard (ev5_nonsense)

Prove the following result using inversion.

Theorem ev5_nonsense :
  ev 5 2 + 2 = 9.
Proof.
  (* FILL IN HERE *) Admitted.
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, more generally, to analyze evidence for inductively defined propositions. As examples, we'll use it to reprove 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.)

Theorem inversion_ex1 : (n m o : nat),
  [n; m] = [o; o]
  [n] = [m].
Proof.
  intros n m o H. inversion H. reflexivity. Qed.

Theorem inversion_ex2 : (n : nat),
  S n = O
  2 + 2 = 5.
Proof.
  intros n contra. inversion contra. Qed.
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 exact, 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 all equations into the proof context that must hold of the arguments given to P (e.g., S (S 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.

Lemma ev_even_firsttry : n,
  ev n even n.
Proof.
  (* WORKED IN CLASS *)
We could try to proceed by case analysis or induction on n. But since ev is mentioned in a premise, this strategy would probably lead to a dead end, 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.

  intros n E. inversion E as [EQ' | n' E' EQ'].
  - (* E = ev_0 *)
     0. reflexivity.
  - (* E = ev_SS n' E' *) simpl.
Unfortunately, the second case is harder. We need to show k, S (S n') = double k, 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 can finish our proof by showing 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 suffices.
Unforunately, now we are stuck. To make that apparent, let's move E' back into the goal from the hypotheses.

    generalize dependent E'.
Now it is clear we are trying to prove another instance of the same theorem we set out to prove. This instance is with n', instead of n, where n' is a smaller natural number than n.
Abort.

Induction on Evidence

If this looks familiar, it is no coincidence: We've 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 a property of n holds for all numbers 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 our current lemma again:

Lemma ev_even : n,
  ev n even n.
Proof.
  intros n E.
  induction E as [|n' E' IH].
  - (* E = ev_0 *)
     0. reflexivity.
  - (* E = ev_SS n' E'
       with IH : even E' *)

    unfold even in IH.
    destruct IH as [k Hk].
    rewrite Hk. (S k). simpl. reflexivity.
Qed.
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.

Theorem ev_even_iff : n,
  ev n even n.
Proof.
  intros n. split.
  - (* -> *) apply ev_even.
  - (* <- *) unfold even. intros [k Hk]. rewrite Hk. apply ev_double.
Qed.
As we will see in later chapters, induction on evidence is a recurring technique across many areas, and in particular when formalizing the semantics of programming languages, where many properties of interest are defined inductively.
The following exercises provide simple examples of this technique, to help you familiarize yourself with it.

Exercise: 2 stars, standard (ev_sum)

Theorem ev_sum : n m, ev n ev m ev (n + m).
Proof.
  (* FILL IN HERE *) Admitted.

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:

Inductive ev' : nat Prop :=
| ev'_0 : ev' 0
| ev'_2 : ev' 2
| ev'_sum n m (Hn : ev' n) (Hm : ev' m) : ev' (n + m).
Prove that this definition is logically equivalent to the old one. To streamline the proof, use the technique (from Logic) of applying theorems to arguments, and note that the same technique works with constructors of inductively defined propositions.

Theorem ev'_ev : n, ev' n ev n.
Proof.
 (* FILL IN HERE *) Admitted.

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

There are two pieces of evidence you could attempt to induct upon here. If one doesn't work, try the other.

Theorem ev_ev__ev : n m,
  ev (n+m) ev n ev m.
Proof.
  (* FILL IN HERE *) Admitted.

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?

Theorem ev_plus_plus : n m p,
  ev (n+m) ev (n+p) ev (m+p).
Proof.
  (* FILL IN HERE *) Admitted.

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.

Module Playground.
Just like properties, relations can be defined inductively. One useful example is the "less than or equal to" relation on numbers.
The following definition should be fairly intuitive. It says that there are two ways to give evidence that one number is less than or equal to another: either observe that they are the same number, or give evidence that the first is less than or equal to the predecessor of the second.

Inductive le : nat nat Prop :=
  | le_n (n : nat) : le n n
  | le_S (n m : nat) (H : le n m) : le n (S m).

Notation "n <= m" := (le n m).
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.)

Theorem test_le1 :
  3 3.
Proof.
  (* WORKED IN CLASS *)
  apply le_n. Qed.

Theorem test_le2 :
  3 6.
Proof.
  (* WORKED IN CLASS *)
  apply le_S. apply le_S. apply le_S. apply le_n. Qed.

Theorem test_le3 :
  (2 1) 2 + 2 = 5.
Proof.
  (* WORKED IN CLASS *)
  intros H. inversion H. inversion H2. Qed.
The "strictly less than" relation n < m can now be defined in terms of le.

Definition lt (n m:nat) := le (S n) m.

Notation "m < n" := (lt m n).

End Playground.
Here are a few more simple relations on numbers:

Inductive square_of : nat nat Prop :=
  | sq n : square_of n (n × n).

Inductive next_nat : nat nat Prop :=
  | nn n : next_nat n (S n).

Inductive next_ev : nat nat Prop :=
  | ne_1 n (H: ev (S n)) : next_ev n (S n)
  | ne_2 n (H: ev (S (S n))) : next_ev n (S (S n)).

Exercise: 2 stars, standard, optional (total_relation)

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

(* FILL IN HERE *)

Exercise: 2 stars, standard, optional (empty_relation)

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

(* FILL IN HERE *)
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'.

Exercise: 3 stars, standard, optional (le_exercises)

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.

Lemma le_trans : m n o, m n n o m o.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem O_le_n : n,
  0 n.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem n_le_m__Sn_le_Sm : n m,
  n m S n S m.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem Sn_le_Sm__n_le_m : n m,
  S n S m n m.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem le_plus_l : a b,
  a a + b.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem plus_le : n1 n2 m,
  n1 + n2 m
  n1 m n2 m.
Proof.
 (* FILL IN HERE *) Admitted.
Hint: the next one may be easiest to prove by induction on n.

Theorem add_le_cases : n m p q,
    n + m p + q n p m q.
Proof.
(* FILL IN HERE *) Admitted.

Theorem lt_S : n m,
  n < m
  n < S m.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem plus_lt : n1 n2 m,
  n1 + n2 < m
  n1 < m n2 < m.
Proof.
(* FILL IN HERE *) Admitted.

Theorem leb_complete : n m,
  n <=? m = true n m.
Proof.
  (* FILL IN HERE *) Admitted.
Hint: The next one may be easiest to prove by induction on m.

Theorem leb_correct : n m,
  n m
  n <=? m = true.
Proof.
  (* FILL IN HERE *) Admitted.
Hint: The next one can easily be proved without using induction.

Theorem leb_true_trans : n m o,
  n <=? m = true m <=? o = true n <=? o = true.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (leb_iff)

Theorem leb_iff : n m,
  n <=? m = true n m.
Proof.
  (* FILL IN HERE *) Admitted.

Module R.

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:

Inductive R : nat nat nat Prop :=
   | c1 : R 0 0 0
   | c2 m n o (H : R m n o) : R (S m) n (S o)
   | c3 m n o (H : R m n o) : R m (S n) (S o)
   | c4 m n o (H : R (S m) (S n) (S (S o))) : R m n o
   | c5 m n o (H : R m n o) : R n m o.
  • 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.

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_R_provability : option (nat×string) := None.

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?

Definition fR : nat nat nat
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem R_equiv_fR : m n o, R m n o fR m n = o.
Proof.
(* FILL IN HERE *) Admitted.

End R.

Exercise: 2 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.
  • (Optional, 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. Hint: choose your induction carefully!

Inductive subseq : list nat list nat Prop :=
(* FILL IN HERE *)
.

Theorem subseq_refl : (l : list nat), subseq l l.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem subseq_app : (l1 l2 l3 : list nat),
  subseq l1 l2
  subseq l1 (l2 ++ l3).
Proof.
  (* FILL IN HERE *) Admitted.

Theorem subseq_trans : (l1 l2 l3 : list nat),
  subseq l1 l2
  subseq l2 l3
  subseq l1 l3.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (R_provability2)

Suppose we give Coq the following definition:
    Inductive R : natlist natProp :=
      | 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]

(* FILL IN HERE *)

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:

Inductive reg_exp (T : Type) : Type :=
  | EmptySet
  | EmptyStr
  | Char (t : T)
  | App (r1 r2 : reg_exp T)
  | Union (r1 r2 : reg_exp T)
  | Star (r : reg_exp T).

Arguments EmptySet {T}.
Arguments EmptyStr {T}.
Arguments Char {T} _.
Arguments App {T} _ _.
Arguments Union {T} _ _.
Arguments Star {T} _.
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.
(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 our 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; by "reserving" the notation before defining the Inductive, we can use it in the definition!

Reserved Notation "s =~ re" (at level 80).

Inductive exp_match {T} : list T reg_exp T Prop :=
  | MEmpty : [] =~ EmptyStr
  | MChar x : [x] =~ (Char x)
  | MApp s1 re1 s2 re2
             (H1 : s1 =~ re1)
             (H2 : s2 =~ re2)
           : (s1 ++ s2) =~ (App re1 re2)
  | MUnionL s1 re1 re2
                (H1 : s1 =~ re1)
              : s1 =~ (Union re1 re2)
  | MUnionR re1 s2 re2
                (H2 : s2 =~ re2)
              : s2 =~ (Union re1 re2)
  | MStar0 re : [] =~ (Star re)
  | MStarApp s1 s2 re
                 (H1 : s1 =~ re)
                 (H2 : s2 =~ (Star re))
               : (s1 ++ s2) =~ (Star re)
  where "s =~ re" := (exp_match s re).

Lemma quiz : T (s:list T), ~(s =~ EmptySet).
Proof. intros T s Hc. inversion Hc. Qed.
Again, for readability, we display this definition using inference-rule notation.
   (MEmpty)  

[] =~ EmptyStr
   (MChar)  

[x] =~ Char x
s1 =~ re1    s2 =~ re2 (MApp)  

s1 ++ s2 =~ App re1 re2
s1 =~ re1 (MUnionL)  

s1 =~ Union re1 re2
s2 =~ re2 (MUnionR)  

s2 =~ Union re1 re2
   (MStar0)  

[] =~ Star re
s1 =~ re    s2 =~ Star re (MStarApp)  

s1 ++ s2 =~ Star 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.

Example reg_exp_ex1 : [1] =~ Char 1.
Proof.
  apply MChar.
Qed.

Example reg_exp_ex2 : [1; 2] =~ App (Char 1) (Char 2).
Proof.
  apply (MApp [1]).
  - apply MChar.
  - apply MChar.
Qed.
(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:

Example reg_exp_ex3 : ¬ ([1; 2] =~ Char 1).
Proof.
  intros H. inversion H.
Qed.
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:

Fixpoint reg_exp_of_list {T} (l : list T) :=
  match l with
  | []EmptyStr
  | x :: l'App (Char x) (reg_exp_of_list l')
  end.

Example reg_exp_ex4 : [1; 2; 3] =~ reg_exp_of_list [1; 2; 3].
Proof.
  simpl. apply (MApp [1]).
  { apply MChar. }
  apply (MApp [2]).
  { apply MChar. }
  apply (MApp [3]).
  { apply MChar. }
  apply MEmpty.
Qed.
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.

Lemma MStar1 :
   T s (re : reg_exp T) ,
    s =~ re
    s =~ Star re.
Proof.
  intros T s re H.
  rewrite <- (app_nil_r _ s).
  apply MStarApp.
  - apply H.
  - apply MStar0.
Qed.
(Note the use of app_nil_r to change the goal of the theorem to exactly the same 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.

Lemma empty_is_empty : T (s : list T),
  ¬ (s =~ EmptySet).
Proof.
  (* FILL IN HERE *) Admitted.

Lemma MUnion' : T (s : list T) (re1 re2 : reg_exp T),
  s =~ re1 s =~ re2
  s =~ Union re1 re2.
Proof.
  (* FILL IN HERE *) Admitted.
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.

Lemma MStar' : T (ss : list (list T)) (re : reg_exp T),
  ( s, In s ss s =~ re)
  fold app ss [] =~ Star re.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 4 stars, standard, optional (reg_exp_of_list_spec)

Prove that reg_exp_of_list satisfies the following specification:

Lemma reg_exp_of_list_spec : T (s1 s2 : list T),
  s1 =~ reg_exp_of_list s2 s1 = s2.
Proof.
  (* FILL IN HERE *) Admitted.
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 that we wanted 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 theorem, we first define a function re_chars that lists all characters that occur in a regular expression:

Fixpoint re_chars {T} (re : reg_exp T) : list T :=
  match re with
  | EmptySet[]
  | EmptyStr[]
  | Char x[x]
  | App re1 re2re_chars re1 ++ re_chars re2
  | Union re1 re2re_chars re1 ++ re_chars re2
  | Star rere_chars re
  end.
We can then phrase our theorem as follows:

Theorem in_re_match : T (s : list T) (re : reg_exp T) (x : T),
  s =~ re
  In x s
  In x (re_chars re).
Proof.
  intros T s re x Hmatch Hin.
  induction Hmatch
    as [| x'
        | s1 re1 s2 re2 Hmatch1 IH1 Hmatch2 IH2
        | s1 re1 re2 Hmatch IH | re1 s2 re2 Hmatch IH
        | re | s1 s2 re Hmatch1 IH1 Hmatch2 IH2].
  (* WORKED IN CLASS *)
  - (* MEmpty *)
    simpl in Hin. destruct Hin.
  - (* MChar *)
    simpl. simpl in Hin.
    apply Hin.
  - (* MApp *)
    simpl.
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).

    simpl. rewrite In_app_iff in ×.
    destruct Hin as [Hin | Hin].
    + (* In x s1 *)
      left. apply (IH1 Hin).
    + (* In x s2 *)
      right. apply (IH2 Hin).
  - (* MUnionL *)
    simpl. rewrite In_app_iff.
    left. apply (IH Hin).
  - (* MUnionR *)
    simpl. rewrite In_app_iff.
    right. apply (IH Hin).
  - (* MStar0 *)
    destruct Hin.
  - (* MStarApp *)
    simpl.
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.

    rewrite In_app_iff in Hin.
    destruct Hin as [Hin | Hin].
    + (* In x s1 *)
      apply (IH1 Hin).
    + (* In x s2 *)
      apply (IH2 Hin).
Qed.

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.

Fixpoint re_not_empty {T : Type} (re : reg_exp T) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Lemma re_not_empty_correct : T (re : reg_exp T),
  ( s, s =~ re) re_not_empty re = true.
Proof.
  (* FILL IN HERE *) Admitted.

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:

Lemma star_app: T (s1 s2 : list T) (re : reg_exp T),
  s1 =~ Star re
  s2 =~ Star re
  s1 ++ s2 =~ Star re.
Proof.
  intros T s1 s2 re H1.
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:

  generalize dependent s2.
  induction H1
    as [|x'|s1 re1 s2' re2 Hmatch1 IH1 Hmatch2 IH2
        |s1 re1 re2 Hmatch IH|re1 s2' re2 Hmatch IH
        |re''|s1 s2' re'' Hmatch1 IH1 Hmatch2 IH2].
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...

  - (* MEmpty *)
    simpl. intros s2 H. apply H.
... but most cases get stuck. For MChar, for instance, we must show that
    s2 =~ Char x'x' :: s2 =~ Char x',
which is clearly impossible.

  - (* MChar. *) intros s2 H. simpl. (* Stuck... *)
Abort.
The problem 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, such as Star re.
(In this respect, induction on evidence behaves more like destruct-without-eqn: than like inversion.)
An awkward way to solve this problem is "manually generalizing" over the problematic expressions by adding explicit equality hypotheses to the lemma:

Lemma star_app: T (s1 s2 : list T) (re re' : reg_exp T),
  re' = Star re
  s1 =~ re'
  s2 =~ Star re
  s1 ++ s2 =~ Star re.
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 idiom is so common that Coq provides a tactic to automatically generate such equations for us, avoiding thus the need for changing the statements of our theorems.

Abort.
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:

Lemma star_app: T (s1 s2 : list T) (re : reg_exp T),
  s1 =~ Star re
  s2 =~ Star re
  s1 ++ s2 =~ Star re.
Proof.
  intros T s1 s2 re H1.
  remember (Star re) as re'.
We now have Heqre' : re' = Star re.

  generalize dependent s2.
  induction H1
    as [|x'|s1 re1 s2' re2 Hmatch1 IH1 Hmatch2 IH2
        |s1 re1 re2 Hmatch IH|re1 s2' re2 Hmatch IH
        |re''|s1 s2' re'' Hmatch1 IH1 Hmatch2 IH2].
The Heqre' is contradictory in most cases, allowing us to conclude immediately.

  - (* MEmpty *) discriminate.
  - (* MChar *) discriminate.
  - (* MApp *) discriminate.
  - (* MUnionL *) discriminate.
  - (* MUnionR *) discriminate.
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.

  - (* MStar0 *)
    injection Heqre' as Heqre''. intros s H. apply H.

  - (* MStarApp *)
    injection Heqre' as Heqre''.
    intros s2 H1. rewrite <- app_assoc.
    apply MStarApp.
    + apply Hmatch1.
    + apply IH2.
      × rewrite Heqre''. reflexivity.
      × apply H1.
Qed.

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.

Lemma MStar'' : T (s : list T) (re : reg_exp T),
  s =~ Star re
   ss : list (list T),
    s = fold app ss []
     s', In s' ss s' =~ re.
Proof.
  (* FILL IN HERE *) Admitted.

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.)
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."

Module Pumping.

Fixpoint pumping_constant {T} (re : reg_exp T) : nat :=
  match re with
  | EmptySet ⇒ 1
  | EmptyStr ⇒ 1
  | Char _ ⇒ 2
  | App re1 re2
      pumping_constant re1 + pumping_constant re2
  | Union re1 re2
      pumping_constant re1 + pumping_constant re2
  | Star rpumping_constant r
  end.
You may find these lemmas about the pumping constant useful when proving the pumping lemma below.

Lemma pumping_constant_ge_1 :
   T (re : reg_exp T),
    pumping_constant re 1.
Proof.
  intros T re. induction re.
  - (* Emptyset *)
    apply le_n.
  - (* EmptyStr *)
    apply le_n.
  - (* Char *)
    apply le_S. apply le_n.
  - (* App *)
    simpl.
    apply le_trans with (n:=pumping_constant re1).
    apply IHre1. apply le_plus_l.
  - (* Union *)
    simpl.
    apply le_trans with (n:=pumping_constant re1).
    apply IHre1. apply le_plus_l.
  - (* Star *)
    simpl. apply IHre.
Qed.

Lemma pumping_constant_0_false :
   T (re : reg_exp T),
    pumping_constant re = 0 False.
Proof.
  intros T re H.
  assert (Hp1 : pumping_constant re 1).
  { apply pumping_constant_ge_1. }
  inversion Hp1 as [Hp1'| p Hp1' Hp1''].
  - rewrite H in Hp1'. discriminate Hp1'.
  - rewrite H in Hp1''. discriminate Hp1''.
Qed.
Next, it is useful to define an auxiliary function that repeats a string (appends it to itself) some number of times.

Fixpoint napp {T} (n : nat) (l : list T) : list T :=
  match n with
  | 0 ⇒ []
  | S n'l ++ napp n' l
  end.
This auxiliary lemma might also be useful in your proof of the pumping lemma.

Lemma napp_plus: T (n m : nat) (l : list T),
  napp (n + m) l = napp n l ++ napp m l.
Proof.
  intros T n m l.
  induction n as [|n IHn].
  - reflexivity.
  - simpl. rewrite IHn, app_assoc. reflexivity.
Qed.

Lemma napp_star :
   T m s1 s2 (re : reg_exp T),
    s1 =~ re s2 =~ Star re
    napp m s1 ++ s2 =~ Star re.
Proof.
  intros T m s1 s2 re Hs1 Hs2.
  induction m.
  - simpl. apply Hs2.
  - simpl. rewrite <- app_assoc.
    apply MStarApp.
    + apply Hs1.
    + apply IHm.
Qed.
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.

Lemma weak_pumping : T (re : reg_exp T) s,
  s =~ re
  pumping_constant re length s
   s1 s2 s3,
    s = s1 ++ s2 ++ s3
    s2 []
     m, s1 ++ napp m s2 ++ s3 =~ re.
You are to fill in the proof. Several of the lemmas about le that were in an optional exercise earlier in this chapter may be useful.
Proof.
  intros T re s Hmatch.
  induction Hmatch
    as [ | x | s1 re1 s2 re2 Hmatch1 IH1 Hmatch2 IH2
       | s1 re1 re2 Hmatch IH | re1 s2 re2 Hmatch IH
       | re | s1 s2 re Hmatch1 IH1 Hmatch2 IH2 ].
  - (* MEmpty *)
    simpl. intros contra. inversion contra.
  (* FILL IN HERE *) Admitted.

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.

Lemma pumping : T (re : reg_exp T) s,
  s =~ re
  pumping_constant re length s
   s1 s2 s3,
    s = s1 ++ s2 ++ s3
    s2 []
    length s1 + length s2 pumping_constant re
     m, s1 ++ napp m s2 ++ s3 =~ re.
You may want to copy your proof of weak_pumping below.
Proof.
  intros T re s Hmatch.
  induction Hmatch
    as [ | x | s1 re1 s2 re2 Hmatch1 IH1 Hmatch2 IH2
       | s1 re1 re2 Hmatch IH | re1 s2 re2 Hmatch IH
       | re | s1 s2 re Hmatch1 IH1 Hmatch2 IH2 ].
  - (* MEmpty *)
    simpl. intros contra. inversion contra.
  (* FILL IN HERE *) Admitted.

End Pumping.

Case Study: Improving Reflection

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

Theorem filter_not_empty_In : n l,
  filter (fun xn =? x) l []
  In n l.
Proof.
  intros n l. induction l as [|m l' IHl'].
  - (* l =  *)
    simpl. intros H. apply H. reflexivity.
  - (* l = m :: l' *)
    simpl. destruct (n =? m) eqn:H.
    + (* n =? m = true *)
      intros _. rewrite eqb_eq in H. rewrite H.
      left. reflexivity.
    + (* n =? m = false *)
      intros H'. right. apply IHl'. apply H'.
Qed.
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 by defining an inductive proposition that yields a better case-analysis principle for n =? m. Instead of generating an equation such as (n =? m) = true, which is generally not directly useful, this principle gives us right away the assumption we really need: n = m.

Inductive reflect (P : Prop) : bool Prop :=
| ReflectT (H : P) : reflect P true
| ReflectF (H : ¬ P) : reflect P false.
The reflect property takes two arguments: a proposition P and a boolean b. Intuitively, it states that the property P is reflected in (i.e., 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 it should be possible to extract evidence for P from a proof of reflect P true. Similarly, the only way to show reflect P false is by combining evidence for ¬ P with the ReflectF constructor.
It is easy to formalize this intuition and show that the statements P b = true and reflect P b are indeed equivalent. First, the left-to-right implication:

Theorem iff_reflect : P b, (P b = true) reflect P b.
Proof.
  (* WORKED IN CLASS *)
  intros P b H. destruct b.
  - apply ReflectT. rewrite H. reflexivity.
  - apply ReflectF. rewrite H. intros H'. discriminate.
Qed.
Now you prove the right-to-left implication:

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

Theorem reflect_iff : P b, reflect P b (P b = true).
Proof.
  (* FILL IN HERE *) Admitted.
The advantage of reflect over the normal "if and only if" connective 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).

Lemma eqbP : n m, reflect (n = m) (n =? m).
Proof.
  intros n m. apply iff_reflect. rewrite eqb_eq. reflexivity.
Qed.
A smoother proof of filter_not_empty_In now goes as follows. Notice how the calls to destruct and rewrite are combined into a single call to destruct.
(To see this clearly, look at 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.)

Theorem filter_not_empty_In' : n l,
  filter (fun xn =? x) l []
  In n l.
Proof.
  intros n l. induction l as [|m l' IHl'].
  - (* l =  *)
    simpl. intros H. apply H. reflexivity.
  - (* l = m :: l' *)
    simpl. destruct (eqbP n m) as [H | H].
    + (* n = m *)
      intros _. rewrite H. left. reflexivity.
    + (* n <> m *)
      intros H'. right. apply IHl'. apply H'.
Qed.

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

Use eqbP as above to prove the following:

Fixpoint count n l :=
  match l with
  | [] ⇒ 0
  | m :: l'(if n =? m then 1 else 0) + count n l'
  end.

Theorem eqbP_practice : n l,
  count n l = 0 ~(In n l).
Proof.
  (* FILL IN HERE *) Admitted.
This small example shows how reflection gives us a small gain in convenience; in larger developments, using reflect consistently can often lead to noticeably shorter and clearer proof scripts. We'll see many more examples in later chapters and in Programming Language Foundations.
The use of the reflect property has been popularized by SSReflect, a Coq library that has been used to formalize important results in mathematics, including as the 4-color theorem and the Feit-Thompson theorem. The name SSReflect stands for small-scale reflection, i.e., the pervasive use of reflection to simplify small proof steps with boolean computations.

Additional Exercises

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

Formulating inductive definitions of properties is an important skill you'll need in this course. Try to solve this exercise without any help at all.
We say that a list "stutters" if it repeats the same element consecutively. (This is different from not containing duplicates: the sequence [1;4;1] repeats the element 1 but does not stutter.) The property "nostutter mylist" means that mylist does not stutter. Formulate an inductive definition for nostutter.

Inductive nostutter {X:Type} : list X Prop :=
 (* FILL IN HERE *)
.
Make sure each of these tests succeeds, but feel free to change the suggested proof (in comments) if the given one doesn't work for you. Your definition might be different from ours and still be correct, in which case the examples might need a different proof. (You'll notice that the suggested proofs use a number of tactics we haven't talked about, to make them more robust to different possible ways of defining nostutter. You can probably just uncomment and use them as-is, but you can also prove each example with more basic tactics.)

Example test_nostutter_1: nostutter [3;1;4;1;5;6].
(* FILL IN HERE *) Admitted.
(* 
  Proof. repeat constructor; apply eqb_neq; auto.
  Qed.
*)


Example test_nostutter_2: nostutter (@nil nat).
(* FILL IN HERE *) Admitted.
(* 
  Proof. repeat constructor; apply eqb_neq; auto.
  Qed.
*)


Example test_nostutter_3: nostutter [5].
(* FILL IN HERE *) Admitted.
(* 
  Proof. repeat constructor; auto. Qed.
*)


Example test_nostutter_4: not (nostutter [3;1;1;4]).
(* FILL IN HERE *) Admitted.
(* 
  Proof. intro.
  repeat match goal with
    h: nostutter _ ⊢ _ => inversion h; clear h; subst
  end.
  contradiction; auto. Qed.
*)


(* Do not modify the following line: *)
Definition manual_grade_for_nostutter : option (nat×string) := None.

Exercise: 4 stars, advanced (filter_challenge)

Let's prove that our definition of filter from the Poly chapter matches an abstract specification. Here is the specification, written out informally in English:
A list l is an "in-order merge" of l1 and l2 if it contains all the same elements as l1 and l2, in the same order as l1 and l2, but possibly interleaved. For example,
    [1;4;6;2;3]
is an in-order merge of
    [1;6;2]
and
    [4;3].
Now, suppose we have a set X, a function test: Xbool, and a list l of type list X. Suppose further that l is an in-order merge of two lists, l1 and l2, such that every item in l1 satisfies test and no item in l2 satisfies test. Then filter test l = l1.
Translate this specification into a Coq theorem and prove it. (You'll need to begin by defining what it means for one list to be a merge of two others. Do this with an inductive relation, not a Fixpoint.)

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_filter_challenge : option (nat×string) := None.

Exercise: 5 stars, advanced, optional (filter_challenge_2)

A different way to characterize the behavior of filter goes like this: Among all subsequences of l with the property that test evaluates to true on all their members, filter test l is the longest. Formalize this claim and prove it.

(* FILL IN HERE *)

Exercise: 4 stars, standard, optional (palindromes)

A palindrome is a sequence that reads the same backwards as forwards.
  • Define an inductive proposition pal on list X that captures what it means to be a palindrome. (Hint: You'll need three cases. Your definition should be based on the structure of the list; just having a single constructor like
            c : l, l = rev lpal l
    may seem obvious, but will not work very well.)
  • Prove (pal_app_rev) that
            l, pal (l ++ rev l).
  • Prove (pal_rev that)
            l, pal ll = rev l.

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_pal_pal_app_rev_pal_rev : option (nat×string) := None.

Exercise: 5 stars, standard, optional (palindrome_converse)

Again, the converse direction is significantly more difficult, due to the lack of evidence. Using your definition of pal from the previous exercise, prove that
      l, l = rev lpal l.

(* FILL IN HERE *)

Exercise: 4 stars, advanced, optional (NoDup)

Recall the definition of the In property from the Logic chapter, which asserts that a value x appears at least once in a list l:

(* Fixpoint In (A : Type) (x : A) (l : list A) : Prop :=
   match l with
   |  => False
   | x' :: l' => x' = x \/ In A x l'
   end *)

Your first task is to use In to define a proposition disjoint X l1 l2, which should be provable exactly when l1 and l2 are lists (with elements of type X) that have no elements in common.

(* FILL IN HERE *)
Next, use In to define an inductive proposition NoDup X l, which should be provable exactly when l is a list (with elements of type X) where every member is different from every other. For example, NoDup nat [1;2;3;4] and NoDup bool [] should be provable, while NoDup nat [1;2;1] and NoDup bool [true;true] should not be.

(* FILL IN HERE *)
Finally, state and prove one or more interesting theorems relating disjoint, NoDup and ++ (list append).

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_NoDup_disjoint_etc : option (nat×string) := None.

Exercise: 4 stars, advanced, optional (pigeonhole_principle)

The pigeonhole principle states a basic fact about counting: if we distribute more than n items into n pigeonholes, some pigeonhole must contain at least two items. As often happens, this apparently trivial fact about numbers requires non-trivial machinery to prove, but we now have enough...
First prove an easy useful lemma.

Lemma in_split : (X:Type) (x:X) (l:list X),
  In x l
   l1 l2, l = l1 ++ x :: l2.
Proof.
  (* FILL IN HERE *) Admitted.
Now define a property repeats such that repeats X l asserts that l contains at least one repeated element (of type X).

Inductive repeats {X:Type} : list X Prop :=
  (* FILL IN HERE *)
.

(* Do not modify the following line: *)
Definition manual_grade_for_check_repeats : option (nat×string) := None.
Now, here's a way to formalize the pigeonhole principle. Suppose list l2 represents a list of pigeonhole labels, and list l1 represents the labels assigned to a list of items. If there are more items than labels, at least two items must have the same label -- i.e., list l1 must contain repeats.
This proof is much easier if you use the excluded_middle hypothesis to show that In is decidable, i.e., x l, (In x l) ¬ (In x l). However, it is also possible to make the proof go through without assuming that In is decidable; if you manage to do this, you will not need the excluded_middle hypothesis.

Theorem pigeonhole_principle: (X:Type) (l1 l2:list X),
   excluded_middle
   ( x, In x l1 In x l2)
   length l2 < length l1
   repeats l1.
Proof.
   intros X l1. induction l1 as [|x l1' IHl1'].
  (* FILL IN HERE *) Admitted.

Extended Exercise: A Verified Regular-Expression Matcher

We have now defined a match relation over regular expressions and polymorphic lists. We can use such a definition to manually prove that a given regex matches a given string, but it does not give us a program that we can run to determine a match autmatically.
It would be reasonable to hope that we can translate the definitions of the inductive rules for constructing evidence of the match relation into cases of a recursive function that reflects the relation by recursing on a given regex. However, it does not seem straightforward to define such a function in which the given regex is a recursion variable recognized by Coq. As a result, Coq will not accept that the function always terminates.
Heavily-optimized regex matchers match a regex by translating a given regex into a state machine and determining if the state machine accepts a given string. However, regex matching can also be implemented using an algorithm that operates purely on strings and regexes without defining and maintaining additional datatypes, such as state machines. We'll implemement such an algorithm, and verify that its value reflects the match relation.
We will implement a regex matcher that matches strings represented as lists of ASCII characters:
Require Import Coq.Strings.Ascii.

Definition string := list ascii.
The Coq standard library contains a distinct inductive definition of strings of ASCII characters. However, we will use the above definition of strings as lists as ASCII characters in order to apply the existing definition of the match relation.
We could also define a regex matcher over polymorphic lists, not lists of ASCII characters specifically. The matching algorithm that we will implement needs to be able to test equality of elements in a given list, and thus needs to be given an equality-testing function. Generalizing the definitions, theorems, and proofs that we define for such a setting is a bit tedious, but workable.
The proof of correctness of the regex matcher will combine properties of the regex-matching function with properties of the match relation that do not depend on the matching function. We'll go ahead and prove the latter class of properties now. Most of them have straightforward proofs, which have been given to you, although there are a few key lemmas that are left for you to prove.
Each provable Prop is equivalent to True.
Lemma provable_equiv_true : (P : Prop), P (P True).
Proof.
  intros.
  split.
  - intros. constructor.
  - intros _. apply H.
Qed.
Each Prop whose negation is provable is equivalent to False.
Lemma not_equiv_false : (P : Prop), ¬P (P False).
Proof.
  intros.
  split.
  - apply H.
  - intros. destruct H0.
Qed.
EmptySet matches no string.
Lemma null_matches_none : (s : string), (s =~ EmptySet) False.
Proof.
  intros.
  apply not_equiv_false.
  unfold not. intros. inversion H.
Qed.
EmptyStr only matches the empty string.
Lemma empty_matches_eps : (s : string), s =~ EmptyStr s = [ ].
Proof.
  split.
  - intros. inversion H. reflexivity.
  - intros. rewrite H. apply MEmpty.
Qed.
EmptyStr matches no non-empty string.
Lemma empty_nomatch_ne : (a : ascii) s, (a :: s =~ EmptyStr) False.
Proof.
  intros.
  apply not_equiv_false.
  unfold not. intros. inversion H.
Qed.
Char a matches no string that starts with a non-a character.
Lemma char_nomatch_char :
   (a b : ascii) s, b a (b :: s =~ Char a False).
Proof.
  intros.
  apply not_equiv_false.
  unfold not.
  intros.
  apply H.
  inversion H0.
  reflexivity.
Qed.
If Char a matches a non-empty string, then the string's tail is empty.
Lemma char_eps_suffix : (a : ascii) s, a :: s =~ Char a s = [ ].
Proof.
  split.
  - intros. inversion H. reflexivity.
  - intros. rewrite H. apply MChar.
Qed.
App re0 re1 matches string s iff s = s0 ++ s1, where s0 matches re0 and s1 matches re1.
Lemma app_exists : (s : string) re0 re1,
    s =~ App re0 re1
     s0 s1, s = s0 ++ s1 s0 =~ re0 s1 =~ re1.
Proof.
  intros.
  split.
  - intros. inversion H. s1, s2. split.
    × reflexivity.
    × split. apply H3. apply H4.
  - intros [ s0 [ s1 [ Happ [ Hmat0 Hmat1 ] ] ] ].
    rewrite Happ. apply (MApp s0 _ s1 _ Hmat0 Hmat1).
Qed.

Exercise: 3 stars, standard, optional (app_ne)

App re0 re1 matches a::s iff re0 matches the empty string and a::s matches re1 or s=s0++s1, where a::s0 matches re0 and s1 matches re1.
Even though this is a property of purely the match relation, it is a critical observation behind the design of our regex matcher. So (1) take time to understand it, (2) prove it, and (3) look for how you'll use it later.
Lemma app_ne : (a : ascii) s re0 re1,
    a :: s =~ (App re0 re1)
    ([ ] =~ re0 a :: s =~ re1)
     s0 s1, s = s0 ++ s1 a :: s0 =~ re0 s1 =~ re1.
Proof.
  (* FILL IN HERE *) Admitted.
s matches Union re0 re1 iff s matches re0 or s matches re1.
Lemma union_disj : (s : string) re0 re1,
    s =~ Union re0 re1 s =~ re0 s =~ re1.
Proof.
  intros. split.
  - intros. inversion H.
    + left. apply H2.
    + right. apply H1.
  - intros [ H | H ].
    + apply MUnionL. apply H.
    + apply MUnionR. apply H.
Qed.

Exercise: 3 stars, standard, optional (star_ne)

a::s matches Star re iff s = s0 ++ s1, where a::s0 matches re and s1 matches Star re. Like app_ne, this observation is critical, so understand it, prove it, and keep it in mind.
Hint: you'll need to perform induction. There are quite a few reasonable candidates for Prop's to prove by induction. The only one that will work is splitting the iff into two implications and proving one by induction on the evidence for a :: s =~ Star re. The other implication can be proved without induction.
In order to prove the right property by induction, you'll need to rephrase a :: s =~ Star re to be a Prop over general variables, using the remember tactic.

Lemma star_ne : (a : ascii) s re,
    a :: s =~ Star re
     s0 s1, s = s0 ++ s1 a :: s0 =~ re s1 =~ Star re.
Proof.
  (* FILL IN HERE *) Admitted.
The definition of our regex matcher will include two fixpoint functions. The first function, given regex re, will evaluate to a value that reflects whether re matches the empty string. The function will satisfy the following property:
Definition refl_matches_eps m :=
   re : reg_exp ascii, reflect ([ ] =~ re) (m re).

Exercise: 2 stars, standard, optional (match_eps)

Complete the definition of match_eps so that it tests if a given regex matches the empty string:
Fixpoint match_eps (re: reg_exp ascii) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Exercise: 3 stars, standard, optional (match_eps_refl)

Now, prove that match_eps indeed tests if a given regex matches the empty string. (Hint: You'll want to use the reflection lemmas ReflectT and ReflectF.)
Lemma match_eps_refl : refl_matches_eps match_eps.
Proof.
  (* FILL IN HERE *) Admitted.
We'll define other functions that use match_eps. However, the only property of match_eps that you'll need to use in all proofs over these functions is match_eps_refl.
The key operation that will be performed by our regex matcher will be to iteratively construct a sequence of regex derivatives. For each character a and regex re, the derivative of re on a is a regex that matches all suffixes of strings matched by re that start with a. I.e., re' is a derivative of re on a if they satisfy the following relation:

Definition is_der re (a : ascii) re' :=
   s, a :: s =~ re s =~ re'.
A function d derives strings if, given character a and regex re, it evaluates to the derivative of re on a. I.e., d satisfies the following property:
Definition derives d := a re, is_der re a (d a re).

Exercise: 3 stars, standard, optional (derive)

Define derive so that it derives strings. One natural implementation uses match_eps in some cases to determine if key regex's match the empty string.
Fixpoint derive (a : ascii) (re : reg_exp ascii) : reg_exp ascii
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
The derive function should pass the following tests. Each test establishes an equality between an expression that will be evaluated by our regex matcher and the final value that must be returned by the regex matcher. Each test is annotated with the match fact that it reflects.
Example c := ascii_of_nat 99.
Example d := ascii_of_nat 100.
"c" =~ EmptySet:
Example test_der0 : match_eps (derive c (EmptySet)) = false.
Proof.
  (* FILL IN HERE *) Admitted.
"c" =~ Char c:
Example test_der1 : match_eps (derive c (Char c)) = true.
Proof.
  (* FILL IN HERE *) Admitted.
"c" =~ Char d:
Example test_der2 : match_eps (derive c (Char d)) = false.
Proof.
  (* FILL IN HERE *) Admitted.
"c" =~ App (Char c) EmptyStr:
Example test_der3 : match_eps (derive c (App (Char c) EmptyStr)) = true.
Proof.
  (* FILL IN HERE *) Admitted.
"c" =~ App EmptyStr (Char c):
Example test_der4 : match_eps (derive c (App EmptyStr (Char c))) = true.
Proof.
  (* FILL IN HERE *) Admitted.
"c" =~ Star c:
Example test_der5 : match_eps (derive c (Star (Char c))) = true.
Proof.
  (* FILL IN HERE *) Admitted.
"cd" =~ App (Char c) (Char d):
Example test_der6 :
  match_eps (derive d (derive c (App (Char c) (Char d)))) = true.
Proof.
  (* FILL IN HERE *) Admitted.
"cd" =~ App (Char d) (Char c):
Example test_der7 :
  match_eps (derive d (derive c (App (Char d) (Char c)))) = false.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 4 stars, standard, optional (derive_corr)

Prove that derive in fact always derives strings.
Hint: one proof performs induction on re, although you'll need to carefully choose the property that you prove by induction by generalizing the appropriate terms.
Hint: if your definition of derive applies match_eps to a particular regex re, then a natural proof will apply match_eps_refl to re and destruct the result to generate cases with assumptions that the re does or does not match the empty string.
Hint: You can save quite a bit of work by using lemmas proved above. In particular, to prove many cases of the induction, you can rewrite a Prop over a complicated regex (e.g., s =~ Union re0 re1) to a Boolean combination of Prop's over simple regex's (e.g., s =~ re0 s =~ re1) using lemmas given above that are logical equivalences. You can then reason about these Prop's naturally using intro and destruct.
Lemma derive_corr : derives derive.
Proof.
  (* FILL IN HERE *) Admitted.
We'll define the regex matcher using derive. However, the only property of derive that you'll need to use in all proofs of properties of the matcher is derive_corr.
A function m matches regexes if, given string s and regex re, it evaluates to a value that reflects whether s is matched by re. I.e., m holds the following property:
Definition matches_regex m : Prop :=
   (s : string) re, reflect (s =~ re) (m s re).

Exercise: 2 stars, standard, optional (regex_match)

Complete the definition of regex_match so that it matches regexes.
Fixpoint regex_match (s : string) (re : reg_exp ascii) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Exercise: 3 stars, standard, optional (regex_refl)

Finally, prove that regex_match in fact matches regexes.
Hint: if your definition of regex_match applies match_eps to regex re, then a natural proof applies match_eps_refl to re and destructs the result to generate cases in which you may assume that re does or does not match the empty string.
Hint: if your definition of regex_match applies derive to character x and regex re, then a natural proof applies derive_corr to x and re to prove that x :: s =~ re given s =~ derive x re, and vice versa.
Theorem regex_refl : matches_regex regex_match.
Proof.
  (* FILL IN HERE *) Admitted.

(* 2020-08-25 12:53 *)