# IndPropInductively Defined Propositions

Require Export Logic.

# Inductively Defined Propositions

*inductive definitions*.

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

*Inference rules*are one such notation:

(ev_0) | |

ev 0 |

ev n | (ev_SS) |

ev (S (S n)) |

*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.

*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 below.
ev 0

------ (ev_SS)

ev 2

------ (ev_SS)

ev 4

Inductive ev : nat → Prop :=

| ev_0 : ev 0

| ev_SS : ∀n : nat, ev n → ev (S (S n)).

This definition is different in one crucial respect from
previous uses of Inductive: its result is not a Type, but
rather a function from nat to Prop — that is, a property of
numbers. Note that we've already seen other inductive definitions
that result in functions, such as list, whose type is Type →
Type. What is new here is that, because the nat argument of
ev appears
In contrast, the definition of list names the X parameter

*unnamed*, to the*right*of the colon, 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.*globally*, to the*left*of the colon, forcing the result of nil and cons to be the same (list X). Had we tried to bring nat to the left 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 : ∀n, wrong_ev n → wrong_ev (S (S n)).

(* ===> Error: A parameter of an inductive type n is not

allowed to be used as a bound variable in the type

of its constructor. *)

("Parameter" here is Coq jargon for an argument on the left of the
colon in an Inductive definition; "index" is used to refer to
arguments on the right of the colon.)
We can think of the definition of ev as defining a Coq property
ev : nat → Prop, together with theorems ev_0 : ev 0 and
ev_SS : ∀ n, ev n → ev (S (S n)). Such "constructor
theorems" 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.

Theorem ev_double : ∀n,

ev (double n).

Proof.

(* FILL IN HERE *) Admitted.

ev (double n).

Proof.

(* FILL IN HERE *) Admitted.

☐

# Using Evidence in Proofs

*constructing*evidence that numbers are even, we can also

*reason about*such evidence.

*only*ways to build evidence that numbers are even (in the sense of ev). In other words, if someone gives us evidence E for the assertion ev n, then we know that E must have one of two shapes:

- 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').

*induction*and

*case analysis*on such evidence. Let's look at a few examples to see what this means in practice.

## Inversion on Evidence

Theorem evenb_minus2: ∀n,

evenb n = true → evenb (pred (pred n)) = true.

Proof.

intros [ | [ | n' ] ].

- (* n = 0 *) reflexivity.

- (* n = 1; contradiction *) intros H. inversion H.

- (* n = n' + 2 *) simpl. intros H. apply H.

Qed.

We can state the same claim in terms of ev, but this quickly
leads us to an obstacle: Since ev is defined inductively —
rather than as a function — Coq doesn't know how to simplify a
goal involving ev n after case analysis on n. As a
consequence, the same proof strategy fails:

Theorem ev_minus2: ∀n,

ev n → ev (pred (pred n)).

Proof.

intros [ | [ | n' ] ].

- (* n = 0 *) simpl. intros _. apply ev_0.

- (* n = 1; we're stuck! *) simpl.

Abort.

The solution is to perform case analysis on the evidence that ev
n
We can invoke this kind of argument in Coq using the inversion
tactic. Besides allowing us to reason about equalities involving
constructors, inversion provides a case-analysis principle for
inductively defined propositions. When used in this way, its
syntax is similar to destruct: We pass it a list of identifiers
separated by | characters to name the arguments to each of the
possible constructors. For instance:

*directly*. By the definition of ev, there are two cases to consider:- If that evidence is of the form ev_0, we know that n = 0.
Therefore, it suffices to show that ev (pred (pred 0)) holds.
By the definition of pred, this is equivalent to showing that
ev 0 holds, which directly follows from ev_0.
- Otherwise, that evidence must have the form ev_SS n' E', where n = S (S n') and E' is evidence for ev n'. We must then show that ev (pred (pred (S (S n')))) holds, which, after simplification, follows directly from E'.

Theorem ev_minus2 : ∀n,

ev n → ev (pred (pred n)).

Proof.

intros n E.

inversion E as [| n' E'].

- (* E = ev_0 *) simpl. apply ev_0.

- (* E = ev_SS n' E' *) simpl. apply E'. Qed.

Note that, in this particular case, it is also possible to replace
inversion by destruct:

Theorem ev_minus2' : ∀n,

ev n → ev (pred (pred n)).

Proof.

intros n E.

destruct E as [| n' E'].

- (* E = ev_0 *) simpl. apply ev_0.

- (* E = ev_SS n' E' *) simpl. apply E'. Qed.

The difference between the two forms is that inversion is more
convenient when used on a hypothesis that consists of an inductive
property applied to a complex expression (as opposed to a single
variable). Here's is a concrete example. Suppose that we wanted
to prove the following variation of ev_minus2:

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'].

- (* 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.
The inversion tactic, on the other hand, can detect (1) that the
first case does not apply, and (2) that the n' that appears on
the ev_SS case must be the same as n. This allows us to
complete the proof:

Theorem evSS_ev : ∀n,

ev (S (S n)) → ev n.

Proof.

intros n E.

inversion E as [| n' E'].

(* We are in the E = ev_SS n' E' case now. *)

apply E'.

Qed.

By using inversion, we can also apply the principle of explosion
to "obviously contradictory" hypotheses involving inductive
properties. For example:

Theorem one_not_even : ¬ ev 1.

Proof.

intros H. inversion H. Qed.

Theorem SSSSev__even : ∀n,

ev (S (S (S (S n)))) → ev n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem even5_nonsense :

ev 5 → 2 + 2 = 9.

Proof.

(* FILL IN HERE *) Admitted.

☐
The way we've used inversion here may seem a bit
mysterious at first. Until now, we've only used inversion on
equality propositions, to utilize injectivity of constructors or
to discriminate between different constructors. But we see here
that inversion can also be applied to analyzing evidence for
inductively defined propositions.
Here's how inversion works in general. Suppose the name I
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 I generates a subgoal in which
I 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).

## Induction on Evidence

Lemma ev_even : ∀n,

ev n → ∃k, n = double k.

Proof.

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, as in the previous section. Thus, it seems
better to first try inversion on the evidence for ev. Indeed,
the first case can be solved trivially.

intros n E. inversion E as [| n' E'].

- (* 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 an similar
one that involves a

*different*piece of evidence for ev: 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.
assert (I : (∃k', n' = double k') →

(∃k, S (S n') = double k)).

{ intros [k' Hk']. rewrite Hk'. ∃(S k').

reflexivity. }

apply I. (* reduce the original goal to the new one *)

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 hypotheses for each recursive occurrence of
the property in question.
Let's try our current lemma again:

Abort.

Lemma ev_even : ∀n,

ev n → ∃k, n = double k.

Proof.

intros n E.

induction E as [|n' E' IH].

- (* E = ev_0 *)

∃0. reflexivity.

- (* E = ev_SS n' E'

with IH : exists k', n' = double k' *)

destruct IH as [k' Hk'].

rewrite Hk'. ∃(S k'). 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 ↔ ∃k, n = double k.

Proof.

intros n. split.

- (* -> *) apply ev_even.

- (* <- *) intros [k Hk]. rewrite Hk. apply ev_double.

Qed.

As we will see in later chapters, induction on evidence is a
recurring technique when studying 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 (ev_sum)

Theorem ev_sum : ∀n m, ev n → ev m → ev (n + m).

Proof.

(* FILL IN HERE *) Admitted.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 4 stars, advanced (ev_alternate)

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, ev' n → ev' m → ev' (n + m).

Prove that this definition is logically equivalent to
the old one.

Theorem ev'_ev : ∀n, ev' n ↔ ev n.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars, advanced, recommended (ev_ev__ev)

Finding the appropriate thing to do induction on is a bit tricky here:Theorem ev_ev__ev : ∀n m,

ev (n+m) → ev n → ev m.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars, optional (ev_plus_plus)

This exercise just requires applying existing lemmas. No induction or even case analysis is needed, though some of the rewriting may be tedious.Theorem ev_plus_plus : ∀n m p,

ev (n+m) → ev (n+p) → ev (m+p).

Proof.

(* FILL IN HERE *) Admitted.

☐

# Inductive Relations

*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.

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, le n n

| le_S : ∀n m, (le n m) → (le n (S m)).

Notation "m ≤ n" := (le m n).

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).

Here are a few more simple relations on numbers:

Inductive square_of : nat → nat → Prop :=

sq : ∀n:nat, square_of n (n × n).

Inductive next_nat : nat → nat → Prop :=

| nn : ∀n:nat, next_nat n (S n).

Inductive next_even : nat → nat → Prop :=

| ne_1 : ∀n, ev (S n) → next_even n (S n)

| ne_2 : ∀n, ev (S (S n)) → next_even n (S (S n)).

#### Exercise: 2 stars, recommended (total_relation)

Define an inductive binary relation total_relation that holds between every pair of natural numbers.(* FILL IN HERE *)

☐

#### Exercise: 2 stars (empty_relation)

Define an inductive binary relation empty_relation (on numbers) that never holds.(* FILL IN HERE *)

☐

#### Exercise: 3 stars, 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_lt : ∀n

_{1}n

_{2}m,

n

_{1}+ n

_{2}< m →

n

_{1}< m ∧ n

_{2}< m.

Proof.

unfold lt.

(* FILL IN HERE *) Admitted.

Theorem lt_S : ∀n m,

n < m →

n < S m.

Proof.

(* FILL IN HERE *) Admitted.

Theorem leb_complete : ∀n m,

leb 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 →

leb n m = true.

Proof.

(* FILL IN HERE *) Admitted.

Hint: This theorem can easily be proved without using induction.

Theorem leb_true_trans : ∀n m o,

leb n m = true → leb m o = true → leb n o = true.

Proof.

(* FILL IN HERE *) Admitted.

Theorem leb_iff : ∀n m,

leb n m = true ↔ n ≤ m.

Proof.

(* FILL IN HERE *) Admitted.

leb n m = true ↔ n ≤ m.

Proof.

(* FILL IN HERE *) Admitted.

☐

Module R.

#### Exercise: 3 stars, recommended (R_provability2)

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, R m n o → R (S m) n (S o)

| c3 : ∀m n o, R m n o → R m (S n) (S o)

| c4 : ∀m n o, R (S m) (S n) (S (S o)) → R m n o

| c5 : ∀m n o, 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.

☐

#### Exercise: 3 stars, 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 :=

(* FILL IN HERE *) admit.

Theorem R_equiv_fR : ∀m n o, R m n o ↔ fR m n = o.

Proof.

(* FILL IN HERE *) Admitted.

☐

End R.

#### Exercise: 4 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 [1;1;1;2;2;3]

[1;2;7;3]

[5;6;1;9;9;2;7;3;8]

*not*a subsequence of any of the lists

[1;2]

[1;3]

[5;6;2;1;7;3;8]

[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!

(* FILL IN HERE *)

☐
☐

#### Exercise: 2 stars, optional (R_provability)

Suppose we give Coq the following definition:
Inductive R : nat → list nat → Prop :=

| c1 : R 0 []

| c2 : ∀n l, R n l → R (S n) (n :: l)

| c3 : ∀n l, R (S n) l → R n l.

Which of the following propositions are provable?
| c1 : R 0 []

| c2 : ∀n l, R n l → R (S n) (n :: l)

| c3 : ∀n l, R (S n) l → R n l.

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

# Case Study: Regular Expressions

*regular expressions*.

Inductive reg_exp (T : Type) : Type :=

| EmptySet : reg_exp T

| EmptyStr : reg_exp T

| Char : T → reg_exp T

| App : reg_exp T → reg_exp T → reg_exp T

| Union : reg_exp T → reg_exp T → reg_exp T

| Star : reg_exp T → 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
We connect regular expressions and strings via the following
rules, which define when a regular expression
We can easily translate this informal definition into an
Inductive one as follows:

*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.)*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 s
_{1}, and re2 matches s_{2}, then App re1 re2 matches s_{1}++ s_{2}. - 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. (As a special case, the sequence of strings may be empty, so Star re always matches the empty string [] no matter what re is.)

Inductive exp_match {T} : list T → reg_exp T → Prop :=

| MEmpty : exp_match [] EmptyStr

| MChar : ∀x, exp_match [x] (Char x)

| MApp : ∀s

_{1}re1 s

_{2}re2,

exp_match s

_{1}re1 →

exp_match s

_{2}re2 →

exp_match (s

_{1}++ s

_{2}) (App re1 re2)

| MUnionL : ∀s

_{1}re1 re2,

exp_match s

_{1}re1 →

exp_match s

_{1}(Union re1 re2)

| MUnionR : ∀re1 s

_{2}re2,

exp_match s

_{2}re2 →

exp_match s

_{2}(Union re1 re2)

| MStar0 : ∀re, exp_match [] (Star re)

| MStarApp : ∀s

_{1}s

_{2}re,

exp_match s

_{1}re →

exp_match s

_{2}(Star re) →

exp_match (s

_{1}++ s

_{2}) (Star re).

Once again, for readability, we can also display this definition
using inference-rule notation. At the same time, let's introduce
a more readable infix notation.

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

(MEmpty) | |

[] =~ EmptyStr |

(MChar) | |

[x] =~ Char x |

s_{1} =~ re1 s_{2} =~ re2 |
(MApp) |

s_{1} ++ s_{2} =~ App re1 re2 |

s_{1} =~ re1 |
(MUnionL) |

s_{1} =~ Union re1 re2 |

s_{2} =~ re2 |
(MUnionR) |

s_{2} =~ Union re1 re2 |

(MStar0) | |

[] =~ Star re |

s_{1} =~ re s_{2} =~ Star re |
(MStarApp) |

s_{1} ++ s_{2} =~ Star re |

*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.")

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] _ [2]).

- apply MChar.

- apply MChar.

Qed.

(Notice how the last example applies MApp to the strings [1]
and [2] 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 to help write 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 s [] re).

- 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 (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 s

_{1}, ..., 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 (reg_exp_of_list)

Prove that reg_exp_of_list satisfies the following specification:Lemma reg_exp_of_list_spec : ∀T (s

_{1}s

_{2}: list T),

s

_{1}=~ reg_exp_of_list s

_{2}↔ s

_{1}= s

_{2}.

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 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 re2 ⇒ re_chars re1 ++ re_chars re2

| Union re1 re2 ⇒ re_chars re1 ++ re_chars re2

| Star re ⇒ re_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'

|s

_{1}re1 s

_{2}re2 Hmatch1 IH1 Hmatch2 IH2

|s

_{1}re1 re2 Hmatch IH|re1 s

_{2}re2 Hmatch IH

|re|s

_{1}s

_{2}re Hmatch1 IH1 Hmatch2 IH2].

(* WORKED IN CLASS *)

- (* MEmpty *)

apply Hin.

- (* MChar *)

apply Hin.

- simpl. rewrite in_app_iff in ×.

destruct Hin as [Hin | Hin].

+ (* In x s

_{1}*)

left. apply (IH1 Hin).

+ (* In x s

_{2}*)

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.

Something interesting happens in the MStarApp case. We obtain

*two*induction hypotheses: One that applies when x occurs in s_{1}(which matches re), and a second one that applies when x occurs in s_{2}(which matches Star re). This is a good illustration of why we need induction on evidence for exp_match, as opposed to 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 s_{2}.- (* MStarApp *)

simpl. rewrite in_app_iff in Hin.

destruct Hin as [Hin | Hin].

+ (* In x s

_{1}*)

apply (IH1 Hin).

+ (* In x s

_{2}*)

apply (IH2 Hin).

Qed.

#### Exercise: 4 stars (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} (re : reg_exp T) : bool :=

(* FILL IN HERE *) admit.

Lemma re_not_empty_correct : ∀T (re : reg_exp T),

(∃s, s =~ re) ↔ re_not_empty re = true.

Proof.

(* FILL IN HERE *) Admitted.

☐
One potentially confusing feature of the induction tactic is
that it happily lets you try to set up an induction over a term
that isn't sufficiently general. The net effect of this will be
to lose information (much as destruct can do), and leave you
unable to complete the proof. Here's an example:

## The remember Tactic

Lemma star_app: ∀T (s

_{1}s

_{2}: list T) (re : reg_exp T),

s

_{1}=~ Star re →

s

_{2}=~ Star re →

s

_{1}++ s

_{2}=~ Star re.

Proof.

intros T s

_{1}s

_{2}re H1.

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

induction H1

as [|x'|s

_{1}re1 s

_{2}' re2 Hmatch1 IH1 Hmatch2 IH2

|s

_{1}re1 re2 Hmatch IH|re1 s

_{2}' re2 Hmatch IH

|re''|s

_{1}s

_{2}' re'' Hmatch1 IH1 Hmatch2 IH2].

But now, although we get seven cases (as we would expect from the
definition of exp_match), we lost a very important bit of
information from H1: the fact that s

_{1}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 H. apply H.

... but most of them get stuck. For MChar, for instance, we
must show that

s

which is clearly impossible.
_{2}=~ Char x' → x' :: s_{2}=~ Char x',- (* MChar. 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 it
behaves more like destruct than like inversion.
We can solve this problem by generalizing over the problematic
expressions with an explicit equality:

Lemma star_app: ∀T (s

_{1}s

_{2}: list T) (re re' : reg_exp T),

s

_{1}=~ re' →

re' = Star re →

s

_{2}=~ Star re →

s

_{1}++ s

_{2}=~ 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. Calling
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:

Abort.

Lemma star_app: ∀T (s

_{1}s

_{2}: list T) (re : reg_exp T),

s

_{1}=~ Star re →

s

_{2}=~ Star re →

s

_{1}++ s

_{2}=~ Star re.

Proof.

intros T s

_{1}s

_{2}re H1.

remember (Star re) as re'.

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

generalize dependent s

_{2}.

induction H1

as [|x'|s

_{1}re1 s

_{2}' re2 Hmatch1 IH1 Hmatch2 IH2

|s

_{1}re1 re2 Hmatch IH|re1 s

_{2}' re2 Hmatch IH

|re''|s

_{1}s

_{2}' re'' Hmatch1 IH1 Hmatch2 IH2].

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

- (* MEmpty *) inversion Heqre'.

- (* MChar *) inversion Heqre'.

- (* MApp *) inversion Heqre'.

- (* MUnionL *) inversion Heqre'.

- (* MUnionR *) inversion Heqre'.

In the interesting cases (those that correspond to Star), we can
proceed as usual. 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 *)

inversion Heqre'. intros s H. apply H.

- (* MStarApp *)

inversion Heqre'. rewrite H0 in IH2, Hmatch1.

intros s

_{2}H1. rewrite ← app_assoc.

apply MStarApp.

+ apply Hmatch1.

+ apply IH2.

× reflexivity.

× apply H1.

Qed.

#### Exercise: 4 stars (exp_match_ex2)

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.

☐

Module Pumping.

One of the first interesting theorems in the theory of regular
expressions is the so-called
To begin, 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."

*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.Fixpoint pumping_constant {T} (re : reg_exp T) : nat :=

match re with

| EmptySet ⇒ 0

| EmptyStr ⇒ 1

| Char _ ⇒ 2

| App re1 re2 ⇒

pumping_constant re1 + pumping_constant re2

| Union re1 re2 ⇒

pumping_constant re1 + pumping_constant re2

| Star _ ⇒ 1

end.

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.

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.

Now, the 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 s

_{1}++ s_{2}++ s_{3}in such a way that s_{2}can be repeated any number of times and the result, when combined with s_{1}and s_{3}will still match re. Since s_{2}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 pumping : ∀T (re : reg_exp T) s,

s =~ re →

pumping_constant re ≤ length s →

∃s

_{1}s

_{2}s

_{3},

s = s

_{1}++ s

_{2}++ s

_{3}∧

s

_{2}≠ [] ∧

∀m, s

_{1}++ napp m s

_{2}++ s

_{3}=~ re.

To streamline the proof (which you are to fill in), the omega
tactic, which is enabled by the following Require, is helpful in
several places for automatically completing tedious low-level
arguments involving equalities or inequalities over natural
numbers. We'll return to omega in a later chapter, but feel
free to experiment with it now if you like. The first case of the
induction gives an example of how it is used.

Require Import Coq.omega.Omega.

Proof.

intros T re s Hmatch.

induction Hmatch

as [ | x | s

_{1}re1 s

_{2}re2 Hmatch1 IH1 Hmatch2 IH2

| s

_{1}re1 re2 Hmatch IH | re1 s

_{2}re2 Hmatch IH

| re | s

_{1}s

_{2}re Hmatch1 IH1 Hmatch2 IH2 ].

- (* MEmpty *)

simpl. omega.

(* FILL IN HERE *) Admitted.

End Pumping.

☐

# Improving Reflection

Theorem filter_not_empty_In : ∀n l,

filter (beq_nat n) 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 (beq_nat n m) eqn:H.

+ (* beq_nat n m = true *)

intros _. rewrite beq_nat_true_iff in H. rewrite H.

left. reflexivity.

+ (* beq_nat n m = false *)

intros H'. right. apply IHl'. apply H'.

Qed.

In the first branch after destruct, we explicitly
apply the beq_nat_true_iff lemma to the equation generated by
destructing beq_nat n m, to convert the assumption beq_nat n m
= true into the assumption n = m, which is what we need to
complete this case.
We can streamline this proof by defining an inductive proposition
that yields a better case-analysis principle for beq_nat n
m. Instead of generating an equation such as beq_nat n m =
true, which is not directly useful, this principle gives us right
away the assumption we need: n = m. We'll actually define
something a bit more general, which can be used with arbitrary
properties (and not just equalities):

Inductive reflect (P : Prop) : bool → Prop :=

| ReflectT : P → reflect P true

| ReflectF : ¬ 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
It is easy to formalize this intuition and show that the two
statements are indeed equivalent:

*reflected*in (i.e., equivalent to) the boolean b: P holds if and only if b = true. To see this, notice that, by definition, the only way we can produce evidence that reflect P true holds is by showing that P is true and 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. Conversely, the only way to show reflect P false is by combining evidence for ¬ P with the ReflectF constructor.Theorem iff_reflect : ∀P b, (P ↔ b = true) → reflect P b.

Proof.

intros P [] H.

- apply ReflectT. rewrite H. reflexivity.

- apply ReflectF. rewrite H. intros H'. inversion H'.

Qed.

Theorem reflect_iff : ∀P b, reflect P b → (P ↔ b = true).

Proof.

(* FILL IN HERE *) Admitted.

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).
To use reflect to produce a better proof of
filter_not_empty_In, we begin by recasting the
beq_nat_iff_true lemma into a more convenient form in terms of
reflect:

Lemma beq_natP : ∀n m, reflect (n = m) (beq_nat n m).

Proof.

intros n m.

apply iff_reflect. rewrite beq_nat_true_iff. reflexivity.

Qed.

The new proof of filter_not_empty_In now goes as follows.
Notice how the calls to destruct and apply are combined into a
single call to destruct. (To see this clearly, look at the two
proofs of filter_not_empty_In in your Coq browser 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 (beq_nat n) 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 (beq_natP n m) as [H | H].

+ (* n = m *)

intros _. rewrite H. left. reflexivity.

+ (* n <> m *)

intros H'. right. apply IHl'. apply H'.

Qed.

Although this technique arguably gives us only a small gain
in convenience for this particular proof, using reflect
consistently often leads to shorter and clearer proofs. We'll see
many more examples where reflect comes in handy in later
chapters.
The use of the reflect property was 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: 4 stars, recommended (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
c : ∀l, l = rev l → pal lmay seem obvious, but will not work very well.)
- Prove pal_app_rev that
∀l, pal (l ++ rev l).
- Prove pal_rev that
∀l, pal l → l = rev l.

(* FILL IN HERE *)

☐

#### Exercise: 5 stars, 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 l → pal l.

(* FILL IN HERE *)

☐
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,
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.)

#### 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:
[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: X→bool, 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.
(* FILL IN HERE *)

☐

#### 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, advanced (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 *)

☐
We say that a list "stutters" if it repeats the same element
consecutively. The property "nostutter mylist" means that
mylist does not stutter. Formulate an inductive definition for
nostutter. (This is different from the NoDup property in the
exercise above; the sequence 1;4;1 repeats but does not
stutter.)

#### Exercise: 3 stars, recommended (nostutter)

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.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 beq_nat_false_iff; auto.

Qed.

*)

Example test_nostutter_2: nostutter (@nil nat).

(* FILL IN HERE *) Admitted.

(*

Proof. repeat constructor; apply beq_nat_false_iff; auto.

Qed.

*)

Example test_nostutter_3: nostutter [5].

(* FILL IN HERE *) Admitted.

(*

Proof. repeat constructor; apply beq_nat_false; 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 H1; auto. Qed.

*)

☐
First prove an easy useful lemma.

#### Exercise: 4 stars, advanced (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...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 *)

.

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.

☐