# IndPropInductively Defined Propositions

# 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

ev 0

------ (ev_SS)

ev 2

------ (ev_SS)

ev 4

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...

... or we can use function application syntax:

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.

☐

# 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).

- 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

*directly*.

- If the evidence is of the form ev_0, we know that n = 0.
- Otherwise, the evidence must have the form ev_SS n' E', where n = S (S n') and E' is evidence for ev n'.

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.

In words, here is how the inversion reasoning works in this proof:
This particular proof also works if we replace inversion by
destruct:

- If the 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, the 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.

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:

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

We could try to proceed by case analysis or induction on n. But
since ev is mentioned in a premise, this strategy 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 *)

Admitted.

## Induction on Evidence

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 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 (ev_sum)

☐

#### Exercise: 4 stars, advanced, optional (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. (You may want to look at the previous theorem when you get
to the induction step.)

☐

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

Finding the appropriate thing to do induction on is a bit tricky here:
☐

#### 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.
☐

# 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_le

_{1}:

3 ≤ 3.

Proof.

(* WORKED IN CLASS *)

apply le_n. Qed.

Theorem test_le

_{2}:

3 ≤ 6.

Proof.

(* WORKED IN CLASS *)

apply le_S. apply le_S. apply le_S. apply le_n. Qed.

Theorem test_le

_{3}:

(2 ≤ 1) → 2 + 2 = 5.

Proof.

(* WORKED IN CLASS *)

intros H. inversion H. inversion H

_{2}. Qed.

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

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, optional (total_relation)

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

☐

#### Exercise: 2 stars, optional (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.

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.

☐

#### Exercise: 3 stars, recommendedM (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 :=

| c

_{1}: R 0 0 0

| c

_{2}: ∀m n o, R m n o → R (S m) n (S o)

| c

_{3}: ∀m n o, R m n o → R m (S n) (S o)

| c

_{4}: ∀m n o, R (S m) (S n) (S (S o)) → R m n o

| c

_{5}: ∀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 c
_{5}from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer. - If we dropped constructor c
_{4}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

(* 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.

☐

#### 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 l
_{1}, l_{2}, and l_{3}, if l_{1}is a subsequence of l_{2}, then l_{1}is also a subsequence of l_{2}++ l_{3}. - (Optional, harder) Prove subseq_trans that subsequence is
transitive — that is, if l
_{1}is a subsequence of l_{2}and l_{2}is a subsequence of l_{3}, then l_{1}is a subsequence of l_{3}. Hint: choose your induction carefully!

(* FILL IN HERE *)

☐
☐

#### Exercise: 2 stars, optionalM (R_provability2)

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

| c

| c

| c

Which of the following propositions are provable?
| c

_{1}: R 0 []| c

_{2}: ∀n l, R n l → R (S n) (n :: l)| c

_{3}: ∀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 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
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.*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 re
_{1}matches s_{1}, and re_{2}matches s_{2}, then App re_{1}re_{2}matches s_{1}++ s_{2}. - If at least one of re
_{1}and re_{2}matches s, then Union re_{1}re_{2}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.

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

| MEmpty : exp_match [] EmptyStr

| MChar : ∀x, exp_match [x] (Char x)

| MApp : ∀s

_{1}re

_{1}s

_{2}re

_{2},

exp_match s

_{1}re

_{1}→

exp_match s

_{2}re

_{2}→

exp_match (s

_{1}++ s

_{2}) (App re

_{1}re

_{2})

| MUnionL : ∀s

_{1}re

_{1}re

_{2},

exp_match s

_{1}re

_{1}→

exp_match s

_{1}(Union re

_{1}re

_{2})

| MUnionR : ∀re

_{1}s

_{2}re

_{2},

exp_match s

_{2}re

_{2}→

exp_match s

_{2}(Union re

_{1}re

_{2})

| 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).

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.

(MEmpty) | |

[] =~ EmptyStr |

(MChar) | |

[x] =~ Char x |

s_{1} =~ re_{1} s_{2} =~ re_{2} |
(MApp) |

s_{1} ++ s_{2} =~ App re_{1} re_{2} |

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

s_{1} =~ Union re_{1} re_{2} |

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

s_{2} =~ Union re_{1} re_{2} |

(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.")

_{1}and exp_match_ex

_{2}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.)

Example reg_exp_ex

_{1}: [1] =~ Char 1.

Proof.

apply MChar.

Qed.

Example reg_exp_ex

_{2}: [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:
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_ex

_{4}: [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_ex

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

#### Exercise: 3 stars (exp_match_ex_{1})

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) (re

_{1}re

_{2}: reg_exp T),

s =~ re

_{1}∨ s =~ re

_{2}→

s =~ Union re

_{1}re

_{2}.

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 re

_{1}re

_{2}⇒ re_chars re

_{1}++ re_chars re

_{2}

| Union re

_{1}re

_{2}⇒ re_chars re

_{1}++ re_chars re

_{2}

| 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}re

_{1}s

_{2}re

_{2}Hmatch1 IH

_{1}Hmatch2 IH

_{2}

|s

_{1}re

_{1}re

_{2}Hmatch IH|re

_{1}s

_{2}re

_{2}Hmatch IH

|re|s

_{1}s

_{2}re Hmatch1 IH

_{1}Hmatch2 IH

_{2}].

(* 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 (IH

_{1}Hin).

+ (* In x s

_{2}*)

right. apply (IH

_{2}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 (IH

_{1}Hin).

+ (* In x s

_{2}*)

apply (IH

_{2}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 : 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.

☐
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 effect of this is 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 H

_{1}.

Just doing an inversion on H

_{1}won't get us very far in the recursive cases. (Try it!). So we need induction. Here is a naive first attempt:induction H

_{1}

as [|x'|s

_{1}re

_{1}s

_{2}' re

_{2}Hmatch1 IH

_{1}Hmatch2 IH

_{2}

|s

_{1}re

_{1}re

_{2}Hmatch IH|re

_{1}s

_{2}' re

_{2}Hmatch IH

|re''|s

_{1}s

_{2}' re'' Hmatch1 IH

_{1}Hmatch2 IH

_{2}].

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 H

_{1}: 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 cases 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, induction on evidence 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.
Invoking 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:

Abort.

Lemma star_app: ∀T (s

s

s

s

Proof.

intros T s

remember (Star re) as re'.

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 H_{1}.remember (Star re) as re'.

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

generalize dependent s

_{2}.

induction H

_{1}

as [|x'|s

_{1}re

_{1}s

_{2}' re

_{2}Hmatch1 IH

_{1}Hmatch2 IH

_{2}

|s

_{1}re

_{1}re

_{2}Hmatch IH|re

_{1}s

_{2}' re

_{2}Hmatch IH

|re''|s

_{1}s

_{2}' re'' Hmatch1 IH

_{1}Hmatch2 IH

_{2}].

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'.

The interesting cases are those that correspond to Star. Note
that the induction hypothesis IH

_{2}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 H

_{0}in IH

_{2}, Hmatch1.

intros s

_{2}H

_{1}. rewrite ← app_assoc.

apply MStarApp.

+ apply Hmatch1.

+ apply IH

_{2}.

* reflexivity.

* apply H

_{1}.

Qed.

#### Exercise: 4 stars (exp_match_ex_{2})

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.

☐
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."

#### Exercise: 5 stars, advanced (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.Module Pumping.

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

match re with

| EmptySet ⇒ 0

| EmptyStr ⇒ 1

| Char _ ⇒ 2

| App re

_{1}re

_{2}⇒

pumping_constant re

_{1}+ pumping_constant re

_{2}

| Union re

_{1}re

_{2}⇒

pumping_constant re

_{1}+ pumping_constant re

_{2}

| 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}re

_{1}s

_{2}re

_{2}Hmatch1 IH

_{1}Hmatch2 IH

_{2}

| s

_{1}re

_{1}re

_{2}Hmatch IH | re

_{1}s

_{2}re

_{2}Hmatch IH

| re | s

_{1}s

_{2}re Hmatch1 IH

_{1}Hmatch2 IH

_{2}].

- (* MEmpty *)

simpl. omega.

(* FILL IN HERE *) Admitted.

End Pumping.

☐

# Case Study: 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; 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 beq_nat n m.
Instead of generating an equation such as beq_nat n m = true,
which is generally not directly useful, this principle gives us
right away the assumption we really need: n = m.
We'll actually define something a bit more general, which can be
used with arbitrary properties (and not just equalities):

Module FirstTry.

Inductive reflect : Prop → bool → Prop :=

| ReflectT : ∀(P:Prop), P → reflect P true

| ReflectF : ∀(P:Prop), ¬ P → reflect P false.

Before explaining this, let's rearrange it a little: Since the
types of both ReflectT and ReflectF begin with
∀ (P:Prop), we can make the definition a bit more readable
and easier to work with by making P a parameter of the whole
Inductive declaration.

End FirstTry.

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.

(* WORKED IN CLASS *)

intros P b H. destruct b.

- apply ReflectT. rewrite H. reflexivity.

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

Qed.

☐
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 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 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 (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.

Fixpoint count n l :=

match l with

| [] ⇒ 0

| m :: l' ⇒ (if beq_nat n m then 1 else 0) + count n l'

end.

Theorem beq_natP_practice : ∀n l,

count n l = 0 → ~(In n l).

Proof.

(* FILL IN HERE *) Admitted.

☐
This technique gives us only a small gain in convenience for
the proofs we've seen here, but using reflect consistently often
leads to noticeably shorter and clearer scripts as proofs get
larger. We'll see many more examples 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: 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.
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 H

_{1}; auto. Qed.

*)

☐
A list l is an "in-order merge" of l
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}and l_{2}if it contains all the same elements as l_{1}and l_{2}, in the same order as l_{1}and l_{2}, 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: X→bool, and a
list l of type list X. Suppose further that l is an
in-order merge of two lists, l_{1}and l_{2}, such that every item in l_{1}satisfies test and no item in l_{2}satisfies test. Then filter test l = l_{1}.(* 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, 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 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 *)

☐

#### 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
l

_{1}l_{2}, which should be provable exactly when l_{1}and l_{2}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 *)

☐
First prove an easy useful lemma.

#### 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...Lemma in_split : ∀(X:Type) (x:X) (l:list X),

In x l →

∃l

_{1}l

_{2}, l = l

_{1}++ x :: l

_{2}.

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).

Now, here's a way to formalize the pigeonhole principle. Suppose
list l
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

_{2}represents a list of pigeonhole labels, and list l_{1}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 l_{1}must contain repeats.*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) (l

_{1}l

_{2}:list X),

excluded_middle →

(∀x, In x l

_{1}→ In x l

_{2}) →

length l

_{2}< length l

_{1}→

repeats l

_{1}.

Proof.

intros X l

_{1}. induction l

_{1}as [|x l

_{1}' IHl1'].

(* FILL IN HERE *) Admitted.

☐