# 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.
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.
Such definitions are often presented using inference rules, they consist of a line separating the premises above from the conclusion below:
 (ev_0) even 0
 even n (ev_SS) even (S (S n))
We can use a proof tree to display the reasoning steps demonstrating that 4 is even:
--------  (ev_0)
even 0
-------- (ev_SS)
even 2
-------- (ev_SS)
even 4

## 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 even : natProp :=
| ev_0 : even 0
| ev_SS (n : nat) (H : even n) : even (S (S n)).
We can think of the definition of even as defining a Coq property even : nat Prop, together with primitive theorems ev_0 : even 0 and ev_SS : n, even n even (S (S n)).
That definition can also be written as follows...
Inductive even : nat → Prop :=
| ev_0 : even 0
| ev_SS : neven n → even (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 even for particular numbers...
Theorem ev_4 : even 4.
Proof. apply ev_SS. apply ev_SS. apply ev_0. Qed.
... or we can use function application syntax:
Theorem ev_4' : even 4.
Proof. apply (ev_SS 2 (ev_SS 0 ev_0)). Qed.
We can also prove theorems that have hypotheses involving even.
Theorem ev_plus4 : n, even neven (4 + n).
Proof.
intros n. simpl. intros Hn.
apply ev_SS. apply ev_SS. apply Hn.
Qed.

# Using Evidence in Proofs

Besides constructing evidence that numbers are even, we can also reason about such evidence.
Introducing even 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 even, but also that these two constructors are the only ways to build evidence that numbers are even (in the sense of even).
In other words, if someone gives us evidence E for the assertion even 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 even n').
This suggests that it should be possible to do induction and case analysis on evidence of evenness...

## Inversion on Evidence

We can prove our characterization of evidence for even n, using destruct.
Theorem ev_inversion :
(n : nat), even n
(n = 0) ∨ (n', n = S (S n') ∧ even n').
Proof.
intros n E.
destruct E as [ | n' E'].
- (* E = ev_0 : even 0 *)
left. reflexivity.
- (* E = ev_SS n' E' : even (S (S n')) *)
right. n'. split. reflexivity. apply E'.
Qed.
Which tactics are needed to prove this goal?
n : nat
E : even n
F : n = 1
======================
true = false
(1) destruct
(2) discriminate
(3) destruct, discriminate
(4) These tactics are not sufficient to solve the goal.

The following theorem can easily be proved using destruct on evidence.
Theorem ev_minus2 : n,
even neven (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.
However, this variation cannot easily be handled with destruct.
Theorem evSS_ev : n,
even (S (S n)) → even n.
Proof.
intros n E.
destruct E as [| n' E'].
- (* E = ev_0. *)
(* We must prove that n is even from no assumptions! *)
Abort.
The proof is straightforward using our inversion lemma.
Theorem evSS_ev : n, even (S (S n)) → even n.
Proof. intros n H. apply ev_inversion in H. destruct H.
- discriminate H.
- destruct H as [n' [Hnm Hev]]. injection Hnm.
intro Heq. rewrite Heq. apply Hev.
Qed.
Coq provides a tactic called inversion, which does the work of our inversion lemma and more besides.
Theorem evSS_ev' : n,
even (S (S n)) → even n.
Proof.
intros n E.
inversion E as [| n' E'].
(* 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 inductive properties, something that takes a bit more work using our inversion lemma. For example:
Theorem one_not_even : ¬even 1.
Proof.
intros H. apply ev_inversion in H.
destruct H as [ | [m [Hm _]]].
- discriminate H.
- discriminate Hm.
Qed.

Theorem one_not_even' : ¬even 1.
intros H. inversion H. Qed.
We can use inversion to reprove some theorems from Tactics.v.
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.
The tactic inversion actually works on any H : P where P is defined by Inductive:
• For each constructor of P, make a subgoal and replace H by how exactly this constructor could have been used to prove P.
• Generate auxiliary equalities (as with ev_SS above).
Which tactics are needed to prove this goal, in addition to simpl and apply?
n : nat
E : even (n + 2)
=====================
even n
(1) inversion
(2) inversion, discriminate
(3) inversion, rewrite plus_comm
(4) inversion, rewrite plus_comm, discriminate
(5) These tactics are not sufficient to prove the goal.
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,
even nk, n = double k.
Proof.
(* WORK IN CLASS *) Admitted.

## Induction on Evidence

Let's try our current lemma again:
Lemma ev_even : n,
even nk, 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.
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.

# Inductive Relations

Just as a single-argument proposition defines a property, a two-argument proposition defines a relation.
One useful example is the "less than or equal to" relation on numbers.
Inductive le : natnatProp :=
| le_n n : le n n
| le_S n m (H : le n m) : le n (S m).

Notation "m ≤ n" := (le m n).
Some sanity checks...
Theorem test_le1 :
3 ≤ 3.
Proof.
(* WORK IN CLASS *) Admitted.

Theorem test_le2 :
3 ≤ 6.
Proof.
(* WORK IN CLASS *) Admitted.

Theorem test_le3 :
(2 ≤ 1) → 2 + 2 = 5.
Proof.
(* WORK IN CLASS *) Admitted.
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 : natnatProp :=
| sq n : square_of n (n * n).

Inductive next_nat : natnatProp :=
| nn n : next_nat n (S n).
Define next_even : nat nat Prop such that next_even a b holds iff b is the first even natural number that is greater than a.

Inductive next_even : natnatProp :=
| ne_1 n : even (S n) → next_even n (S n)
| ne_2 n (H : even (S (S n))) : next_even n (S (S n)).

# Case Study: 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)
| Union (r1 r2 : reg_exp)
| Star (r : reg_exp).
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.
As a special case, 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:
Inductive exp_match {T} : list Treg_expProp :=
| MEmpty : exp_match [] EmptyStr
| MChar x : exp_match [x] (Char x)
| MApp s1 re1 s2 re2
(H1 : exp_match s1 re1)
(H2 : exp_match s2 re2) :
exp_match (s1 ++ s2) (App re1 re2)
| MUnionL s1 re1 re2
(H1 : exp_match s1 re1) :
exp_match s1 (Union re1 re2)
| MUnionR re1 s2 re2
(H2 : exp_match s2 re2) :
exp_match s2 (Union re1 re2)
| MStar0 re : exp_match [] (Star re)
| MStarApp s1 s2 re
(H1 : exp_match s1 re)
(H2 : exp_match s2 (Star re)) :
exp_match (s1 ++ s2) (Star re).
Notice that this clause in our informal definition...
• The expression EmptySet does not match any string.
... is not explicitly reflected in the above definition. Do we need to add something?
(1) Yes, we should add a rule for this.
(2) No, one of the other rules already covers this case.
(3) No, the lack of a rule actually gives us the behavior we want.

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
 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
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.

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.
Something more interesting:
Lemma MStar1 :
T s (re : @reg_exp T) ,
s =~ re
s =~ Star re.
(* WORK IN CLASS *) Admitted.
Naturally, proofs about exp_match often require induction.
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) : 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.

Theorem in_re_match : T (s : list T) (re : reg_exp) (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].
(* WORK IN CLASS *) 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.
A naive first attempt at setting up the induction. (Note that we are performing induction on evidence!)
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].
We can get through the first case...
- (* MEmpty *)
simpl. intros 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. 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.
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),
re' = Star re
s1 =~ re'
s2 =~ Star re
s1 ++ s2 =~ Star re.
This works, but it requires making the statement of the lemma a bit ugly, which is a shame. Fortunately, there is a better way:
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),
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'. intros Heqre'' s H. apply H.

- (* MStarApp *)
injection Heqre'. intros H0.
intros s2 H1. rewrite <- app_assoc.
apply MStarApp.
+ apply Hmatch1.
+ apply IH2.
* rewrite H0. reflexivity.
* apply H1.
Qed.

# Case Study: Improving Reflection

We've seen that we often need to relate boolean computations to statements in Prop. Unfortunately, this can sometimes result in tedious proof scripts. Consider:
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.
The first subcase (where n =? m = true) is a bit awkward.
It would be annoying to have to do this kind of thing all the time.
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) : boolProp :=
| ReflectT (H : P) : reflect P true
| ReflectF (H : ¬P) : reflect P false.
The only way to produce evidence for reflect P true is by showing P and using the ReflectT constructor.
If we invert this reasoning, it says we can extract evidence for P from evidence for reflect P true.
Theorem iff_reflect : P b, (Pb = true) → reflect P b.
Proof.
(* WORK IN CLASS *) Admitted.
(The right-to-left implication is left as an exercise.)
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 eqb_iff_true lemma in terms of reflect:
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 apply are combined into a single call to 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.
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.