# IndPropInductively Defined Propositions

Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From LF Require Export Logic.
From Coq Require Import Lia.

# Inductively Defined Propositions

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

## The Collatz Conjecture

The Collatz Conjecture is a famous open problem in number theory.
Its statement is surprisingly simple. First, we define a function f on numbers, as follows:
Fixpoint div2 (n : nat) :=
match n with
0 ⇒ 0
| 1 ⇒ 0
| S (S n) ⇒ S (div2 n)
end.

Definition f (n : nat) :=
if even n then div2 n
else (3 × n) + 1.
Next, we look at what happens when we repeatedly apply f to some given starting number. For example, f 12 is 6, and f 6 is 3, so by repeatedly applying f we get the sequence 12, 6, 3, 10, 5, 16, 8, 4, 2, 1.
Similarly, if we start with 19, we get the longer sequence 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
Both of these sequences eventually reach 1. The question posed by Collatz was: Does the sequence starting from any natural number eventually reach 1?
To formalize this question in Coq, we might try to define a recursive function that computes the total number of steps that it takes for such a sequence to reach 1.
Fail Fixpoint reaches_1_in (n : nat) :=
if n =? 1 then 0
else 1 + reaches_1_in (f n).
This definition is rejected by Coq's termination checker, since the argument to the recursive call, f n, is not "obviously smaller" than n.
Indeed, this isn't just a silly limitation of the termination checker. Functions in Coq are required to be total, and checking that this particular function is total would be equivalent to settling the Collatz conjecture!
Fortunately, there is another way to do it: We can express the concept "reaches 1 eventually" as an inductively defined property of numbers:
Inductive reaches_1 : natProp :=
| term_done : reaches_1 1
| term_more (n : nat) : reaches_1 (f n) → reaches_1 n.
The details of such definitions are written will be explained below; for the moment, the way to read this one is: "The number 1 reaches 1, and any number n reaches 1 if f n does."
The Collatz conjecture then states that the sequence beginning from any number reaches 1:
Conjecture collatz : n, reaches_1 n.
If you succeed in proving this conjecture, you've got a bright future as a number theorist. But don't spend too long on it -- it's been open since 1937!

## Transitive Closure

A binary relation on a set X is a family of propositions parameterized by two elements of X -- i.e., a proposition about pairs of elements of X.
For example, a familiar binary relation on nat is le, the less-than-or-equal-to relation.
Module LePlayground.

Inductive le : natnatProp :=
| le_n (n : nat) : le n n
| le_S (n m : nat) : le n mle n (S m).

End LePlayground.
The transitive closure of a relation R is the smallest relation that contains R and that is transitive.
Inductive clos_trans {X: Type} (R: XXProp) : XXProp :=
| t_step (x y : X) :
R x y
clos_trans R x y
| t_trans (x y z : X) :
clos_trans R x y
clos_trans R y z
clos_trans R x z.

## Permutations

The familiar mathematical concept of permutation also has an elegant formulation as an inductive relation. For simplicity, let's focus on permutations of lists with exactly three elements.
Inductive Perm3 {X : Type} : list Xlist XProp :=
| perm3_swap12 (a b c : X) :
Perm3 [a;b;c] [b;a;c]
| perm3_swap23 (a b c : X) :
Perm3 [a;b;c] [a;c;b]
| perm3_trans (l1 l2 l3 : list X) :
Perm3 l1 l2Perm3 l2 l3Perm3 l1 l3.
This definition says:
• If l2 can be obtained from l1 by swapping the first and second elements, then l2 is a permutation of l1.
• If l2 can be obtained from l1 by swapping the second and third elements, then l2 is a permutation of l1.
• If l2 is a permutation of l1 and l3 is a permutation of l2, then l3 is a permutation of l1.

## Evenness (yet again)

We've already seen two ways of stating a proposition that a number n is even: We can say
(1) even n = true, or
(2) k, n = double k.
A third possibility, which we'll use as a running example for the rest of this chapter, is to say that n is even if we can establish its evenness from the following rules:
• The number 0 is even.
• If n is even, then S (S n) is even.
We can translate the informal definition of evenness from above into a formal Inductive declaration, where each "way that a number can be even" corresponds to a separate constructor:
Inductive ev : natProp :=
| ev_0 : ev 0
| ev_SS (n : nat) (H : ev n) : ev (S (S n)).
We can think of this as defining a Coq property ev : nat Prop, together with "evidence constructors" ev_0 : ev 0 and ev_SS : n, ev n ev (S (S n)).
These evidence constructors can be thought of as "primitive evidence of evenness", and they can be used just like proven theorems. In particular, we can use Coq's apply tactic with the constructor names to obtain evidence for ev of particular numbers...
Theorem ev_4 : ev 4.
Proof. apply ev_SS. apply ev_SS. apply ev_0. Qed.
... or we can use function application syntax to combine several constructors:
Theorem ev_4' : ev 4.
Proof. apply (ev_SS 2 (ev_SS 0 ev_0)). Qed.
In this way, we can also prove theorems that have hypotheses involving ev.
Theorem ev_plus4 : n, ev nev (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 destruct such evidence, reasoning about how it could have been built.
Introducing ev with an Inductive declaration tells Coq not only that the constructors ev_0 and ev_SS are valid ways to build evidence that some number is ev, but also that these two constructors are the only ways to build evidence that numbers are ev.
In other words, if someone gives us evidence E for the assertion ev n, then we know that E must be one of two things:
• E is ev_0 (and n is O), or
• E is ev_SS n' E' (and n is S (S n'), where E' is evidence for ev n').
This suggests that it should be possible to do case analysis and even induction on evidence of evenness...

## Inversion on Evidence

We can prove our characterization of evidence for ev n, using destruct.
Theorem ev_inversion : (n : nat),
ev n
(n = 0) ∨ ( n', n = S (S n') ∧ ev n').
Proof.
intros n E. destruct E as [ | n' E'] eqn:EE.
- (* E = ev_0 : ev 0 *)
left. reflexivity.
- (* E = ev_SS n' E' : ev (S (S n')) *)
right. n'. split. reflexivity. apply E'.
Qed.
Facts like this are often called "inversion lemmas" because they allow us to "invert" some given information to reason about all the different ways it could have been derived.
Here, there are two ways to prove ev n, and the inversion lemma makes this explicit.
Which tactics are needed to prove this goal?
n : nat
E : ev n
F : n = 1
======================
true = false
(1) destruct
(2) discriminate
(3) both destruct and discriminate
(4) These tactics are not sufficient to solve the goal.
Lemma quiz_1_not_ev : n, ev nn = 1 → true = false.
Proof.
intros n E F. destruct E as [| n' E'] eqn:EE.
- discriminate F.
- discriminate F.
Qed.
We can use the inversion lemma that we proved above to help structure proofs:
Theorem evSS_ev : n, ev (S (S n)) → ev n.
Proof.
intros n H. apply ev_inversion in H. destruct H as [H0|H1].
- discriminate H0.
- destruct H1 as [n' [Hnm Hev]]. injection Hnm as Heq.
rewrite Heq. apply Hev.
Qed.
Coq provides a handy tactic called inversion that does the work of our inversion lemma and more besides.
Theorem evSS_ev' : n,
ev (S (S n)) → ev n.
Proof.
intros n E. inversion E as [| n' E' Heq].
(* We are in the E = ev_SS n' E' case now. *)
apply E'.
Qed.
We can use inversion to re-prove some theorems from Tactics.v. (Note that inversion also works on equality propositions.)
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 Inductively:
• For each constructor of P, make a subgoal where H is constrained by the form of this constructor.
• 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 : ev (n + 2)
=====================
ev n
(1) inversion
(2) inversion, discriminate
(5) These tactics are not sufficient to prove the goal.
Lemma quiz_ev_plus_2 : n, ev (n + 2) → ev n.
Proof.
intros n E. rewrite add_comm in E. inversion E. apply H0.
Qed.
Let's try to show that our new notion of evenness implies our earlier notion (the one based on double).
Lemma ev_Even_firsttry : n,
ev nEven n.
Proof.
(* WORK IN CLASS *) Admitted.

## Induction on Evidence

If this story feels 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!
Let's try proving that lemma again:
Lemma ev_Even : n,
ev nEven n.
Proof.
intros n E.
induction E as [|n' E' IH].
- (* E = ev_0 *)
unfold Even. 0. reflexivity.
- (* E = ev_SS n' E'
with IH : Even E' *)

unfold Even in IH.
destruct IH as [k Hk].
rewrite Hk.
unfold Even. (S k). simpl. reflexivity.
Qed.
As we will see in later chapters, induction on evidence is a recurring technique across many areas -- in particular for formalizing the semantics of programming languages.

# Inductive Relations

Just as a single-argument proposition defines a property, a two-argument proposition defines a relation.
Module Playground.
Just like properties, relations can be defined inductively. One useful example is the "less than or equal to" relation on numbers that we briefly saw above.
Inductive le : natnatProp :=
| le_n (n : nat) : le n n
| le_S (n m : nat) (H : le n m) : le n (S m).

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

End Playground.

# A Digression on Notation

There are several equivalent ways of writing inductive types.  We've mostly seen this style...
Module bin1.
Inductive bin : Type :=
| Z
| B0 (n : bin)
| B1 (n : bin).
End bin1.
... which omits the result types because they are all bin.
It is completely equivalent to this...
Module bin2.
Inductive bin : Type :=
| Z : bin
| B0 (n : bin) : bin
| B1 (n : bin) : bin.
End bin2.
... where we fill them in, and this...
Module bin3.
Inductive bin : Type :=
| Z : bin
| B0 : binbin
| B1 : binbin.
End bin3.
... where we put everything on the right of the colon.
For inductively defined propositions, we need to explicitly give the result type for each constructor (because they are not all the same), so the first style doesn't make sense, but we can use either the second or the third interchangeably.

# Case Study: Regular Expressions

Regular expressions are a simple language for describing sets of strings. Their syntax is defined as follows:
Inductive reg_exp (T : Type) : Type :=
| EmptySet
| EmptyStr
| Char (t : T)
| App (r1 r2 : reg_exp T)
| Union (r1 r2 : reg_exp T)
| Star (r : reg_exp T).

Arguments EmptySet {T}.
Arguments EmptyStr {T}.
Arguments Char {T} _.
Arguments App {T} _ _.
Arguments Union {T} _ _.
Arguments Star {T} _.
We connect regular expressions and strings via the following rules, which define when a regular expression matches some string:
• The expression EmptySet does not match any string.
• The expression EmptyStr matches the empty string [].
• The expression Char x matches the one-character string [x].
• If re1 matches s1, and re2 matches s2, then App re1 re2 matches s1 ++ s2.
• If at least one of re1 and re2 matches s, then Union re1 re2 matches s.
• Finally, if we can write some string s as the concatenation of a sequence of strings s = s_1 ++ ... ++ s_k, and the expression re matches each one of the strings s_i, then Star re matches s.
In particular, the sequence of strings may be empty, so Star re always matches the empty string [] no matter what re is.
We can easily translate this informal definition into an Inductive one as follows. We use the notation s =~ re in place of exp_match s re. (By "reserving" the notation before defining the Inductive, we can use it in the definition.)
Reserved Notation "s =~ re" (at level 80).

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

where "s =~ re" := (exp_match s re).

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

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

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

Something more interesting:
Lemma MStar1 :
T s (re : reg_exp T) ,
s =~ re
s =~ Star re.
(* 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!) (We can begin by generalizing s2, since it's pretty clear that we are going to have to walk over both s1 and s2 in parallel.)
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].
We can get through the first case...
- (* MEmpty *)
simpl. intros s2 H. apply H.
... but most cases get stuck. For MChar, for instance, we must show
s2 =~ Char x'
x'::s2 =~ Char x'
which is clearly impossible.
- (* MChar. *) intros s2 H. simpl. (* Stuck... *)
Abort.
The problem here is that induction over a Prop hypothesis only works properly with hypotheses that are "completely general," i.e., ones in which all the arguments are variables, as opposed to more complex expressions like Star re.
A possible, but awkward, way to solve this problem is "manually generalizing" over the problematic expressions by adding explicit equality hypotheses to the lemma:
Lemma star_app: T (s1 s2 : list T) (re re' : reg_exp T),
re' = Star re
s1 =~ re'
s2 =~ Star re
s1 ++ s2 =~ Star re.
This works, but it makes the statement of the lemma a bit ugly. 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 T),
s1 =~ Star re
s2 =~ Star re
s1 ++ s2 =~ Star re.
Proof.
intros T s1 s2 re H1.
remember (Star re) as re'.
We now have Heqre' : re' = Star re.
generalize dependent s2.
induction H1
as [|x'|s1 re1 s2' re2 Hmatch1 IH1 Hmatch2 IH2
|s1 re1 re2 Hmatch IH|re1 s2' re2 Hmatch IH
|re''|s1 s2' re'' Hmatch1 IH1 Hmatch2 IH2].
The Heqre' is contradictory in most cases, allowing us to conclude immediately.
- (* MEmpty *) discriminate.
- (* MChar *) discriminate.
- (* MApp *) discriminate.
- (* MUnionL *) discriminate.
- (* MUnionR *) discriminate.
The interesting cases are those that correspond to Star. Note that the induction hypothesis IH2 on the MStarApp case mentions an additional premise Star re'' = Star re, which results from the equality generated by remember.
- (* MStar0 *)
injection Heqre' as Heqre''. intros s H. apply H.

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

# Case Study: Improving Reflection

We've seen that we often need to relate boolean computations to statements in Prop. However, this can result in some tedium in 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 awkward because we have to explicitly "switch worlds."
It would be annoying to have to do this kind of thing all the time.
We can streamline this sort of reasoning by defining an inductive proposition that yields a better case-analysis principle for n =? m. Instead of generating the assumption (n =? m) = true, which usually requires some massaging before we can use it, this principle gives us right away the assumption we really need: n = m.
Following the terminology introduced in Logic, we call this the "reflection principle for equality on numbers," and we say that the boolean n =? m is reflected in the proposition n = m.
Inductive reflect (P : Prop) : boolProp :=
| ReflectT (H : P) : reflect P true
| ReflectF (H : ¬P) : reflect P false.
Notice that the only way to produce evidence for reflect P true is by showing P and then using the ReflectT constructor.
If we play this reasoning backwards, it says we can extract evidence for P from evidence for reflect P true.
To put this observation to work, we first prove that the statements P b = true and reflect P b are indeed equivalent. First, the left-to-right implication:
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.)
We can think of reflect as a kind of variant of the usual "if and only if" connective; the advantage of reflect is that, by destructing a hypothesis or lemma of the form reflect P b, we can perform case analysis on b while at the same time generating appropriate hypothesis in the two branches (P in the first subgoal and ¬ P in the second).
Let's use reflect to produce a smoother proof of filter_not_empty_In.
We begin by recasting the eqb_eq lemma in terms of reflect:
Lemma eqbP : n m, reflect (n = m) (n =? m).
Proof.
intros n m. apply iff_reflect. rewrite eqb_eq. reflexivity.
Qed.
The proof of filter_not_empty_In now goes as follows. Notice how the calls to destruct and rewrite in the earlier proof of this theorem are combined here into a single call to destruct.
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 reflection giving 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.
This use of reflect was popularized by SSReflect, a Coq library that has been used to formalize important results in mathematics, including 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 by turning them into boolean computations.