# RelProperties of Relations

This short (and optional) chapter develops some basic definitions and a few theorems about binary relations in Coq. The key definitions are repeated where they are actually used (in the Smallstep chapter of Programming Language Foundations), so readers who are already comfortable with these ideas can safely skim or skip this chapter. However, relations are also a good source of exercises for developing facility with Coq's basic reasoning facilities, so it may be useful to look at this material just after the IndProp chapter.

# Relations

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.
Definition relation (X: Type) := XXProp.
An example relation on nat is le, the less-than-or-equal-to relation, which we usually write n1 n2.
Print le.
(* ====> Inductive le (n : nat) : nat -> Prop :=
le_n : n <= n
| le_S : forall m : nat, n <= m -> n <= S m *)

Check le : natnatProp.
Check le : relation nat.
(Why did we write it this way instead of starting with Inductive le : relation nat...? Because we wanted to put the first nat to the left of the :, which makes Coq generate a somewhat nicer induction principle for reasoning about .)

# Basic Properties

### Partial Functions

A relation R on a set X is a partial function if, for every x, there is at most one y such that R x y — i.e., R x y1 and R x y2 together imply y1 = y2.
Definition partial_function {X: Type} (R: relation X) :=
x y1 y2 : X, R x y1R x y2y1 = y2.
For example, the next_nat relation defined earlier is a partial function.
Print next_nat.
(* ====> Inductive next_nat (n : nat) : nat -> Prop :=
nn : next_nat n (S n) *)

Check next_nat : relation nat.

Theorem next_nat_partial_function :
partial_function next_nat.
Proof.
unfold partial_function.
intros x y1 y2 H1 H2.
inversion H1. inversion H2.
reflexivity. Qed.
However, the relation on numbers is not a partial function. (Assume, for a contradiction, that is a partial function. But then, since 0 0 and 0 1, it follows that 0 = 1. This is nonsense, so our assumption was contradictory.)
Theorem le_not_a_partial_function :
¬(partial_function le).
Proof.
unfold not. unfold partial_function. intros Hc.
assert (0 = 1) as Nonsense. {
apply Hc with (x := 0).
- apply le_n.
- apply le_S. apply le_n. }
discriminate Nonsense. Qed.

### Reflexive Relations

A reflexive relation on a set X is one for which every element of X is related to itself.
Definition reflexive {X: Type} (R: relation X) :=
a : X, R a a.

Theorem le_reflexive :
reflexive le.
Proof.
unfold reflexive. intros n. apply le_n. Qed.

### Transitive Relations

A relation R is transitive if R a c holds whenever R a b and R b c do.
Definition transitive {X: Type} (R: relation X) :=
a b c : X, (R a b) → (R b c) → (R a c).

Theorem le_trans :
transitive le.
Proof.
intros n m o Hnm Hmo.
induction Hmo.
- (* le_n *) apply Hnm.
- (* le_S *) apply le_S. apply IHHmo. Qed.
Reflexivity and transitivity are the main concepts we'll need for later chapters, but, for a bit of additional practice working with relations in Coq, let's look at a few other common ones...

### Symmetric and Antisymmetric Relations

A relation R is symmetric if R a b implies R b a.
Definition symmetric {X: Type} (R: relation X) :=
a b : X, (R a b) → (R b a).
A relation R is antisymmetric if R a b and R b a together imply a = b — that is, if the only "cycles" in R are trivial ones.
Definition antisymmetric {X: Type} (R: relation X) :=
a b : X, (R a b) → (R b a) → a = b.

### Equivalence Relations

A relation is an equivalence if it's reflexive, symmetric, and transitive.
Definition equivalence {X:Type} (R: relation X) :=
(reflexive R) ∧ (symmetric R) ∧ (transitive R).

### Partial Orders and Preorders

A relation is a partial order when it's reflexive, anti-symmetric, and transitive. In the Coq standard library it's called just "order" for short.
Definition order {X:Type} (R: relation X) :=
(reflexive R) ∧ (antisymmetric R) ∧ (transitive R).
A preorder is almost like a partial order, but doesn't have to be antisymmetric.
Definition preorder {X:Type} (R: relation X) :=
(reflexive R) ∧ (transitive R).

# Reflexive, Transitive Closure

The reflexive, transitive closure of a relation R is the smallest relation that contains R and that is both reflexive and transitive. Formally, it is defined like this in the Relations module of the Coq standard library:
Inductive clos_refl_trans {A: Type} (R: relation A) : relation A :=
| rt_step x y (H : R x y) : clos_refl_trans R x y
| rt_refl x : clos_refl_trans R x x
| rt_trans x y z
(Hxy : clos_refl_trans R x y)
(Hyz : clos_refl_trans R y z) :
clos_refl_trans R x z.
For example, the reflexive and transitive closure of the next_nat relation coincides with the le relation.
Theorem next_nat_closure_is_le : n m,
(nm) ↔ ((clos_refl_trans next_nat) n m).
Proof.
intros n m. split.
- (* -> *)
intro H. induction H.
+ (* le_n *) apply rt_refl.
+ (* le_S *)
apply rt_trans with m. apply IHle. apply rt_step.
apply nn.
- (* <- *)
intro H. induction H.
+ (* rt_step *) inversion H. apply le_S. apply le_n.
+ (* rt_refl *) apply le_n.
+ (* rt_trans *)
apply le_trans with y.
apply IHclos_refl_trans1.
apply IHclos_refl_trans2. Qed.
The above definition of reflexive, transitive closure is natural: it says, explicitly, that the reflexive and transitive closure of R is the least relation that includes R and that is closed under rules of reflexivity and transitivity. But it turns out that this definition is not very convenient for doing proofs, since the "nondeterminism" of the rt_trans rule can sometimes lead to tricky inductions. Here is a more useful definition:
Inductive clos_refl_trans_1n {A : Type}
(R : relation A) (x : A)
: AProp :=
| rt1n_refl : clos_refl_trans_1n R x x
| rt1n_trans (y z : A)
(Hxy : R x y) (Hrest : clos_refl_trans_1n R y z) :
clos_refl_trans_1n R x z.
Our new definition of reflexive, transitive closure "bundles" the rt_step and rt_trans rules into the single rule step. The left-hand premise of this step is a single use of R, leading to a much simpler induction principle.
Before we go on, we should check that the two definitions do indeed define the same relation...
First, we prove two lemmas showing that clos_refl_trans_1n mimics the behavior of the two "missing" clos_refl_trans constructors.
Lemma rsc_R : (X:Type) (R:relation X) (x y : X),
R x yclos_refl_trans_1n R x y.
Proof.
intros X R x y H.
apply rt1n_trans with y. apply H. apply rt1n_refl. Qed.

#### Exercise: 2 stars, standard, optional (rsc_trans)

Lemma rsc_trans :
(X:Type) (R: relation X) (x y z : X),
clos_refl_trans_1n R x y
clos_refl_trans_1n R y z
clos_refl_trans_1n R x z.
Proof.
(* FILL IN HERE *) Admitted.
Then we use these facts to prove that the two definitions of reflexive, transitive closure do indeed define the same relation.

#### Exercise: 3 stars, standard, optional (rtc_rsc_coincide)

Theorem rtc_rsc_coincide :
(X:Type) (R: relation X) (x y : X),
clos_refl_trans R x yclos_refl_trans_1n R x y.
Proof.
(* FILL IN HERE *) Admitted.