# RelProperties of Relations

(* $Date: 2011-03-21 10:44:46 -0400 (Mon, 21 Mar 2011) $ *)

This short chapter develops some basic definitions that will be
needed when we come to working with small-step operational
semantics in Smallstep.v. It can be postponed until just before
Smallstep.v, but it is also a good source of good exercises for
developing facility with Coq's basic reasoning facilities, so it
may be useful to look at it just after Logic.v.

A
A relation

*relation*is just a parameterized proposition. As you know from your undergraduate discrete math course, there are a lot of ways of discussing and describing relations*in general*— ways of classifying relations (are they reflexive, transitive, etc.), theorems that can be proved generically about classes of relations, constructions that build one relation from another, etc. Let us pause here to review a few that will be useful in what follows.*on*a set X is a proposition parameterized by two Xs — i.e., it is a logical assertion involving two values from the set X.
Somewhat confusingly, the Coq standard library hijacks the generic
term "relation" for this specific instance. To maintain
consistency with the library, we will do the same. So, henceforth
the Coq identifier relation will always refer to a binary
relation between some set and itself, while the English word
"relation" can refer either to the specific Coq concept or the
more general concept of a relation between any number of possibly
different sets. The context of the discussion should always make
clear which is meant.
A relation R on a set X is a

# Basic Properties of Relations

*partial function*if, for every x, there is at most one y such that R x y — i.e., if R x y1 and R x y2 together imply y1 = y2.
For example, the next_nat relation defined in Logic.v is a
partial function.

Theorem next_nat_partial_function :

partial_function next_nat.

Proof.

unfold partial_function.

intros x y1 y2 P Q.

inversion P. inversion Q.

reflexivity. Qed.

However, the <= relation on numbers is not a partial function.
This can be shown by contradiction. In short: 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 H.

assert (0 = 1) as Nonsense.

Case "Proof of assertion".

apply H with 0.

apply le_n.

apply le_S. apply le_n.

inversion Nonsense. Qed.

#### Exercise: 2 stars, optional

Show that the total_relation defined in Logic.v is not a partial function.(* FILL IN HERE *)

☐

#### Exercise: 2 stars, optional

Show that the empty_relation defined in Logic.v is a partial function.(* FILL IN HERE *)

☐
A

*reflexive*relation on a set X is one that holds for every element of X.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.

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.

Case "le_n". apply Hnm.

Case "le_S". apply le_S. apply IHHmo. Qed.

Theorem lt_trans:

transitive lt.

Proof.

unfold lt. unfold transitive.

intros n m o Hnm Hmo.

apply le_S in Hnm.

apply le_trans with (a := (S n)) (b := (S m)) (c := o).

apply Hnm.

apply Hmo. Qed.

#### Exercise: 2 stars, optional

We can also prove lt_trans more laboriously by induction, without using le_trans. Do this.Theorem lt_trans' :

transitive lt.

Proof.

(* Prove this by induction on evidence that m is less than o. *)

unfold lt. unfold transitive.

intros n m o Hnm Hmo.

induction Hmo as [| m' Hm'o].

(* FILL IN HERE *) Admitted.

Theorem lt_trans'' :

transitive lt.

Proof.

unfold lt. unfold transitive.

intros n m o Hnm Hmo.

induction o as [| o'].

(* FILL IN HERE *) Admitted.

☐
The transitivity of le, in turn, can be used to prove some facts
that will be useful later (e.g., for the proof of antisymmetry
below)...

Theorem le_Sn_le : ∀ n m, S n <= m → n <= m.

Proof.

intros n m H. apply le_trans with (S n).

apply le_S. apply le_n.

apply H. Qed.

☐
Theorem: For every n, ~(S n <= n)
A formal proof of this is an optional exercise below, but try
the informal proof without doing the formal proof first.
Proof:
(* FILL IN HERE *)

☐

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

Provide an informal proof of the following theorem:☐

#### Exercise: 1 star, optional

☐
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, here are a few more common ones.
A relation R is

*symmetric*if R a b implies 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.
☐
A relation is an

*equivalence*if it's reflexive, symmetric, and transitive.
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.
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).

Theorem le_order :

order le.

Proof.

unfold order. split.

Case "refl". apply le_reflexive.

split.

Case "antisym". apply le_antisymmetric.

Case "transitive.". apply le_trans. Qed.

# Reflexive, Transitive Closure

*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, R x y → clos_refl_trans R x y

| rt_refl : ∀ x, clos_refl_trans R x x

| rt_trans : ∀ x y z,

clos_refl_trans R x y → 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,

(n <= m) ↔ ((clos_refl_trans next_nat) n m).

Proof.

intros n m. split.

Case "→".

intro H. induction H.

apply rt_refl.

apply rt_trans with m. apply IHle. apply rt_step. apply nn.

Case "←".

intro H. induction H.

SCase "rt_step". inversion H. apply le_S. apply le_n.

SCase "rt_refl". apply le_n.

SCase "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 — the "nondeterminism" of the rt_trans rule can sometimes
lead to tricky inductions.
Here is a more useful definition...

Inductive refl_step_closure {X:Type} (R: relation X)

: X → X → Prop :=

| rsc_refl : ∀ (x : X),

refl_step_closure R x x

| rsc_step : ∀ (x y z : X),

R x y →

refl_step_closure R y z →

refl_step_closure R x z.

(The following Tactic Notation definitions are explained in
Imp.v. You can ignore them if you haven't read that chapter
yet.)

Tactic Notation "rt_cases" tactic(first) ident(c) :=

first;

[ Case_aux c "rt_step" | Case_aux c "rt_refl"

| Case_aux c "rt_trans" ].

Tactic Notation "rsc_cases" tactic(first) ident(c) :=

first;

[ Case_aux c "rsc_refl" | Case_aux c "rsc_step" ].

Our new definition of reflexive, transitive closure "bundles"
the rtc_R and rtc_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 rsc mimics the behavior
of the two "missing" rtc constructors.

Theorem rsc_R : ∀ (X:Type) (R:relation X) (x y : X),

R x y → refl_step_closure R x y.

Proof.

intros X R x y r.

apply rsc_step with y. apply r. apply rsc_refl. Qed.

Theorem rsc_trans :

∀ (X:Type) (R: relation X) (x y z : X),

refl_step_closure R x y →

refl_step_closure R y z →

refl_step_closure R x z.

Proof.

(* FILL IN HERE *) Admitted.

∀ (X:Type) (R: relation X) (x y z : X),

refl_step_closure R x y →

refl_step_closure R y z →

refl_step_closure 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, optional (rtc_rsc_coincide)

Theorem rtc_rsc_coincide :

∀ (X:Type) (R: relation X) (x y : X),

clos_refl_trans R x y ↔ refl_step_closure R x y.

Proof.

(* FILL IN HERE *) Admitted.

∀ (X:Type) (R: relation X) (x y : X),

clos_refl_trans R x y ↔ refl_step_closure R x y.

Proof.

(* FILL IN HERE *) Admitted.

☐