RelProperties of Relations
This short, 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), 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 Logic chapter.
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.
Somewhat confusingly, the Coq standard library hijacks the generic
term "relation" for this specific instance of the idea. 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.
An example relation on nat is le, the less-than-or-equal-to
relation, which we usually write like this 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 : nat → nat → Prop.
Check le : relation nat.
Basic Properties of Relations
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.
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 Hc.
assert (0 = 1) as Nonsense.
{ (* Proof of assertion *)
apply Hc with (x := 0).
- apply le_n.
- apply le_S. apply le_n. }
inversion Nonsense. Qed.
Exercise: 2 stars, optional
Show that the total_relation defined in earlier is not a partial function.(* FILL IN HERE *)
(* FILL IN HERE *)
☐
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.
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.
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.
- (* refl *) apply le_reflexive.
- split.
+ (* antisym *) apply le_antisymmetric.
+ (* transitive. *) apply le_trans. Qed.
Reflexive, Transitive Closure
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.
- (* -> *)
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 — 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)
: A → Prop :=
| rt1n_refl : clos_refl_trans_1n R x x
| rt1n_trans (y z : A) :
R x y → 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.
Theorem rsc_R : ∀(X:Type) (R:relation X) (x y : X),
R x y → clos_refl_trans_1n R x y.
Proof.
intros X R x y H.
apply rt1n_trans with y. apply H. apply rt1n_refl. Qed.
Theorem 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.
∀(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, optional (rtc_rsc_coincide)
Theorem rtc_rsc_coincide :
∀(X:Type) (R: relation X) (x y : X),
clos_refl_trans R x y ↔ clos_refl_trans_1n R x y.
Proof.
(* FILL IN HERE *) Admitted.
∀(X:Type) (R: relation X) (x y : X),
clos_refl_trans R x y ↔ clos_refl_trans_1n R x y.
Proof.
(* FILL IN HERE *) Admitted.
☐