MapsTotal and Partial Maps
From Coq Require Import Bool.Bool.
From Coq Require Import Logic.FunctionalExtensionality.
From VFA Require Import Perm.
From Coq Require Import Logic.FunctionalExtensionality.
From VFA Require Import Perm.
Total Maps
The function t_empty yields an empty total map, given a
default element; this map always returns the default element when
applied to any id.
More interesting is the update function, which (as before)
takes a map m, a key x, and a value v and returns a new map
that takes x to v and takes every other key to whatever m
does.
Definition t_update {A : Type}
(m : total_map A) (x : nat) (v : A) : total_map A :=
fun x' ⇒ if x =? x' then v else m x'.
(m : total_map A) (x : nat) (v : A) : total_map A :=
fun x' ⇒ if x =? x' then v else m x'.
This definition is a nice example of higher-order programming.
The t_update function takes a function m and yields a new
function fun x' ⇒ ... that behaves like the desired map.
For example, we can build a map taking ids to bools, where Id 3
is mapped to true and every other key is mapped to false, like
this:
This completes the definition of total maps. Note that we don't
need to define a find operation because it is just function
application!
Example update_example1 : examplemap 0 = false.
Example update_example2 : examplemap 1 = false.
Example update_example3 : examplemap 2 = false.
Example update_example4 : examplemap 3 = true.
Proof. reflexivity. Qed.
Example update_example2 : examplemap 1 = false.
Proof. reflexivity. Qed.
Example update_example3 : examplemap 2 = false.
Proof. reflexivity. Qed.
Example update_example4 : examplemap 3 = true.
Proof. reflexivity. Qed.
To use maps in later chapters, we'll need several
fundamental facts about how they behave. Even if you don't work
the following exercises, make sure you thoroughly understand the
statements of the lemmas! (Some of the proofs require the
functional_extensionality axiom, which is discussed in the
Logic chapter.)
Exercise: 1 star, standard (t_apply_empty)
The empty map returns its default element for all keys:Exercise: 1 star, standard (t_update_eq)
If we update a map m at a key x with a new value v and then look up x in the map resulting from the update, we get back v:
Lemma t_update_eq : ∀ A (m: total_map A) x v,
(t_update m x v) x = v.
Proof.
(* FILL IN HERE *) Admitted.
☐
(t_update m x v) x = v.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 1 star, standard (t_update_neq)
If we update a map m at a key x1 and then look up a different key x2 in the resulting map, we get the same result that m would have given:
Theorem t_update_neq : ∀ (X:Type) v x1 x2
(m : total_map X),
x1 ≠ x2 →
(t_update m x1 v) x2 = m x2.
Proof.
(* FILL IN HERE *) Admitted.
☐
(m : total_map X),
x1 ≠ x2 →
(t_update m x1 v) x2 = m x2.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 1 star, standard (t_update_shadow)
If we update a map m at a key x with a value v1 and then update again with the same key x and another value v2, the resulting map behaves the same (gives the same result when applied to any key) as the simpler map obtained by performing just the second update on m:
Lemma t_update_shadow : ∀ A (m: total_map A) v1 v2 x,
t_update (t_update m x v1) x v2
= t_update m x v2.
Proof.
(* FILL IN HERE *) Admitted.
☐
t_update (t_update m x v1) x v2
= t_update m x v2.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 1 star, standard (t_update_same)
If we update a map to assign key x the same value as it already has in m, then the result is equal to m:
Theorem t_update_same : ∀ X x (m : total_map X),
t_update m x (m x) = m.
Proof.
(* FILL IN HERE *) Admitted.
☐
t_update m x (m x) = m.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 1 star, standard (t_update_permute)
If we update a map m at two distinct keys, it doesn't matter in which order we do the updates.
Theorem t_update_permute : ∀ (X:Type) v1 v2 x1 x2
(m : total_map X),
x2 ≠ x1 →
(t_update (t_update m x2 v2) x1 v1)
= (t_update (t_update m x1 v1) x2 v2).
Proof.
(* FILL IN HERE *) Admitted.
☐
(m : total_map X),
x2 ≠ x1 →
(t_update (t_update m x2 v2) x1 v1)
= (t_update (t_update m x1 v1) x2 v2).
Proof.
(* FILL IN HERE *) Admitted.
☐
Partial Maps
Definition partial_map (A : Type) := total_map (option A).
Definition empty {A : Type} : partial_map A :=
t_empty None.
Definition update {A : Type}
(m : partial_map A) (x : nat) (v : A) : partial_map A :=
t_update m x (Some v).
Definition empty {A : Type} : partial_map A :=
t_empty None.
Definition update {A : Type}
(m : partial_map A) (x : nat) (v : A) : partial_map A :=
t_update m x (Some v).
We can easily lift all of the basic lemmas about total maps to
partial maps.
Lemma apply_empty : ∀ A x, @empty A x = None.
Lemma update_eq : ∀ A (m: partial_map A) x v,
(update m x v) x = Some v.
Theorem update_neq : ∀ (X:Type) v x1 x2
(m : partial_map X),
x2 ≠ x1 →
(update m x2 v) x1 = m x1.
Lemma update_shadow : ∀ A (m: partial_map A) v1 v2 x,
update (update m x v1) x v2 = update m x v2.
Theorem update_same : ∀ X v x (m : partial_map X),
m x = Some v →
update m x v = m.
Theorem update_permute : ∀ (X:Type) v1 v2 x1 x2
(m : partial_map X),
x2 ≠ x1 →
(update (update m x2 v2) x1 v1)
= (update (update m x1 v1) x2 v2).
(* 2023-10-01 07:22 *)
Lemma update_eq : ∀ A (m: partial_map A) x v,
(update m x v) x = Some v.
Theorem update_neq : ∀ (X:Type) v x1 x2
(m : partial_map X),
x2 ≠ x1 →
(update m x2 v) x1 = m x1.
Lemma update_shadow : ∀ A (m: partial_map A) v1 v2 x,
update (update m x v1) x v2 = update m x v2.
Theorem update_same : ∀ X v x (m : partial_map X),
m x = Some v →
update m x v = m.
Theorem update_permute : ∀ (X:Type) v1 v2 x1 x2
(m : partial_map X),
x2 ≠ x1 →
(update (update m x2 v2) x1 v1)
= (update (update m x1 v1) x2 v2).
(* 2023-10-01 07:22 *)