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328 lines
9.1 KiB
Coq
328 lines
9.1 KiB
Coq
(** Formal Reasoning About Programs <http://adam.chlipala.net/frap/>
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* Supplementary Coq material: polymorphism and generic data structures
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* Author: Adam Chlipala
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* License: https://creativecommons.org/licenses/by-nc-nd/4.0/ *)
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Require Import Frap.
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Set Implicit Arguments.
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(* This command sets up automatic inference of tedious arguments. *)
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(* One of the recurring ingredients in effective machine formalization is data
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* structures, and, as in general programming, we want to be able to define data
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* structures that are *generic* in the kinds of data they contain. Coq
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* supports such things in roughly the same way as Haskell and OCaml, via
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* *parametric polymorphism*, which is closely related to the idea of *generics*
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* in languages like Java. *)
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(* Our first example: the [option] type family. While Java and friends force
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* all sorts of different types to include the special value [null], in Coq we
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* request that option explicitly by wrapping a type in [option]. Specifically,
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* any value of type [option A], for some type [A], is either [None] (sort of
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* like [null]) or [Some v] for a [v] of type [A]. *)
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Inductive option (A : Set) : Set :=
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| None
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| Some (v : A).
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Arguments None [A].
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(* This command asks Coq to *infer* the [A] type for each specific use of
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* [None]. *)
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(* Here are a few example terms using [option]. *)
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Example no_number : option nat := None.
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Example a_number : option nat := Some 42.
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Example no_number_squared : option (option nat) := None.
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Example no_number_squared_inside : option (option nat) := Some None.
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Example a_number_squared : option (option nat) := Some (Some 42).
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(* Pattern matching is the key ingredient for working with inductive definitions
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* of all sorts. Here are some examples matching on [option]s. *)
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Definition increment_optional (no : option nat) : option nat :=
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match no with
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| None => None
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| Some n => Some (n + 1)
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end.
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(* Here we use type [A * B] of *pairs*, inhabited by values [(a, b)], with
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* [a : A] and [b : B]. *)
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Definition add_optional (po : option (nat * nat)) : option nat :=
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match po with
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| None => None
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| Some (n, m) => Some (n + m)
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end.
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(** * Lists *)
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(* For functional programming (as in Coq), the king of all generic data
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* structures is the *list*, which you explored a bit in the first problem set.
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* Let's recap that type definition. *)
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Inductive list (A : Set) : Set :=
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| nil
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| cons (hd : A) (tl : list A).
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Arguments nil [A].
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(* [nil] is the empty list, while [cons], standing for "construct," extends a
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* list of length [n] into one of length [n+1]. *)
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(* Here are some simple lists. *)
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Example nats0 : list nat := nil.
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Example nats1 : list nat := cons 1 nil.
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Example nats2 : list nat := cons 1 (cons 2 nil).
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(* Coq features a wonderful notation system, to help us write more concise and
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* readable code after introducing new syntactic forms. We will not give a
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* systematic presentation of the notation system, but we will show many
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* examples, from which it is possible to infer generality by scientific
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* induction. And, of course, the interested reader can always check the
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* notations chapter of the Coq reference manual. *)
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(* First, our examples can get more readable with an infix operator for [cons]. *)
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Infix "::" := cons.
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Example nats1' : list nat := 1 :: nil.
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Example nats2' : list nat := 1 :: 2 :: nil.
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(* Getting even more fancy, we declare a notation for list literals. *)
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Notation "[ ]" := nil.
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Notation "[ x1 ; .. ; xN ]" := (cons x1 (.. (cons xN nil) ..)).
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Example nats0'' : list nat := [].
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Example nats1'' : list nat := [1].
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Example nats2'' : list nat := [1; 2].
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Example nats3'' : list nat := [1; 2; 3].
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(* Here are some classic recursive functions that operate over lists.
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* First, here is how to compute the length of a list. Recall that we put
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* *implicit* function arguments in curly braces, asking Coq to infer them at
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* call sites. *)
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Fixpoint length {A} (ls : list A) : nat :=
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match ls with
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| nil => 0
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| _ :: ls' => 1 + length ls'
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end.
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(* The first problem set involved an exercise with list append and reverse
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* operations. To avoid spoiling the proofs there about those functions, we
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* will give their definitions here without proving the classic theorems from
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* the problem set. *)
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Fixpoint app {A} (ls1 ls2 : list A) : list A :=
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match ls1 with
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| nil => ls2
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| x :: ls1' => x :: app ls1' ls2
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end.
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Infix "++" := app.
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Fixpoint rev {A} (ls : list A) : list A :=
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match ls with
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| nil => nil
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| x :: ls' => rev ls' ++ [x]
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end.
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(* One of the classic gotchas in functional-programming class is how slow this
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* naive [rev] is. Each [app] operation requires linear time, so running
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* linearly many [app]s brings us to quadratic time for [rev]. Using a helper
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* function, we can bring [rev] to its optimal linear time. *)
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Fixpoint rev_append {A} (ls acc : list A) : list A :=
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match ls with
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| nil => acc
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| x :: ls' => rev_append ls' (x :: acc)
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end.
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(* This function [rev_append] takes an extra *accumulator* argument, in which we
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* gradually build up the original input in reversed order. The base case just
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* returns the accumulator. Now reversal just needs to do a [rev_append] with
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* an empty initial accumulator. *)
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Definition rev' {A} (ls : list A) : list A :=
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rev_append ls [].
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(* A few test cases can help convince us that this seems to work. *)
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Compute rev [1; 2; 3; 4].
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Compute rev' [1; 2; 3; 4].
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Compute rev ["hi"; "bye"; "sky"].
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Compute rev' ["hi"; "bye"; "sky"].
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(* OK, great. Now it seems worth investing in a correctness proof. We'll
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* discover it interactively in class, but here's a worked-out final
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* answer, with the several lemmas that we discover are useful. *)
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(* List concatenation is associative. *)
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Lemma app_assoc : forall A (ls1 ls2 ls3 : list A),
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(ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3).
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Proof.
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induct ls1; simplify.
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equality.
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rewrite IHls1.
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equality.
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Qed.
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(* The natural correctness condition for [rev_append]: it does what it says on
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* the package, combining reversal with appending! *)
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Lemma rev_append_ok : forall A (ls acc : list A),
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rev_append ls acc = rev ls ++ acc.
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Proof.
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induct ls; simplify.
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equality.
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rewrite IHls.
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simplify.
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rewrite app_assoc.
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simplify.
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equality.
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Qed.
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(* Concatenating the empty list has no effect. *)
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Lemma app_nil : forall A (ls : list A),
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ls ++ [] = ls.
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Proof.
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induct ls; simplify.
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equality.
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equality.
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Qed.
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(* Now we can prove equivalence of [rev'] and [rev], with no new induction. *)
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Theorem rev'_ok : forall A (ls : list A),
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rev' ls = rev ls.
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Proof.
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simplify.
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unfold rev'.
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rewrite rev_append_ok.
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apply app_nil.
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Qed.
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(** ** Zipping and unzipping *)
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(* Another classic pair of list operations is zipping and unzipping.
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* These functions convert between pairs of lists and lists of pairs. *)
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Fixpoint zip {A1 A2} (ls1 : list A1) (ls2 : list A2) : list (A1 * A2) :=
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match ls1, ls2 with
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| x1 :: ls1', x2 :: ls2' => (x1, x2) :: zip ls1' ls2'
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| _, _ => []
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end.
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(* Note how, when passed two lengths of different lists, [zip] drops the
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* mismatched suffix of the longer list. *)
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(* An explicit [Set] annotation is needed here, for obscure type-inference
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* reasons. *)
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Fixpoint unzip {A1 A2 : Set} (ls : list (A1 * A2)) : list A1 * list A2 :=
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match ls with
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| [] => ([], [])
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| (x1, x2) :: ls' =>
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let (ls1, ls2) := unzip ls' in
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(x1 :: ls1, x2 :: ls2)
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end.
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(* A few common-sense properties hold of these definitions. *)
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Theorem length_zip : forall A1 A2 (ls1 : list A1) (ls2 : list A2),
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length (zip ls1 ls2) = min (length ls1) (length ls2).
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Proof.
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induct ls1; simplify.
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linear_arithmetic.
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cases ls2; simplify.
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linear_arithmetic.
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rewrite IHls1.
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linear_arithmetic.
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Qed.
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(* We write [fst] and [snd] for the first and second projection operators on
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* pairs, respectively. *)
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Theorem length_unzip1 : forall (A1 A2 : Set) (ls : list (A1 * A2)),
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length (fst (unzip ls)) = length ls.
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Proof.
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induct ls; simplify.
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equality.
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cases hd.
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(* Note that [cases] allows us to pull apart a pair into its two pieces. *)
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cases (unzip ls).
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simplify.
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equality.
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Qed.
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Theorem length_unzip2 : forall (A1 A2 : Set) (ls : list (A1 * A2)),
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length (snd (unzip ls)) = length ls.
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Proof.
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induct ls; simplify.
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equality.
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cases hd.
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cases (unzip ls).
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simplify.
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equality.
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Qed.
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Theorem zip_unzip : forall (A1 A2 : Set) (ls : list (A1 * A2)),
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(let (ls1, ls2) := unzip ls in zip ls1 ls2) = ls.
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Proof.
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induct ls; simplify.
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equality.
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cases hd.
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cases (unzip ls).
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simplify.
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equality.
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Qed.
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(* There are also interesting interactions with [app] and [rev]. *)
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Theorem unzip_app : forall (A1 A2 : Set) (x y : list (A1 * A2)),
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unzip (x ++ y)
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= (let (x1, x2) := unzip x in
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let (y1, y2) := unzip y in
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(x1 ++ y1, x2 ++ y2)).
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Proof.
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induct x; simplify.
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cases (unzip y).
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equality.
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cases hd.
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cases (unzip x).
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simplify.
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rewrite IHx.
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cases (unzip y).
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equality.
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Qed.
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Theorem unzip_rev : forall (A1 A2 : Set) (ls : list (A1 * A2)),
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unzip (rev ls) = (let (ls1, ls2) := unzip ls in
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(rev ls1, rev ls2)).
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Proof.
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induct ls; simplify.
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equality.
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cases hd.
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cases (unzip ls).
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simplify.
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rewrite unzip_app.
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rewrite IHls.
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simplify.
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equality.
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Qed.
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