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<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<link rel="stylesheet" href="jscoq/node_modules/bootstrap/dist/css/bootstrap.min.css" />
<title>Machine-Checked Mathematics</title>
<link rel="stylesheet" href="local.css" />
<script src='https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-MML-AM_CHTML'
async></script>
<script src="Blob.js" type="text/javascript"></script>
<script src="FileSaver.js" type="text/javascript"></script>
</head>
<body>
<div id="ide-wrapper" class="toggled">
<div id="code-wrapper">
<div id="document">
<p>
Use ALT-(up-arrow) and ALT-(down-arrow) to process this document inside your browser, line-by-line.
Use ALT-(right-arrow) to go to the cursor.
You can
<span class="save-button" onClick="save_coq_snippets()">save your edits</span>
inside your browser and
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Finally, you can
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your working copy of the file, e.g., for sending it to teachers.
<hl />
</p>
<div><textarea id='coq-ta-1'>
(* ignore these directives *)
From mathcomp Require Import mini_ssreflect.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Add Printing Coercion is_true.
Notation "x '= true'" := (is_true x) (x at level 100, at level 0, only printing).
Remove Printing If bool.
</textarea></div>
<div><p>
<h4>
maxn defines the maximum of two numbers
</h4>
</div>
<div><textarea id='coq-ta-2'>
Print maxn.
Search maxn.
</textarea></div>
<div><p>
<h4>
We define the maximum of three number as folllows
</h4>
</div>
<div><textarea id='coq-ta-3'>
Definition max3n a b c :=
if a < b then maxn b c else maxn a c.
</textarea></div>
<div><p>
<h4>
Try to prove the following facts (you may use properties of maxn provided by the library)
</h4>
<p>
<h3>
Exercise 1
</h3>
</div>
<div><textarea id='coq-ta-4'>
Lemma max3n3n a : max3n a a a = a.
Proof. by rewrite /max3n if_same maxnn. Qed.
</textarea></div>
<div><p>
<h3>
Exercise 2
</h3>
</div>
<div><textarea id='coq-ta-5'>
Lemma max3E a b c : max3n a b c = maxn (maxn a b) c.
Proof. by rewrite /max3n /maxn; case: (a < b). Qed.
</textarea></div>
<div><p>
<h3>
Exercise 3
</h3>
</div>
<div><textarea id='coq-ta-6'>
Lemma max3n_213 a b c : max3n a b c = max3n b a c.
Proof. by rewrite max3E (maxnC a) -max3E. Qed.
</textarea></div>
<div><p>
<h3>
Exercise 4
</h3>
</div>
<div><textarea id='coq-ta-7'>
Lemma max3n_132 a b c : max3n a b c = max3n a c b.
Proof. by rewrite max3E -maxnA (maxnC b) maxnA -max3E. Qed.
</textarea></div>
<div><p>
<h3>
Exercise 5
</h3>
</div>
<div><textarea id='coq-ta-8'>
Lemma max3n_231 a b c : max3n a b c = max3n b c a.
Proof. by rewrite max3n_213 max3n_132. Qed.
</textarea></div>
<div><p>
<h3>
</h3>
<p>
The next exercise consists in proving an alternative induction scheme on type list,
whose base case is the last element in the list.
<p>
For this purpose, we first define a concatenation operation between lists
and prove a few lemmas about it.
<p>
The 'rcons' operation is provided by the underlying library, and appends its
last argument to the end of its first, which is a list.
<p>
The exercise is self-contained : you should not use results that you did not
prove yourself.
</div>
<div><textarea id='coq-ta-9'>
About rcons.
Module Seq.
Section Cat.
Variable A : Type.
</textarea></div>
<div><p>
<h3>
Definition of the concatenation operation
</h3>
</div>
<div><textarea id='coq-ta-10'>
Fixpoint cat (s1 s2 : list A) := if s1 is x :: s1' then x :: s1' ++ s2 else s2
where "s1 ++ s2" := (cat s1 s2) : seq_scope.
</textarea></div>
<div><p>
<h3>
Exercise 5
</h3>
</div>
<div><textarea id='coq-ta-11'>
Lemma cat0s (s : list A) : [::] ++ s = s.
Proof. (* fill this in *)
by [].
Qed.
</textarea></div>
<div><p>
<h3>
Exercise 6
</h3>
</div>
<div><textarea id='coq-ta-12'>
Lemma cats0 (s : list A) : s ++ [::] = s.
Proof. (* fill this in *)
elim: s => [| x s ihs] /=.
- by [].
- by rewrite ihs.
Qed.
</textarea></div>
<div><p>
<h3>
Exercise 7
</h3>
</div>
<div><textarea id='coq-ta-13'>
Lemma cats1 (s : list A) (z : A) : s ++ [:: z] = rcons s z.
Proof. (* fill this in *)
elim: s => [| x s ihs] /=.
- by [].
- by rewrite ihs.
Qed.
</textarea></div>
<div><p>
<h3>
Exercise 8
</h3>
</div>
<div><textarea id='coq-ta-14'>
Lemma catA (s1 s2 s3 : list A) : s1 ++ s2 ++ s3 = (s1 ++ s2) ++ s3.
Proof. (* fill this in *)
elim: s1 => [| x s1 ihs1] /=.
- by [].
- by rewrite ihs1.
Qed.
</textarea></div>
<div><p>
<h3>
Exercise 9.
</h3>
<h4>
(Hint: this proof does not require an induction step)
</h4>
</div>
<div><textarea id='coq-ta-15'>
Lemma cat_rcons x s1 s2 : rcons s1 x ++ s2 = s1 ++ x :: s2.
Proof. (* fill this in *)
by rewrite -cats1 -catA.
Qed.
</textarea></div>
<div><p>
<h3>
Exercise 10.
</h3>
<h4>
Prove lemma last_ind using first the intermediate lemma hcat,
</h4>
and then by induction. Hint: use the lemmas you have proved
so far on function cat. </div>
<div><textarea id='coq-ta-16'>
Lemma last_ind (P : list A -> Prop) :
P [::] -> (forall s x, P s -> P (rcons s x)) -> forall s, P s.
Proof.
move=> Hnil Hlast s.
suff hcat : forall m, P m -> P (m ++ s).
(* fill this in *)
rewrite -(cat0s s). apply: hcat.
by [].
(* fill this in *)
elim: s => [|x s2 IHs] m Pm; first by rewrite cats0.
rewrite -cat_rcons.
apply: IHs.
apply: Hlast.
by [].
Qed.
End Cat.
End Seq.
</textarea></div>
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