diff src/InductiveTypes.v @ 370:549d604c3d16

Move exercises out of mainline book
author Adam Chlipala <adam@chlipala.net>
date Fri, 02 Mar 2012 09:58:00 -0500
parents 70e51e8cfce7
children d1276004eec9
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--- a/src/InductiveTypes.v	Fri Dec 16 13:28:11 2011 -0500
+++ b/src/InductiveTypes.v	Fri Mar 02 09:58:00 2012 -0500
@@ -1262,26 +1262,3 @@
 (** The key piece of creativity in this theorem comes in the use of the natural number predecessor function [pred].  Embodied in the implementation of %\coqdockw{%#<tt>#injectivity#</tt>#%}% is a generic recipe for writing such type-specific functions.
 
 The examples in this section illustrate an important aspect of the design philosophy behind Coq.  We could certainly design a Gallina replacement that built in rules for constructor discrimination and injectivity, but a simpler alternative is to include a few carefully chosen rules that enable the desired reasoning patterns and many others.  A key benefit of this philosophy is that the complexity of proof checking is minimized, which bolsters our confidence that proved theorems are really true. *)
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-(** * Exercises *)
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-(** %\begin{enumerate}%#<ol>#
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-%\item%#<li># Define an inductive type [truth] with three constructors, [Yes], [No], and [Maybe].  [Yes] stands for certain truth, [No] for certain falsehood, and [Maybe] for an unknown situation.  Define %``%#"#not,#"#%''% %``%#"#and,#"#%''% and %``%#"#or#"#%''% for this replacement boolean algebra.  Prove that your implementation of %``%#"#and#"#%''% is commutative and distributes over your implementation of %``%#"#or.#"#%''%#</li>#
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-%\item%#<li># Define an inductive type [slist] that implements lists with support for constant-time concatenation.  This type should be polymorphic in a choice of type for data values in lists.  The type [slist] should have three constructors, for empty lists, singleton lists, and concatenation.  Define a function [flatten] that converts [slist]s to [list]s.  (You will want to run [Require Import] %\coqdocconstructor{%#<tt>#List#</tt>#%}%[.] to bring list definitions into scope.)  Finally, prove that [flatten] distributes over concatenation, where the two sides of your quantified equality will use the [slist] and [list] versions of concatenation, as appropriate.  Recall from Chapter 2 that the infix operator [++] is syntactic sugar for the [list] concatenation function [app].#</li>#
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-%\item%#<li># $\star$ Modify the first example language of Chapter 2 to include variables, where variables are represented with [nat].  Extend the syntax and semantics of expressions to accommodate the change.  Your new [expDenote] function should take as a new extra first argument a value of type [var -> nat], where [var] is a synonym for naturals-as-variables, and the function assigns a value to each variable.  Define a constant folding function which does a bottom-up pass over an expression, at each stage replacing every binary operation on constants with an equivalent constant.  Prove that constant folding preserves the meanings of expressions.#</li>#
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-%\item%#<li># $\star$ Reimplement the second example language of Chapter 2 to use mutually inductive types instead of dependent types.  That is, define two separate (non-dependent) inductive types [nat_exp] and [bool_exp] for expressions of the two different types, rather than a single indexed type.  To keep things simple, you may consider only the binary operators that take naturals as operands.  Add natural number variables to the language, as in the last exercise, and add an %``%#"#if#"#%''% expression form taking as arguments one boolean expression and two natural number expressions.  Define semantics and constant-folding functions for this new language.  Your constant folding should simplify not just binary operations (returning naturals or booleans) with known arguments, but also %``%#"#if#"#%''% expressions with known values for their test expressions but possibly undetermined %``%#"#then#"#%''% and %``%#"#else#"#%''% cases.  Prove that constant-folding a natural number expression preserves its meaning.#</li>#
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-%\item%#<li># Define mutually inductive types of even and odd natural numbers, such that any natural number is isomorphic to a value of one of the two types.  (This problem does not ask you to prove that correspondence, though some interpretations of the task may be interesting exercises.)  Write a function that computes the sum of two even numbers, such that the function type guarantees that the output is even as well.  Prove that this function is commutative.#</li>#
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-%\item%#<li># Using a reflexive inductive definition, define a type [nat_tree] of infinitary trees, with natural numbers at their leaves and a countable infinity of new trees branching out of each internal node.  Define a function [increment] that increments the number in every leaf of a [nat_tree].  Define a function [leapfrog] over a natural [i] and a tree [nt].  [leapfrog] should recurse into the [i]th child of [nt], the [i+1]st child of that node, the [i+2]nd child of the next node, and so on, until reaching a leaf, in which case [leapfrog] should return the number at that leaf.  Prove that the result of any call to [leapfrog] is incremented by one by calling [increment] on the tree.#</li>#
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-%\item%#<li># Define a type of trees of trees of trees of (repeat to infinity).  That is, define an inductive type [trexp], whose members are either base cases containing natural numbers or binary trees of [trexp]s.  Base your definition on a parameterized binary tree type [btree] that you will also define, so that [trexp] is defined as a nested inductive type.  Define a function [total] that sums all of the naturals at the leaves of a [trexp].  Define a function [increment] that increments every leaf of a [trexp] by one.  Prove that, for all [tr], [total (increment tr) >= total tr].  On the way to finishing this proof, you will probably want to prove a lemma and add it as a hint using the syntax [Hint Resolve name_of_lemma.].#</li>#
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-%\item%#<li># $\star$ Prove discrimination and injectivity theorems for the [nat_btree] type defined earlier in this chapter.  In particular, without using the tactics [discriminate], [injection], or [congruence], prove that no leaf equals any node, and prove that two equal nodes carry the same natural number.#</li>#
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-#</ol>#%\end{enumerate}% *)