# HG changeset patch # User Adam Chlipala # Date 1317137986 14400 # Node ID d7178fb7732127d300c1a7adff13034aa6c3da11 # Parent cbeccef45f4e9564cbb592b104e6fe79bc957e8b Two exercises for Match diff -r cbeccef45f4e -r d7178fb77321 src/Match.v --- a/src/Match.v Sun Sep 25 13:20:56 2011 -0400 +++ b/src/Match.v Tue Sep 27 11:39:46 2011 -0400 @@ -996,3 +996,112 @@ | [ |- fst ?x = 3 ] => equate x (3, 2) end; reflexivity. Qed. + + +(** * Exercises *) + +(** %\begin{enumerate}%#
    # + + %\item%#
  1. # An anonymous Coq fan from the Internet was excited to come up with this tactic definition shortly after getting started learning Ltac: *) + +Ltac deSome := + match goal with + | [ H : Some _ = Some _ |- _ ] => injection H; clear H; intros; subst; deSome + | _ => reflexivity + end. + +(** Without lifting a finger, exciting theorems can be proved: *) + +Theorem test : forall (a b c d e f g : nat), + Some a = Some b + -> Some b = Some c + -> Some e = Some c + -> Some f = Some g + -> c = a. + intros; deSome. +Qed. + +(** Unfortunately, this tactic exhibits some degenerate behavior. Consider the following example: *) + +Theorem test2 : forall (a x1 y1 x2 y2 x3 y3 x4 y4 x5 y5 x6 y6 : nat), + Some x1 = Some y1 + -> Some x2 = Some y2 + -> Some x3 = Some y3 + -> Some x4 = Some y4 + -> Some x5 = Some y5 + -> Some x6 = Some y6 + -> Some a = Some a + -> x1 = x2. + intros. + Time try deSome. +Abort. + +(* begin hide *) +Reset test. +(* end hide *) + +(** This (failed) proof already takes about one second on my workstation. I hope a pattern in the theorem statement is clear; this is a representative of a class of theorems, where we may add more matched pairs of [x] and [y] variables, with equality hypotheses between them. The running time of [deSome] is exponential in the number of such hypotheses. + + The task in this exercise is twofold. First, figure out why [deSome] exhibits exponential behavior for this class of examples and record your explanation in a comment. Second, write an improved version of [deSome] that runs in polynomial time.#
  2. # + + %\item%#
  3. # Some theorems involving existential quantifiers are easy to prove with [eauto]. *) + +Theorem test1 : exists x, x = 0. + eauto. +Qed. + +(** Others are harder. The problem with the next theorem is that the existentially quantified variable does not appear in the rest of the theorem, so [eauto] has no way to deduce its value. However, we know that we had might as well instantiate that variable to [tt], the only value of type [unit]. *) + +Theorem test2 : exists x : unit, 0 = 0. +(* begin hide *) + eauto. +Abort. +(* end hide *) + +(** We also run into trouble in the next theorem, because [eauto] does not understand the [fst] and [snd] projection functions for pairs. *) + +Theorem test3 : exists x : nat * nat, fst x = 7 /\ snd x = 2 + fst x. +(* begin hide *) + eauto. +Abort. +(* end hide *) + +(** Both problems show up in this monster example. *) + +Theorem test4 : exists x : (unit * nat) * (nat * bool), + snd (fst x) = 7 /\ fst (snd x) = 2 + snd (fst x) /\ snd (snd x) = true. +(* begin hide *) + eauto. +Abort. +(* end hide *) + +(** The task in this problem is to write a tactic that preprocesses such goals so that [eauto] can finish them. Your tactic should serve as a complete proof of each of the above examples, along with the wide class of similar examples. The key smarts that your tactic will bring are: first, it introduces separate unification variables for all the %``%#"#leaf types#"#%''% of compound types built out of pairs; and second, leaf unification variables of type [unit] are simply replaced by [tt]. + + A few hints: The following tactic is more convenient than direct use of the built-in tactic [evar], for generation of new unification variables: *) + +Ltac makeEvar T k := let x := fresh in + evar (x : T); let y := eval unfold x in x in clear x; k y. + +(** remove printing exists *) + +(** This is a continuation-passing style tactic. For instance, when the goal begins with existential quantification over a type [T], the following tactic invocation will create a new unification variable to use as the quantifier instantiation: + +[makeEvar T ltac:(][fun x => exists x)] *) + +(** printing exists $\exists$ *) + +(** Recall that [exists] formulas are desugared to uses of the [ex] inductive family. In particular, a pattern like the following can be used to extract the domain of an [exists] quantifier into variable [T]: + +[| [ |- ex ( A := ?T) _ ] => ...] + + The [equate] tactic used as an example in this chapter will probably be useful, to unify two terms, for instance if the first is a unification variable whose value you want to set. +[[ + Ltac equate E1 E2 := let H := fresh in + assert (H : E1 = E2) by reflexivity; clear H. +]] + + Finally, there are some minor complications surrounding overloading of the [*] operator for both numeric multiplication and Cartesian product for sets (i.e., pair types). To ensure that an Ltac pattern is using the type version, write it like this: + +[| (?T1 * ?T2)%type => ...] + +#
#%\end{enumerate}% *)