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December 7, 2011 19:34
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A simple Coq proof exercise.
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(* Consider lists indexed by length. *) | |
Inductive List (t : Type) : nat -> Type := | |
| Nil : List t 0 | |
| Cons : forall n, t -> List t n -> List t (S n). | |
(* Every non-empty list has a head element. *) | |
Definition Head : forall t n, List t (S n) -> t. | |
intros t n list. inversion list. assumption. Defined. | |
(* In the proof, "inversion" tactic saved the day, eliminating the | |
"Nil" case. However, the body of the proof term is not very | |
pretty. Consider: *) | |
Print Head. | |
(* | |
Head = | |
fun (t : Type) (n : nat) (list : List t (S n)) => | |
let X := | |
match list in (List _ n0) return (n0 = S n -> t) with | |
| Nil => | |
fun H : 0 = S n => | |
let H0 := | |
eq_ind 0 (fun e : nat => match e with | |
| 0 => True | |
| S _ => False | |
end) I (S n) H in | |
False_rect t H0 | |
| Cons n0 X X0 => | |
fun H : S n0 = S n => | |
let H0 := | |
let H0 := | |
f_equal (fun e : nat => match e with | |
| 0 => n0 | |
| S n1 => n1 | |
end) H in | |
eq_rect n (fun n1 : nat => t -> List t n1 -> t) | |
(fun (X1 : t) (_ : List t n) => X1) n0 (eq_sym H0) in | |
H0 X X0 | |
end in | |
X (eq_refl (S n)) | |
: forall (t : Type) (n : nat), List t (S n) -> t | |
*) | |
(* Whoah, what a proof! Can we do better manually? Yes, if we find a | |
way to discharge the "Nil" case. Dependent pattern-matching to the | |
rescue: *) | |
Definition HeadManual t n (list: List t (S n)) : t := | |
match list in List _ k return match k with | |
| O => unit | |
| S _ => t | |
end with | |
| Nil => tt | |
| Cons _ x _ => x | |
end. | |
(* This worked out because "O" case never matches, that is, we | |
exploited the contradiction and were allowed to return unit. The | |
automated proof was more explicit about it, deriving an arbitrary | |
value from "0 = S n" which implies "False". This example is very | |
simple and in a sense we were lucky here. For more involved worked | |
examples see this CPDT chapter: | |
http://adam.chlipala.net/cpdt/html/DataStruct.html *) |
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