作为我遇到的问题的一个最小示例,这里有一个自然数的定义、一个加倍函数和一个由偶数谓词细化的类型:
data Nat' = Z | S Nat' deriving Show
{-@ reflect double' @-}
double' :: Nat' -> Nat'
double' Z = Z
double' (S x) = (S (S (double' x)))
{-@ type Even' = {v:Nat' | even' v} @-}
{-@ reflect even' @-}
even' :: Nat' -> Bool
even' Z = True
even' (S Z) = False
even' (S (S x)) = even' x
我想先声明{-@ double' :: Nat' -> Even' @-}
然后证明这是真的,但我的印象是我必须先写证明然后使用castWithTheorem
(它本身对我有用)这样:
{-@ even_double :: x:Nat' -> {even' (double' x)} @-}
even_double Z = even' (double' Z)
==. even' Z
==. True
*** QED
even_double (S x) = even' (double' (S x))
==. even' (S (S (double' x)))
==. even' (double' x)
? even_double x
==. True
*** QED
{-@ double :: Nat' -> Even' @-}
double x = castWithTheorem (even_double x) (double' x)
但是,这会产生相当难以辨认的错误,例如:
:1:1-1:1: Error
elaborate solver elabBE 177 "lq_anf$##7205759403792806976##d3tK" {lq_tmp$x##1556 : (GHC.Types.$126$$126$ (GHC.Prim.TYPE GHC.Types.LiftedRep) (GHC.Prim.TYPE GHC.Types.LiftedRep) bool bool) | [(lq_tmp$x##1556 = GHC.Types.Eq#)]} failed on:
lq_tmp$x##1556 == GHC.Types.Eq#
with error
Cannot unify (GHC.Types.$126$$126$ (GHC.Prim.TYPE GHC.Types.LiftedRep) (GHC.Prim.TYPE GHC.Types.LiftedRep) bool bool) with func(0 , [(GHC.Prim.$126$$35$ @(42) @(43) @(44) @(45));
(GHC.Types.$126$$126$ @(42) @(43) @(44) @(45))]) in expression: lq_tmp$x##1556 == GHC.Types.Eq#
because
Elaborate fails on lq_tmp$x##1556 == GHC.Types.Eq#
in environment
GHC.Types.Eq# := func(4 , [(GHC.Prim.$126$$35$ @(0) @(1) @(2) @(3));
(GHC.Types.$126$$126$ @(0) @(1) @(2) @(3))])
lq_tmp$x##1556 := (GHC.Types.$126$$126$ (GHC.Prim.TYPE GHC.Types.LiftedRep) (GHC.Prim.TYPE GHC.Types.LiftedRep) bool bool)
我究竟做错了什么?从我的实验来看,这似乎是由于试图证明某些谓词函数对某些论点是正确的。