[Haskell-cafe] Re: a regressive view of support for imperative programming in Haskell

[Stefan O'Rear]

On Mon, Aug 13, 2007 at 05:39:34PM +0200, apfelmus wrote:
> Stefan O'Rear schrieb:
>> On Mon, Aug 13, 2007 at 04:35:12PM +0200, apfelmus wrote:
>>> My assumption is that we have an equivalence
>>>
>>>   forall a,b . m (a -> m b) ~ (a -> m b)
>>>
>>> because any side effect executed by the extra m on the outside can well 
>>> be delayed until we are supplied a value a. Well, at least when all 
>>> arguments are fully applied, for some notion of "fully applied"
>> (\a x -> a >>= ($ x)) ((\f -> return f) X) ==> (β)
>> (\a x -> a >>= ($ x)) (return X)           ==> (β)
>> (\x -> (return X) >>= ($ x))               ==> (monad law)
>> (\x -> ($ x) X)                            ==> (β on the sugar-hidden 
>> 'flip')
>> (\x -> X x)                                ==> (η)
>> X
>> Up to subtle strictness bugs arising from my use of η :), you're safe.
>
> Yes, but that's only one direction :) The other one is the problem:
>
>  return . (\f x -> f >>= ($ x)) =?= id
>
> Here's a counterexample
>
>  f :: IO (a -> IO a)
>  f = writeAHaskellProgram >> return return
>
>  f' :: IO (a -> IO a)
>  f' = return $ (\f x -> f >>= ($ x)) $ f
>  ==> (β)
>   return $ \x -> (writeAHaskellProgram >> return return) >>= ($ x)
>  ==> (BIND)
>   return $ \x -> writeAHaskellProgram >> (return return >>= ($ x))
>  ==> (LUNIT)
>   return $ \x -> writeAHaskellProgram >> (($ x) return)
>  ==> (β)
>   return $ \x -> writeAHaskellProgram >> return x
>
> Those two are different, because
>
>  clever  = f  >> return () = writeAHaskellProgram
>  clever' = f' >> return () = return ()
>
> are clearly different ;)

I figured that wouldn't be a problem since our values don't escape, and
the functions we define all respect the embedding... More formally:

Projections and injections:

proj ma = \x -> ma >>= \ fn' -> fn' x
inj  fn = return fn

Define an equivalence relation:

ma ≡ mb <-> proj ma = proj mb

Projection respects equivalence:

ma ≡ mb -> proj ma = proj mb            (intro ->)
ma ≡ mb => proj ma = proj mb            (equiv def)
proj ma = proj mb => proj ma = proj mb  (assumption)

Application:

(@) ma1 = \x -> join (proj ma1 x)

Application respects equivalence:

ma1 ≡ ma2 -> (@) ma1 = (@) ma2          (intro ->)
ma1 ≡ ma2 => (@) ma1 = (@) ma2          (β)
ma1 ≡ ma2 => (\x -> join (proj ma1 x)) = (\x -> join (proj ma2 x))    (extensionality)
ma1 ≡ ma2 => join (proj ma1 x) = join (proj ma2 x)    (application respects = left)
ma1 ≡ ma2 => proj ma1 x = proj ma2 x    (application respects = right)
ma1 = ma2 => proj ma1 = proj ma2        (lemma)

Stefan
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