Comments on: Optional a -> a (negative proof) http://blog.tmorris.net/optional-a-negative-proof/ The weblog of Tony Morris Sat, 19 May 2012 02:59:47 +0000 http://wordpress.org/?v=2.6 By: Vlad Patryshev http://blog.tmorris.net/optional-a-negative-proof/#comment-171836 Vlad Patryshev Sun, 18 Mar 2012 21:03:30 +0000 http://blog.tmorris.net/?p=779#comment-171836 We could also look at it like this: Optional a -> a turns a into an algebra over Optional (makes a an object with a point actually); in some categories it's okay, but, say, in a topos it would mean 1=0, so the topos would be degenerate. We could also look at it like this:
Optional a -> a turns a into an algebra over Optional (makes a an object with a point actually); in some categories it’s okay, but, say, in a topos it would mean 1=0, so the topos would be degenerate.

]]> By: beroal http://blog.tmorris.net/optional-a-negative-proof/#comment-42487 beroal Wed, 30 Jun 2010 01:12:31 +0000 http://blog.tmorris.net/?p=779#comment-42487 As Brian Howard pointed out, you do not need to harness a catamorphism. (Optional a) corresponds to (a∨⊤). if (a=0), then (a∨⊤→a)=0. As Brian Howard pointed out, you do not need to harness a catamorphism. (Optional a) corresponds to (a∨⊤). if (a=0), then (a∨⊤→a)=0.

]]> By: Brian Howard http://blog.tmorris.net/optional-a-negative-proof/#comment-42388 Brian Howard Mon, 28 Jun 2010 17:01:31 +0000 http://blog.tmorris.net/?p=779#comment-42388 Here's another way, supposing you already have an uninhabited type (your a->b, or Scala's Nothing, for example); call it U. Now, take your supposed function f at type U -- this will have type Optional U -> U. Apply this to Empty, and you've found an inhabitant of U. That's a contradiction, so you must not have had such an f. Here’s another way, supposing you already have an uninhabited type (your a->b, or Scala’s Nothing, for example); call it U. Now, take your supposed function f at type U — this will have type Optional U -> U. Apply this to Empty, and you’ve found an inhabitant of U. That’s a contradiction, so you must not have had such an f.

]]> By: Tony Morris http://blog.tmorris.net/optional-a-negative-proof/#comment-42139 Tony Morris Mon, 21 Jun 2010 09:41:21 +0000 http://blog.tmorris.net/?p=779#comment-42139 Psuedonym, Excellent, I hadn't thought of it that way. Psuedonym, Excellent, I hadn’t thought of it that way.

]]> By: Pseudonym http://blog.tmorris.net/optional-a-negative-proof/#comment-42134 Pseudonym Mon, 21 Jun 2010 08:06:27 +0000 http://blog.tmorris.net/?p=779#comment-42134 Here's an alternative way of looking at it. Introduce: <pre> data Id a = a instance Functor Optional where fmap f (Full a) = Full (f a) fmap f Empty = Empty instance Functor Id where fmap f (Id a) = Id (f a) </pre> We want to find a function: <pre> doesThisExist :: forall a. Optional a -> Id a </pre> doesThisExist is a map between two Functors, and is therefore (in Haskell) a natural transformation and satisfies: <pre> forall types A B. forall g :: A -> B. doesThisExist . fmap g = fmap g . doesThisExist </pre> Note that this is also the free theorem for the type of doesThisExist. For any b :: B, we have: <pre> doesThisExist . fmap (const b) = fmap (const b) . doesThisExist (expand fmap) doesThisExist . (\x -> case x of { Full _ -> Full b; Optional -> Optional }) = const (Id b) => doesThisExist . (\x -> case x of { Full _ -> Full b; Optional -> Optional }) $ Optional = const (Id b) $ Optional => doesThisExist Optional = Id b </pre> That is, doesThisExist Optional is the same as Id b for any b :: B. This is impossible, therefore doesThisExist doesn't exist. Here’s an alternative way of looking at it. Introduce:

data Id a = a

instance Functor Optional where
    fmap f (Full a) = Full (f a)
    fmap f Empty = Empty

instance Functor Id where
    fmap f (Id a) = Id (f a)

We want to find a function:

doesThisExist :: forall a. Optional a -> Id a

doesThisExist is a map between two Functors, and is therefore (in Haskell) a natural transformation and satisfies:

forall types A B. forall g :: A -> B. doesThisExist . fmap g = fmap g . doesThisExist

Note that this is also the free theorem for the type of doesThisExist.

For any b :: B, we have:

 doesThisExist . fmap (const b) = fmap (const b) . doesThisExist
 (expand fmap)
 doesThisExist . (\x ->  case x of { Full _ -> Full b; Optional -> Optional }) = const (Id b)
=>
 doesThisExist . (\x ->  case x of { Full _ -> Full b; Optional -> Optional }) $ Optional = const (Id b) $ Optional
=>
 doesThisExist Optional = Id b

That is, doesThisExist Optional is the same as Id b for any b :: B. This is impossible, therefore doesThisExist doesn’t exist.

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