## antifier}

"for all a:" means that a is a generic type variable. For the two uses of id, a is instantiated to Bool and Int. Compare this with

let id x = x in let f g = (g True, g 1) in f idThis looks similar but f has no legal Hindley-Milner type. If we say

f :: (a -> b) -> (b, b)this would permit g's type to be any instance of (a -> b) rather than requiring it to be at least as general as (a -> b). Furthermore, it constrains both instances of g to have the same result type whereas they do not. The type variables a and b in the above are implicitly quantified at the top level:

f :: for all a: for all b: (a -> b) -> (b, b)so instantiating them (removing the quantifiers) can only be done once, at the top level. To correctly describe the type of f requires that they be locally quantified:

f :: ((for all a: a) -> (for all b: b)) -> (c, d)which means that each time g is applied, a and b may be instantiated differently. f's actual argument must have a type at least as general as ((for all a: a) -> (for all b: b)), and may not be some less general instance of this type. Type variables c and d are still implicitly quantified at the top level and, now that g's result type is a generic type variable, any types chosen for c and d are guaranteed to be instances of it.

This type for f does not express the fact that b only needs to be at least as general as the types c and d. For example, if c and d were both Bool then any function of type (for all a: a -> Bool) would be a suitable argument to f but it would not match the above type for f.

### Nearby terms:

let id x = x in ♦ **antifier}** ♦ ces of g to have ♦ ((for all a: a) -> (for all b: b)) -> (c, d)

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