Infinite types

camarao at camarao at
Sat Dec 6 14:13:51 EST 2003

>>     f n () = (n, f (n + 1))
> In your example, if we assume that "f" has type, say, "a->()->(a,b)", for
> some "a","b", then it is used, in its own definition, with type
> "a->()->b".

Oops, should be: ... then it is used ... with type "a->b" (or Num a=>a->b).

> ... "a->()->b" is not (cannot be) an instance of "a->()->(a,b)" ...

Should be: "a->b" cannot be an instance of "a->()->(a,b)".

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