[Haskell-cafe] IO is not a monad

[Yitzchak Gale]

Lennart Augustsson wrote:
> I'm not sure what you're asking.  The (untyped) lambda calculus is
> Turing complete.
> How could seq improve that?

Obviously, it can't. But how can it hurt?

Classical lambda calculus does not model the
semantics of laziness, so seq is equivalent to
flip const there, just like foldl' is equivalent
to foldl. If we modify the lambda calculus to
model laziness - let's say, by restricting
beta-reduction - then the interesting
properties of seq are revealed.

Why should we treat seq differently in Haskell
just because its interesting properties are not
modeled in the classical lambda calculus?
Haskell is not a classical language, it is
non-strict (among other differences).

Regards,
Yitz