Serge D. Mechveliani wrote:

[...]

People note that in my example of

...

...

(1) (\ x -> (if p x then foo (g x) else foo (h x)) ...)

(2) (\ x -> foo ((if p x then g x else h x)) )

p x may be _|_, and this makes (1) not equivalent to (2).

Yes. I think, the point is that laziness does not only have something to do with execution behavior but also with the meaning of expressions as soon as _|_ values are considered.

Serge Mechveliani

Wolfgang