19 Jun
2014
19 Jun
'14
3:27 a.m.
On Wed, Jun 18, 2014 at 07:40:26PM -0400, William Knop wrote:
f = \x -> x g = \x -> (x,1) h = \x -> fst (g x) i = \x -> case f of f -> True _ -> False
i f => True i h => ?
If g isn't inlined into h and fst optimized out, wouldn't the head normal form of f and h be different, and the comparison fail even though it shouldn't? Or should it? I've taken function equality to be "f and g are equal iff f x == g x, for all x."
You mean 'i = \x -> case x of' ... Anyway, I'm not suggesting using case to *compare* functions, simply force their thunk before proceeding with a single default alternative. Tom