14 Apr
2010
14 Apr
'10
2:55 a.m.
I guess nontermination is a problem (e.g. if one or both functions
fail to terminate for some values, equality will be undecidable).
/Jonas
On 14 April 2010 08:42, Ashley Yakeley
On Wed, 2010-04-14 at 16:11 +1000, Ivan Miljenovic wrote:
but the only way you can "prove" it in Haskell is by comparing the values for the entire domain (which gets computationally expensive)...
It's not expensive if the domain is, for instance, Bool.
-- Ashley Yakeley
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