Robert Dockins wrote:
[snip] From an earlier post:
Now since f and g compute the same results for the same inputs, anywhere in a program that you can use f you could just replace f by g and the observable behaviour of the program would be completely unaffected. This is what referential transparency means.
My essential claim is that the above statement is in error (but in a fairly subtle way).
Ok I see now! :-) I was confusing the concept of referential transparency with a kind of global code equivalence, so the rest of my argument is irrelevant. Thus I should have said: " For particular types T1 and T2, if (f (x::T1))::T2 === g x for all x in T1 then f :: T1->T2 and g ::T1->T2 can be freely substituted since the context T1->T2 cannot tell them apart." Thanks for pointing out the faulty definition, Regards, Brian.