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    Your point about memoization is very good. Note that Haskell strongly emphasizes referential transparency for programming, but the memoization-like behavior of lazy evaluation involves a staggering amount of mutation being done by the language runtime behind the scenes. Commented Apr 25, 2013 at 19:08
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    @C. A. McCann: I think what you say is very important: in a functional language the runtime can use mutation to optimize the computation but there is not construct in the language that allows the programmer to use mutation. Another example is a while loop with a loop variable: in Haskell you can write a tail recursive function that may be implemented with a mutable variable (to avoid using the stack), but what the programmer sees are immutable function arguments that are passed from one call to the next. Commented Apr 26, 2013 at 5:21
  • @Michael Shaw: +1 for "The problem isn't mutability per se, it's a lack of referential transparency." Maybe you can cite the Clean language in which you have uniqueness types: these allow mutability but still guarantee referential transparency. Commented Apr 26, 2013 at 8:33
  • @Giorgio: I don't really know anything about Clean, although I've heard it mentioned from time to time. Maybe I should look into it. Commented Apr 26, 2013 at 12:07
  • @Michael Shaw: I do not know very much about Clean, but I know that it uses uniqueness types to ensure referential transparency. Basically, you can modify a data object provided that after the modification you have no references to the old value. IMO this illustrates your point: referential transparency is the most important point, and immutability is only one possible way of ensuring it. Commented Apr 26, 2013 at 14:51