Perhaps the description was unclear;

F1;f1 gives result r1;r2 (not the same)
F1;f2 gives r1;r2
F2,f1 gives r1;r2

No, you were clear, and Felipe's answer still makes sense. Since f1 and f2 have the same definition, substituting one for the other should not change anything.

Maybe do notation is what is confusing you. Try getting rid of the do notation and writing everything in terms of the more basic (>>) and (>>=) combinators. If (>>) could be any operator, should we expect that f1 = f1 >> f1?

Luke

-------------------------------------------

> -----Original Message-----
> From: Felipe Almeida Lessa [mailto:felipe.lessa@gmail.com]
> Sent: Tuesday, April 19, 2011 2:26 PM
> To: Gregory Guthrie
> Cc: haskell-cafe@haskell.org
> Subject: Re: [Haskell-cafe] Haskell from SML - referrential Transparency?!
>

> On Tue, Apr 19, 2011 at 4:10 PM, Gregory Guthrie <guthrie@mum.edu> wrote:
> > and I get different results from the two executions (f1,f2), even
> > though they have exactly the same definition. Reversing their order,
> > gives the exact same results (i.e. the results are still different,
> > and in the same original order as f2;f1). Even doing (f1;f1) gives two different results.
>
> This shows that referential transparency is working nicely.
>
> --
> Felipe.
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