On Sat, Oct 16, 2010 at 9:21 PM, Ben Franksen <[email protected]> wrote: > I have a formal proof where I am stuck at a certain point. > > Suppose we have a function > > f :: IORef a -> IO b > > I want to prove that > > f r == do > s1 <- readIORef r > r' <- newIORef s1 > x <- f r' > s3 <- readIORef r' > writeIORef r s3 > return x > > What happens here is that the temporary IORef r' takes the place of the > argument r, and after we apply f to it we take its content and store it in > the original r. This should be the same as using r as argument to f in the > first place. > > How can I prove this formally?
You haven't provided us with any information about the formal model you are using and your question is somewhat ambiguously phrased, hence Thomas' response where, I'm pretty sure, he misunderstood what you were asking. At any rate, if you intend to prove this for any arbitrary f, I can't tell you how to prove it formally because it's not true. Regardless, this email has far too little information for anyone to provide you an answer. _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
