> > > do case x of
> > >  -> return 1
> > > (y:ys) -> g y >>= \temp -> f temp
> > See the rule about always binding to the previous line of a do block.
> > This case then violates that.
> I assumed that the example was equivalent to :
> do case x of
>  -> return 1
> (y:ys) -> do f (<- g y)
> Shouldn't the rule work then ?
If the do was inserted, then yes, this would work. Without it, it
doesn't. Perhaps this makes a restriction to not inside
case/let/lambda not that severe, since usually an additional do could
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