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Nitpicks and comments (Score:1)
d does e
I get
d does a
b does e
(from the call to fact).
Inferring
a == d
b == e
leaves
1 in a in Num
1 in b in Num
0 in c in Eq
a does b
a does c
which may not be as tight as you'd like but expresses the constraints reasonab
Re:Nitpicks and comments (Score:1)
> d does e
> I get
> d does a
> b does e
> (from the call to fact).
That's right, good eye.
> 1 in a in Num
> 1 in b in Num
[...]
> which may not be as tight as you'd like but
> expresses the constraints reasonably. It allows,
> for instance,
> a == [1..5]
Actually it doesn't. "a in Num" means that a is a ring. If you say 4 + 3 (which are both in a here), you get 7, which is not in a, so a is not in Num.
What I'm currently stu