Re: Invertible nondeterminism?

From: Sebastian Fischer <>
Date: Fri, 17 Dec 2010 17:08:19 +0900

On Fri, 2010-12-17 at 08:25 +0100, Wolfgang Lux wrote:
> Maybe I'm a bit dense, but using multisets this definition looks
> equivalent to the previous one.

I agree. In fact, this is why I used it initially.

> On the other hand, using sets this
> definition is quite useless, since the right hand side can always be
> satisfied with S = D, where D is the underlying domain and thus (A,X)
> is equivalent to any other value.

Yes, this is what I wanted to express with "every element is equivalent
to failure" (which implies that every element is equivalent to every
other element).

The definition is derived from a more general one for arbitrary
commutative monoids. Specialized for sets (or more generally idempotent
commutative monoids) the resulting abelian group has only one element so
using a set-based notion of non-determinism together with invertible
nondeterminism is boring.


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