On 1/3/2013 8:38 AM, Juan de Nadie wrote:
> Hi Hassan. Thanks for the answer.
>
> This is one of the ways in that I thought. But, if we consider
> that "2^D" denotes the set of all subsets of D, I don't undertand
> what "2^D^n" means (considering D the domain or universe, and n the
> arity of the relation). Which (and how much) tuples I have when I
> consider the set 2^D^n, let D = {a,b,c} and n=2, for example? (01)
The notation D^n is the n-th Cartesian power of the set D; i.e., the set
of n-tuples of elements of D. Hence, the notation 2^(D^n) is the
powerset of the set of n-tuples of D (or if you prefer, the set of
subsets of n-tuples of D). For example, if D = {a,b}, and n=2, then:
D^2 =
{{},
{<a,a>},
{<a,b>},
{<b,a>},
{<b,b>},
{<a,a>, <a,b>},
{<a,a>, <b,a>},
{<a,a>, <b,b>},
{<a,b>, <b,a>},
{<a,b>, <b,b>},
{<b,a>, <b,b>},
{<a,a>, <a,b>, <b,a>},
{<a,a>, <a,b>, <b,b>},
{<a,a>, <b,a>, <b,b>},
{<a,b>, <b,a>, <b,b>},
{<a,a>, <a,b>, <b,a>, <b,b>}
}. (02)
So, |2^(D^2)| = 2^|D^2| = 2^(2^2) = 2^4 = 2x2x2x2 = 16 elements. (03)
-hak (04)
--
http://www.hassan-ait-kaci.net/contactme.html (05)
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