negative sets and commutative and associative addition and subtraction

[Hagen Schwaß]

Please read up

https://groups.google.com/forum/#!topic/sci.math/8sY4Mcgkm1g

if it could help for databases.

This is a quick start:

Set U, e.g. R^2

Set U* set of all subsets of U including U and {}

Definitions:

A,B in U*

AnB {x:x in A and x in B} intersection
AuB {x:x in A or X in B} union
A\B {x:x in A and x not in B} without

Note that traditional set operations bind harder than definitions: (AuB)+(AnB). Further -negation binds hardest: -AuB=(-A)uB.

We are now defining algebraic compatible +/- operators:

A+B=AnB+AuB
A-B=A\B-B\A

A+{}=A
{}=-{}

Proof:

A-B=A+(-B)=An-B+Au-B=A\B-B\A
we choose AnB=A\B and AuB=-(B\A)

-A-B=-(A+B)=-Au-B+-An-B=-(AnB)-(AuB)
we choose -(AnB)=-Au-B and -(AuB)=-An-B

it holds:

-A\-B=B\A

[Hagen Schwaß]

Post by Hagen Schwaß
Please read up
https://groups.google.com/forum/#!topic/sci.math/8sY4Mcgkm1g
if it could help for databases.
Set U, e.g. R^2
Set U* set of all subsets of U including U and {}
A,B in U*
AnB {x:x in A and x in B} intersection
AuB {x:x in A or X in B} union
A\B {x:x in A and x not in B} without
Note that traditional set operations bind harder than definitions: (AuB)+(AnB). Further -negation binds hardest: -AuB=(-A)uB.
A+B=AnB+AuB
A-B=A\B-B\A
A+{}=A
{}=-{}
A-B=A+(-B)=An-B+Au-B=A\B-B\A
we choose AnB=A\B and AuB=-(B\A)
-A-B=-(A+B)=-Au-B+-An-B=-(AnB)-(AuB)
we choose -(AnB)=-Au-B and -(AuB)=-An-B
-A\-B=B\A

We proof associativity and commutativity by using n*n*(n+1)/2 atomic subsets. Please read up

https://groups.google.com/forum/#!topic/de.sci.mathematik/2ndfhuWhwzQ