Forwarded message:

From: Matthew James Gering <mgering@ecosystems.net> Subject: RE: atheism (was: RE: Democracy... (fwd)) (fwd) Date: Mon, 21 Sep 1998 13:09:30 -0700

Jim Choate wrote:

...

As I've demonstrated before (and for the last time):

^(A) = (^A)

f() = belief function A = hypothesis ^ = not

f(^A) != ^f(A)

Prove it. The f in your f() is redundant, but I can work with it. Simply stating it over and over isn't proof. We're doing Boolean Algebra here not Algebra, the rules for transitory functions are different. I'll even do you a favor and list the Boolean Laws and Theorems for you: 1) X+0=X X*1=X 2) X+1=1 X*0=0 3) X+X=X X*X=X 4) (X')'=X 5) X+X'=X X*X'=0 6) X+Y=Y+X X*Y=Y*X 7) (X+Y)+Z=X+(Y+Z) (XY)Z=X(YZ)=XYZ =X+Y+Z 8) X(Y+Z)=XY+XZ X+YZ=(X+Y)*(X+Z) 9) XY+XY'=X (X+Y)*(X+Y')=X 10) X+XY=X X(X+Y)=X 11) (X+Y')Y=XY XY'+Y=X+Y 12) (X+Y+Z+...)'=X'Y'Z'... (XYX...)'=X'+Y'+Z'+... 13) [f(X1, X2, X3,...,0,1,+,*)]'= f(X1', X2', X3',...,1,0,*,+) {The following two are called Duality Theorems, not important here} 14) (X+Y+Z+...)^D = XYZ... (XYZ...)^D=X+Y+Z+... 15) [f(X1, X2, X3, ...,0,1,+,*)]^D=f(X1, X2, X3,...,1,0,*,+) 16) XY+YZ+X'Z=XY+X'Z (X+Y)*(Y+Z)*(X'+Z)=(X+Y)*(X'+Z) 17) (X+Y)(X'+Z)=XZ+X'Y The ones you and your dualist-atheism buddies are confused on are #4, #12, & #13. By #4 we have: ^(X)=(^X) By #12 we have: ^(X)=^X By #13 we have: ^f(X) = f(^X) All are clearly contrary to your assertion. Balls in your court junior. ____________________________________________________________________ The seeker is a finder. Ancient Persian Proverb The Armadillo Group ,::////;::-. James Choate Austin, Tx /:'///// ``::>/|/ ravage@ssz.com www.ssz.com .', |||| `/( e\ 512-451-7087 -====~~mm-'`-```-mm --'- --------------------------------------------------------------------