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RE: atheism (was: RE: Democracy... (fwd)) (fwd)
Forwarded message:
> From: Matthew James Gering <[email protected]>
> 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.
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