From: Jim Choate <ravage@einstein.ssz.com>
Date: Mon, 21 Sep 1998 17:53:19 +0800
To: cypherpunks@einstein.ssz.com (Cypherpunks Distributed Remailer)
Subject: RE: atheism (was: RE: Democracy... (fwd)) (fwd)
Message-ID: <199809212320.SAA19200@einstein.ssz.com>
MIME-Version: 1.0
Content-Type: text



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.


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