DESCRIPTION:
- This web-page contains formal proofs of theorems in Multiple Form Logic, proposed by
other
people, or dictated
by necessity.
- The new version of this page contains important updates,
contributed by Mr. Art Collings in the "Laws of Form Forum"
- These updates
arose from the new
Formal Demonstration (by Art Collings) of Theorem Art-1 (using
only re-write
proof-steps)
- The
latest (more
fundamental) demonstration
of theorem Art-1 by Mr. Art Collings is on-line here:
http://homepage.mac.com/otter/XOR_demo_hybrid.pdf
- A copy of Art Collings' demonstration
is also included here
Current list of theorem proofs:
1) Theorem
ART-1 (about the "#" operator)
2)
An algebraic proof of "Theorem 9"
3) Theorem
ART-1 (alternative "demonstration" by Art Collings)
4) SLIDE-SHOW animated proof of the
Third Axiom of Equality
(in
Tom
Etter’s paper “The Expressive Power of Equality”)
(more to
be included)
LINKS
- (blog
article) The
“Extended XOR Operator” as a Consistent interpretation of George
Spencer-Brown’s "distinction"
- (blog article in Greek): Ενημέρωση
από ΠΟΛΥ πιο σοβαρό Debate: ΠΩΣ Η Πολλαπλότητα εξάγεται από τη
Μοναδική Μορφή
- (blog article) How
Equality is embedded in Multiple Form Logic (Animated GIF Presentation
of a Theorem Proof)
- (blog article)
The
“Axiom of Perception” in Multiple Form Logic (brand new animated GIF
slide-show presentation)
1) Theorem
ART-1 (about the "#" operator)
Recently, Mr. Art Collings expressed some
interesting doubts and objections about certain aspects of Multiple
Form Logic, particularly with regard to the "#" operator (interpreted
as logical "XOR"). E.g. he remarked:
"In
LoF, the typical way to express XOR is via the form ( ( A B
) ( ( A
) ( B ) ) ) [ which basically translates as 'A or B but not A
and B'.
It is easy to verify this expression corresponds to XOR]. In your
notation, the corresponding expression is ( ( A, B)
#1 , ( A#1 ,
B#1)#1 ) #1 . This being the case, we can easily prove
arithmetically
that the following theorem is true:
A#
B = ( ( A, B) #1 , ( A#1 , B#1)#1 ) #1
The
question then becomes whether you can demonstrate this as a consequence
in your algebra. GSB proves that LoF is complete by proving that any
theorem that is true in the arithmetic can be demonstrated as a
consequence in the Algebra. In order for your algebra to be complete in
the same sense as LoF, it must be the case that expression such as this
can be demonstrated as a consequence from your three axioms: (1. A,1 =
1 ; 2. A#A = 0; 3. A, (A, B)#X = A, B#X.
[ i.e., 'demonstrated',
as distinct from 'proved'! ]
ANSWER: Evidently, the
'instinctive guess that such a demonstration is not possible from the
axiom set" (etc.) is wrong.
Here is a complete
algebraic proof: