10) Proofs of some Multiple Form Logic^{™}
Theorems
Hyperlinked
Index of Proofs:
·
Theorem T1) A , B # A = A, B
·
Theorem T3) (in the original paper of 1984…)
·
Theorem T4) George Spencer Brown’s algebraic axiom J1 proved as a theorem
·
Theorem T5) George Spencer Brown’s algebraic axiom J2 proved as a theorem
·
Theorem T5.1) George Spencer Brown’s algebraic axiom J2 can be generalised
·
Theorem T6) Multiple Form Logic™ is equivalent to a Boolean Algebra
·
Theorem T7) Logic inferences are transitive
·
Theorem T8) (the Boolean Multiple Form Logic equivalent forms of “AND”)…
·
Theorem T9) (A # 1,B) = (A , B)#B#1, in Propositional Logic: A > B
·
Theorem T10) Huntington's Axiom is a theorem in Multiple Form Logic
·
Theorem T11) A “generalised Huntington Formula” in Multiple Form Logic
(dozens more to come, in the near future…)
Propositional / Boolean Logic 
Multiple Form Logic^{™} 
Not (A) 
A # 1 
A or B 
A , B 
A and B 
(A # 1, B # 1) # 1 
A = B 
A # B # 1 
A a B 
A # 1, B 
A xor B 
A # B 
Theorem
T1: A , B # A = A,
B
Proof: Axiom 3 states that A, B # (A, C) = A, B # C.
Now, we can replace C
by void (absence of form or distinction).
Therefore A, B # A = A, B.
Theorem
T2: A , A = A
Proof: We have already proved (Theorem
T1) that: A, B #
A = A, B.
Replacing B by void (absence of form
or distinction): A, A = A.
Theorem T3: (in
the original paper of 1984):
1 , X = X (iff “1” is defined as “the
IMPORTANT NOTE 1 (and a “proof”):
This “ theorem” is in reality Axiom 1, in this
presentation of Multiple Forms. However,
in the older paper of 1984 (handed over
to professor Cliff Jones of
Constant “1” was defined “constructively” as the “
George Spencer Brown’s First “Algebraic Initial” (J1)
of the “Primary Algebra” in “Laws of Form” is:
This
corresponds precisely to the Multiple Form Logic theorem: ( p # 1, p ) 1
= (void)
Proof: (p # 1, p) # 1 = (1, p ) # 1(by Theorem T1,
applied to p#1,p)
= 1
# 1(by Axiom 1, or –if you prefer Theorem T3)
= void (by Axiom 2: void # X # X=void, applied to 1 # 1).
Theorem T5: George
Spencer Brown’s algebraic
axiom J2, proved as a theorem:
George Spencer Brown’s Second “Algebraic Initial” (J2)
of the “Primary Algebra” in “Laws of Form” is:
Proof: LHS = ( (p , r) # 1, (q , r) # 1) # 1 , 1 # 1 (adding “1 # 1”, which is void, by
Axiom 2)
= ( (p , r) # 1, (q , r) # 1) # 1
, ( 1 , 1 ) # 1 (changing “1” to “1 , 1”, by Axiom 2)
=( (p , r) # 1, (q , r) # 1) # 1
, ( r # 1 , 1) # 1 (changing “1” to “r # 1, 1”, by Axiom 3)
=( (p , r) # 1, (q , r) # 1) # 1 , ( r # 1 ,( (p , r) # 1, (q , r) # 1) # 1 # 1) # 1 (by Axiom
3)
=( (p , r) # 1, (q , r) # 1) # 1 , ( r # 1 ,( (p , r) # 1,
(q , r) # 1) ) # 1 (by Axiom 2)
=( (p , r) # 1, (q , r) # 1) # 1 , ( r # 1,( (p , 1 ) # 1,
( q , 1) ) # 1 (by Axiom 2)
=( (p , r) # 1, (q , r) # 1) # 1 , ( r # 1,( 1 # 1,1 # 1 ) ) #
1 (by Axiom 1)
= ( (p , r) # 1, (q , r) # 1) #
1 , ( r # 1 ) # 1 (by Axiom 2)
= ( (p , r) #
1, (q , r) # 1) # 1 ,r = RHS (Q.E.D.)
We now demonstrate a generalised version of this “distributive law”:
Theorem T5.1: George Spencer
Brown’s algebraic axiom J2 can be generalised:
If we replace “1” with “X” (an arbitrary form), we get
a Generalised Distributive Law in Multiple Form Logic™:
(((A ,
C) # X), ((B , C) # X)) # X = (A # X , B # X) # X , C
Now, is this
formula valid in Multiple Form Logic? Well, yes, it appears to
be valid iff we also assume Axiom 1. We
shall prove it in two different ways. In the first method, we shall use
the Boolean Substitution Rule:
Proof(1): Examine the
two complementary cases of C = 1, and C = [no form] (or “void”).
Case 0: If
C=0 (or void), (A # X, B #
X) # X=((A # X), (B # X)) # X (true).
Case 1:
If C=1, (A # X, B # X) #
X, 1=(((A, 1) # X), ((B, 1) # X)) # X
Now the lefthand side of this formula needs Axiom 1, to be
reduced to “1”, i.e.
Z , 1 = 1. In
this case, iff we assume Axiom 1, the righthand
side of the expression is:
(((A, 1) # X), ((B, 1) # X)) # X = ((1 # X), (1 # X)) # X (by Axiom 1, for A,1 and B,1)
= (1 # X)) # X (by Theorem T2, applied to (1#X),(1#X) = (1#X))
= 1 # X # X = 1 (by Axiom 2 , which says that: A#X#X=A).
In
the 2nd method we show 1)Left hand side a Right hand side, 2)Right hand side a Left hand
side:
Proof (2a): LHS => RHS, using the transliteration “a
=> b” to “a # 1 , b” (again, assuming Axiom 1):
(( ( ( A , C) # X) , ( ( B , C ) # X ) # X )
# 1 , ( A # X , B # X ) # X , C
=(( ( A # X ) , ( B # X ) # X ) # 1 , ( A
# X , B # X ) # X , C (by Axiom 3)
=1 (“True”) (by Axiom 3, applied to “( A # X , B # X ) #
X ”).
Proof (2b): RHS => LHS, using the transliteration “a
<= b” to “a , b # 1”:
( ( A , C) # X ) , ( ( B , C ) # X ) # X , ( ( A # X ,
B # X ) # X ,C ) # 1
= ( ( A , C) # X) , ( ( B , C ) #
X ) # X , ( ( ( A , C )# X , ( B , C ) # X ) # X ,C ) # 1
= ( ( A , C, C # 1) # X ) , ( ( B
, C , C # 1 ) # X ) # X , C # 1 (by Axiom 3, inserting “ C # 1”)
= ( ( A , 1) # X ) , ( ( B
, 1 ) # X ) # X , C # 1 (by Axiom 3 and Axiom
1, for “C , C # 1”)
= ( 1 # X , 1 # X ) # X , C #
1 = ( 1 # X ) # X , C #
1 = 1 , C # 1
= 1
(“True”) (Q.E.D.)
On reflection, after seeing this, I speculate that perhaps the existence of “1”
forces the distributive law to become valid, i.e. If
we assume there is such a thing as “the All” (or God, or Allah,
whatever you like to call “it”) then we enter the World of Classical Logic,
where the
distributive law holds. Whereas, if we avoid “the All”,
ignoring it, or pretending that it does not exist, etc., then we get a
kind of “Quantum Logic”,
where the distributive law does not hold, and where
childhood and… schizophrenia, Art (etc.), all become possible! Now,
I do not wish to take advantage of Eddie Oshins’s precious time, who
replies to so many students’ queries, but I feel strongly that
it’s worth his
time to deal with this. If this kind of speculation is correct,
then we can combine Classical Logic and Quantum Logic in one system.
If not, I still have work to do, as a programmer! ;)
Theorem T6: Multiple
Form Logic^{™} is equivalent to a Boolean Algebra
NOTE: This is provable only if we assume
all three axioms to be true, having
defined “1” as equal to “all Forms in the Universe”. If on
the other hand if we do not assume axiom
1, then it remains to be seen what actually happens; In the
latter case, the resulting system is not equivalent to a Boolean
Algebra, but perhaps it is equivalent to a “Quantum Logic” of some kind (Dr. Oshins?)
For an unspecified set B of
at least two elements, a binary operation in B, and a unary
operation ’ in B, the following axioms define a Boolean Algebra:

B1 and B2:
True “by definition” (see the “Primordial Theorem
1”).
B3: (Proof) Let A , B # 1=1 = C , C # 1.
Then: A , B = A,
B # 1 # 1 = A, (B # 1, A) # 1 = A , 1 # 1 = A.(QED)
B4: (Proof) Let A , B = A. Then A , B # 1 = A , B , B # 1 = A # 1 = 1.(QED)
The proof is
complete.
Logic inferences are transitive, i.e. if A => B and B => C,
then A => C
Proof: (translating PC into MF) ( (A # 1 , B) # 1 , (B # 1 , C) # 1) # 1 ) #
1) , A # 1 , C =
= ( (A # 1 , B) # 1 , (B #
1 , C) # 1, A # 1, C (by axiom 2, applied to 1#1)
= ( B # 1 , ((B # 1) , C) # 1, A # 1, C (by axiom 3, applying A#1 to A#1,B)
= (B#
1, C # 1, A # 1 , C (by axiom 3, applying B#1 to B#1,C)
= (B# 1,1, A # 1 , C (by axiom 3, applying C to C#1)
= 1 (by axiom 1,
applied to “… , 1 , …”)
I.e. it
reduces to “True”, or “All
possible forms”.
NOTE(1): If you download and run the Prolog
theoremprover “mflogic.exe”, you will see similar proofs. This theorem
is among many examples stored in a knowledge base. All the examples are
shown in a menu, from which you can pick one, to see its proof. However, the
proofs are not stored; Only theorems are stored. (I.e. what you see each
time is a “fresh proof”). You can also type your own theorems
(in another program option) to see other proofs or bugs. ;)
NOTE(2): If you look closely into the above proof, and
into the one automatically generated by the program
(mflogic.exe), you may discover some differences in proof steps
followed. In fact, the program gives a slightly lengthier proof, since
the present algorithm used searches “blindly” to apply the three axioms wherever possible, without using much
intelligence. (Still, it does find the solution, even “blindly”).
NOTE(3): I have sought a generalised version of this
“transitive implication law” which holds in Multiple Form Logic™ and
probably in some “Quantum Logic” as well. The generalised form of this,
has an arbitrary form “X” instead of “1”, and becomes (by successive
applications of Axiom 3):
( A # X , B) # X , (B # X
, C) # X , A # X , C = B # X , (B # X , C) # X , A # X , C
= B # X , C # X , A # X
, C = B # X ,X , A # X , C = B
, X , A , C =
= X , A , B, C (which is the
union of all the logic variables, used in the “law”).
Theorem T8: (the Boolean Multiple Form Logic™equivalent forms of “AND”):
( A # 1, B #
1 ) # 1 = ( A , B ) # A # B= (in Boolean Algebra:A&B )
Proof: In this proof, we shall use the following Boolean Substitution Rule:
If a logic
equation reduces to identical lefthand and righthand sides for each case of
substituting (1) a particular variable by “1”, and (2) the same variable by
“0”(void), then the equation is valid. (A proof
of this can be found in most Boolean Algebra textbooks). If we use the three axioms,
then the Multiple Form Logic system is
equivalent to a Boolean Algebra, as
shown by theorem T6. So this rule is also
valid in (a Boolean) Multiple Form Logic, and can be used in derivations:
LHS = (1 # 1, B # 1 )
# 1 = B # 1 # 1 (by
axiom 2 applied to 1#1)
= B (by axiom 2, applied to 1#1).
RHS = (1 , B ) #
1 # B = 1 # 1 # B (by axiom 1
applied to 1,B)
= B (by axiom 2 applied to 1#1
Therefore: for the case A =
1, both LHS and RHS
are the same.
LHS = ( 1 ,B #
1 ) # 1 = 1 # 1 (by axiom
1, applied
to 1 , B#1)
= void (by axiom 2, applied to 1#1).
RHS = ( B )
# B = (void) (by axiom
2, applied
to B#B)
= void
(by axiom 2, applied to 1#1).
Therefore:
for the case A = void, LHS and RHS are again the same, so the proof is
complete.
The importance of
this theorem, philosophically
as well as derivationwise, is that we can express logical conjunctions (AND) without using “1” at all.
Of course we have already used Axiom 1, in the above proof, and the equation proved is
still Boolean. However, we slowly begin to realize that Multiple Form Logic
is a superior system to Boolean Algebra, for more than one reason;
and one (small) reason is this.
( A # 1, B) = ( A , B ) # B # 1
Proof: The Boolean “Substitution Rule” can be used, together
with the Axioms:
Case of A = 1: LHS = (1 # 1, B) = B, RHS = ( 1 , B ) # B # 1 = 1 # B # 1 = B (= LHS).
Case of A = 0: LHS = (1, B) = 1, RHS = ( B ) # B # 1 = 1 (= LHS). (Q.E.D.)
Theorem T10: Huntington's Axiom is a theorem in Multiple Form Logic™
Huntington's Axiom, in George Spencer
Brown's notation, is:
( ( A B ) ( A ( B ) ) ) = A
It translates to the Multiple Form Logic formula:
( (A , B) # 1, (A , B # 1) # 1 ) # 1 = A
Proof: LHS = ( ( A, B ) #
1, ( A , B # 1 ) # 1 ) # 1
= ( ( A, B ) # 1, ( A , B # ( A , B ) # 1 # 1 ) # 1 ) # 1 (by Axiom 3, insertion)
= ( ( A, B ) # 1, ( A , B # ( A , B ) # 1 # 1 ) # 1 )
# 1
(ready for Axiom 2...)
= (( A, B ) # 1, ( A , B # ( A, B ) ) # 1 ) # 1
(by Axiom 2 applied to "1 # 1")
= (( A, B ) # 1, ( A, B # B ) # 1 ) # 1 (by Axiom
3, cancellation)
= (( A, B ) # 1, A# 1 ) # 1 (by Axiom 2, on "B # B")
= ( ( A, A # 1 , B ) # 1, A # 1 ) # 1 (by Axiom
3, insertion)
= ( ( A, 1 , B ) # 1, A # 1 ) # 1 (by Axiom 3, cancellation of
"A", in "A # 1")
= ( 1 # 1, A # 1 ) # 1 (by Axiom 1, since "1" is in
"A
# 1 # C")
= A # 1 # 1 (by Axiom 2 on "1 # 1")
Like the "Generalised Distributive Law" (Theorem T5.1), there is a "generalised version” of
Huntington's formula, where every occurrence of "1" has been changed
to an arbitrary Form X. However, the Right hand side of this
formula is slightly different now; It is not "A", but "A or (B or X)
xor X".
As far as I know, this theorem is a kind of small novelty in
Logic (like Theorem T5.1). Here it is:
( (A , B) # X, (A , B # X) # X ) # X = A, (B, X) # X
Proof (by the use of the "Boolean Substitution Rule"):
LHS = ( ( ( A, B ) #
1, ( A , B # 1 ) # 1 ) # 1 ) # 1 = A (by Theorem T10, Huntington)
RHS = A, ( B , 1 ) # 1 = A , 1 # 1 (by Axiom 1 applied to "B,1" = "1")
= A
= LHS (by Axiom
2 applied to "1 # 1")
2) Case of X = void: LHS =( (
( A, B ), ( A , B ) ) )= A , B (Omitting "X", and using Theorem T2, or Axiom 3)
RHS =A, ( B
) (Omitting "X") = LHS
The Three Axioms (for quick hyperlink reference):
(1)
Oneness 1 , X = 1
(2) Reflection A # X # X = A
(3) Perception A , X # ( A , B)=A , X # B
Next
Section: A bitcrunching Expert System Deduction
Algorithm
Another relevant Section: More Theorems of Multiple Form Logic