Hyper-linked Index of Proofs:
· The Three Axioms of Multiple Form Logic™ (for quick reference)
· More Theorems (in another web page)
· A bit-crunching theorem prover for Expert Systems (in another web page)
(dozens more to come, in the near future…)
Propositional / Boolean Logic
Multiple Form Logic™
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.
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 “
NOTE 2: From an A.I. programmer’s point of view, it is not so important if we see this as a theorem (T3), or an axiom (axiom 1) if our aim is to prove theorems automatically. However, philosophically speaking, I prefer the older (1984) formulation, since it assumes less and creates more. Furthermore, recently (after writing the first part of this presentation) I browsed Eddie Oshins’s site about Quantum Psychology, and speculate whether axiom 1 should (again) be discarded “as an axiom” and kept only as a theorem (T3). In this case, perhaps we can use the other two axioms, creating a “Quantum Logic” of some kind, compatible with Quantum Psychology, e.g. perhaps through some additional axioms or constructions. (I am curious for Eddie Oshins’s comments about all this (in the LoF forum), and will shortly provide a hyperlink to such comments (here), if he does bother to comment).
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)
George Spencer Brown’s Second “Algebraic Initial” (J2) of the “Primary Algebra” in “Laws of Form” is:
This corresponds precisely to the Multiple Form Logic theorem:
( ( p , r ) # 1 , ( q , r ) # 1 ) # 1 = ( p # 1, q # 1 ) # 1, r
Proof: LHS = ( (p , r) # 1, (q , r) # 1) # 1 , 1 # 1 (adding “1 # 1”, which is void, by Axiom 2)
If we replace “1” with “X” (an arbitrary form), we get
a Generalised Distributive Law in Multiple Form Logic™:
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: “If a logic equation reduces to identical left-hand and right-hand sides, when substituting (1) a variable by “1” and (2) the same variable by “0”, then the equation is valid”. (A proof of this rule can be found in most Boolean Algebra textbooks). If we use the three axioms, the Multiple Form Logic system is “equivalent to a Boolean Algebra”, as proved in theorem T6. So, this rule is also valid in (the Boolean form of) Multiple Form Logic™. So it can be used in derivations, too:
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
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)
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
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! ;)
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. is a commutative operation.
If we choose the operation <> to be <,> (“OR”), and the operation <’> to be the result of a “XOR” with the Universal Distinction “1” (X’ = X # 1), then:
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.
Proof: (translating PC into MF) ( (A # 1 , B) # 1 , (B # 1 , C) # 1) # 1 ) # 1) , A # 1 , C =
NOTE(1): If you download and run the Prolog theorem-prover “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):
If a logic equation reduces to identical left-hand and right-hand 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)
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)
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 derivation-wise, 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.
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.)
Huntington's Axiom, in George Spencer Brown's notation, is: ( ( A B ) ( A ( B ) ) ) = A
It translates to the Multiple Form Logic formula:
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
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")
The Three Axioms (for quick hyperlink reference):
Next Section: A bit-crunching Expert System Deduction Algorithm
Another relevant Section: More Theorems of Multiple Form Logic