Type III TAG Results: Online Resources
Analogous to Type II TAG is the fact that CR3 is nested into the denominator of CR2 and CR2 is
nested into the denominator of CR1 for Type III TAG, producing a three-level deep nested construct.
Specifically, the sum of the numerator and denominator of CR3 make up the denominator of CR2, and the
sum of the numerator and denominator of CR2 make up the [DAG]+ that is in the denominator of CR1.
Furthermore, the sum of the numerator and the denominator of CR1 makes up the sum of all ions.
The nested equations can readily be observed by examining the Case 1.1.1 and Case 1.1.2 solutions.
When [AB]+ is 1 (=100%), indicated by Case 1.1.1, the [BC]+/[AB]+ ratio, CR3, becomes simply [BC]+/1,
and the sum of [AB]++[BC]+ is simply the Unit Simulacrum solution 1+([BC]+/[AB]+), or 1+CR3. And of
course, [AB]++[BC]+ is the denominator of CR2, the [AC]+/([AB]++[BC]+) ratio. So, when
[AC]+/([AB]++[BC]+) is multiplied by 1+([BC]+/[AB]+)= [AB]++[BC]+, the denominator of CR2 cancels
out giving simply [AC]+, as seen in Scheme IIIA. In this way, CR3 is nested into the denominator of CR2,
so that [AC]+=CR2x(1+CR3).
For Case 1.1.2, the situation is similar, except now [BC]+ =1, and the sum of [AB]++[BC]+ is the Unit
Simulacrum inverse solution 1+1/([BC]+/[AB]+), or 1+1/CR3. When this solution for [AB]++[BC]+ is
multiplied times CR2, [AC]+/([AB]++[BC]+), the denominator again cancels out, giving
[AC]+=CR2x(1+1/CR3). Thus, the sum of [AB]++[BC]+ is either 1+ratio or 1+1/ratio, and the Unit
Simulacrum sum of [AB]++[BC]+ is 1+CR3 or 1+1/CR3, depending on Case 1.1.1 or Case 1.1.2. Either
way, [AC]+ is arrived at by multiplying CR2, the [AC]+/([AB]++[BC]+) ratio, times the sum in the
denominator, in the form of the Unit Simulacrum sum solution, to cancel out the denominator and give
[AC]+. Thus, it can be seen that the Unit Simulacrum sum of CR3 (1+CR3 or 1+1/CR3) is the sum that is
nested into the denominator of CR2.
Similarly, the Unit Simulacrum sum of CR2 is nested into the denominator of CR1. It is easiest and
most direct to observe this relationship in the Case 1.2.n examples. In both Case 1.2.1 and Case 1.2.2,
[AC]+=1. In those Cases, the [DAG]+ is simply 1+1/CR2, or 1+1/([AC]+/([AB]++[BC]+)), which
becomes 1+1/(1/([AB]++[BC]+)), which simplifies to 1+([AB]++[BC]+). When [AC]+=1, the sum of all
[DAG]+ fragments, [AC]++[AB]++[BC]+ is 1+([AB]++[BC]+). Then, the value of [MH]+ is obtained by
multiplying CR1x[DAG]+, in the form of [MH]+/[DAG]+x(1+([AB]++[BC]+), where (1+([AB]++[BC]+)
is the Unit Simulacrum inverse solution 1+1/([AC]+/([AB]++[BC]+)), or 1+1/CR2. To summarize Case
1.2.n, [MH]+=CR1x(1+1/CR2), where 1+1/CR2 is equal to the sum in the denominator of CR1.
When [AC]+ is not equal to 1, for instance Case 1.1.1 and 1.1.2, the sum of all [DAG]+ fragments is
not simply 1+1/CR2, but is instead 1+1/CR2 times the Unit Simulacrum solution (1+ratio or 1+1/ratio) of
CR3, as follows: (1+CR2)x((1+CR3) or (1+1/CR3)). Specifically, for Case 1.1.1, the [MH]+ is found by
multiplying [MH]+/[DAG]+ times [DAG]+, where [DAG]+ is given by
(1+([AC]+/([AB]++[BC]+)))x(1+([BC]+/[AB]+)). As discussed above, for Case 1.1.1 [AB]+=1, and the
denominator of CR2 is given as the Unit Simulacrum solution 1+CR3. Therefore, the denominator
([AB]++[BC]+) cancels out when multiplied by (1+([BC]+/1), since ([AB]++[BC]+)= (1+([BC]+/1), leaving
1x(1+([BC]+/1))+[AC]+, which equals [AB]++[BC]++[AC]+, or [DAG]+. To summarize Case 1.1.1,
[MH]+ is given by CR1x(1+CR2)x(1+CR3). Of course, when [BC]+=1 instead of [AB]+, the solution
contains the inverse of CR3, or [MH]+=CR1x(1+CR2)x(1+1/CR3). In both these Cases, the denominator
of CR2 is cancelled out by the Unit Simulacrum solution of CR3, either 1+CR3 or 1+1/CR3, and the
denominator of CR1 is cancelled out by the Unit Simulacrum solution for CR2 times the Unit
Simulacrum solution for CR3, [DAG]+=(1+CR2)x((1+CR3) or (1+1/CR3)).
In the solutions discussed above, the answers depended on which fragment was “carrying the unit”, or
which fragment was =1 (100%). When [AC]+ “carried the unit”, Case 1.2.n, the [DAG]+ was found
using CR2 by itself, and the [MH]+ was found by multiplying [MH]+/[DAG]+ x [DAG]+, or
CR1x(1+1/CR2). When [AB]+ or [BC]+ carried the unit, as in Case 1.1.1 and 1.1.2, the [DAG]+ was
found by multiplying (1+CR2)x((1+CR3) or (1+1/CR3)). In all Case 2.n.n solutions, the [MH]+ carries
the unit and has a value of 1.
 1  B 
Simulacrum Sum (1,B) = SimSum(1,B) =
2-0-1- B
Possibilities to Observe:
1
B
1
 
 B
 B
 
1
1
 B
Case 1 :1  B, B  1,    1 ,    1 ;
 B
1




1 

1 1

 1 
 

B


 B 
1  1    
 1 

or

 1 
B  1    
 B 

or




1 

B 1

 B
 

 1 

1
 B
Case 2 :1  B, B  1,    1 ,    1 ;
 B
1




1 

1 1

 1 
 

B


 B 
1  1    
 1 

or

 1 
B  1    
 B 

or




1 

B 1

 B
 

 1 

Online Resource 1 The Unit Simulacrum (US), or First Specified Form of a Simulacrum (FSFS) where
one value is specified to be 1, and the other value, B, can represent any value. The default Case 1 has 1 
B because this simulacrum came from the FGFS (Fig. 1) having “A” set to 1. A simulacrum is composed
of the Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions
 1  1 
Simulacrum Sum (1,1) = SimSum(1,1) =
2-0-1-1
Possibilities to Observe:
1
1
1
 
1
1
 
1
1
1
Case 1 :1  1, 1  1,    1 ,    1 ;
1
1




1 

1 1

1 
 

1 


1
1  1    
1

or

1
1  1    
1

or




1 

1 1

1 
 

1 

1
1
Case 2 :1  1, 1  1,    1 ,    1 ;
1
1




1 

1 1

1 
 

1 


1
1  1    
1

or

1
1  1    
1

or




1 

1 1

1 
 

1 

Online Resource 2 The Unit Unit Simulacrum (UUS), where both values are specified to be 1 and the
default Case 1 has 1  1. A simulacrum is composed of the Simulacrum Sum, Possibilities to Observe,
two Cases, and eight solutions
 0  1 
Simulacrum Sum (0,1) = SimSum(0,1) =
2-0-0-1
Possibilities to Observe:
0
1
0
 
1
1
 
0
0
1
Case 1 : 0  1, 1  0,    1 ,    1 ;
1
0




1 

0 1

0
 

1


 1 
0  1    
 0 

or

 0 
1  1    
 1 

or




1 

1 1

1
 

0

0
1
Case 2 : 0  1, 1  0,    1 ,    1 ;
1
0




1 

0 1

0
 

1


 1 
0  1    
 0 

or

 0 
1  1    
 1 

or




1 

1 1

1
 

0

Online Resource 3 The Unit Identity Simulacrum (UIS), where one value is specified to be 1 and the
other value is specified to be 0. The default Case 1 has 0  1. A simulacrum is composed of the
Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions
Simulacrum Sum (A,B) = SimSum(A,B) =
 A  B 
2-0- A - B
Possibilities to Observe:
A
B
A
 
B
B
 
A
A
B
Case 1 : A  B, B  A,    1 ,    1 ;
B
A




1 

A 1

A
 

 B


 B 
A  1    
 A 

or

 A 
B  1    
 B 

or




1 

B 1

 B
 

A

A
B
Case 2 : A  B, B  A,    1 ,    1 ;
B
A




1 

A 1

A
 

 B


 B 
A  1    
 A 

or

 A 
B  1    
 B 

or




1 

B 1

 B
 

A

Online Resource 4 The General Form of the Anti Simulacrum (GFAS), where A and B can represent any
values, and the default Case 1 has A  B. A simulacrum is composed of the Simulacrum Sum,
Possibilities to Observe, two Cases, and eight solutions
Simulacrum Sum (1, ) = SimSum(1,) =
 1   
2-0-1-
Possibilities to Observe:
1

1
 


 
1

1
Case 1 :   1, 1  ,    1 ,    1 ;
1





1 

 1


 

 1 


 1 
  1    
  

or

  
1  1    
 1 

or




1 

1 1

 1 
 




1
Case 2 :   1, 1  ,    1 ,    1 ;
1





1 

 1


 

 1 


 1 
  1    
  

or

  
1  1    
 1 

or




1 

1 1

 1 
 



Online Resource 5 The Infinity Unit Simulacrum (IUS), where one value is specified to be 1 and the
other value is specified to be . This is Case 2 by default, since   . A simulacrum is composed of the
Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions
(a)
(b)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Online Resource 6 a The infinitely nested First Decrement or First Deconstruction, which is a First
Increment in the denominator, 1/(1+1). This produces a continued fraction. b 1+ the infinitely nested First
Decrement, or the First Increment of the infinitely nested First Decrement, 1+1/(1+1…), which produces
another continued fraction. The first gives lesser Fibonacci ratios (<1), the second gives greater Fibonacci
ratios (>1). The lesser Fibonacci ratios converge to 1/, while the greater Fibonacci ratios converge to .
  1
1

 1
1
1
1
 1
1
1

1
1
1
  1
1

1
1
1
1
1
1
1
1
1
1
 1

1
  SimSum 1,    1  
 
 
1
1
1
1
1
1
1
Online Resource 7 The First Increment of (1/), which gives the Phi Unit Simulacrum, and the once,
twice, and infinitely nested Phi continued fractions. Phi is the Golden Ratio, or Golden Mean,  =
1.618033989… (=(Sqrt(5)+1)/2)
1
