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IQ-Power
2007
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Compound Nesting (Advanced Theory)

Introduction

Compound nesting completes the application of Set Theory to logical systems as it allows for the inclusion and assessment of overlapping functions. With compound nesting, there is more than one class or type of bracket, and thus there can and often is, overlap of nested functions. It is also presents the most difficult logic problems when the overlap of bracketed sub-functions is both deep and compound.

 

Simple Overlap

Simple overlap occurs when only two types or classes of bracket overlap. For the overall expression to be TRUE, the value of the contents of the overlap, and the value of the contents of both bracketed sub functions must be identical. For the overall expression to be FALSE, the value of all sub bracketed functions must be equal to each other but unequal to the value of the overlap. For the overall expression to be UNKNOWN, the value of the overlap is equal to at least the value of one of the bracketed sub functions but unequal to the value of at least one of the others. For example:

[0 AND 2 OR (1 AND 1] MAYBE 0):
[0 AND 2 OR 1 AND 1] = 1
(1 AND 1 MAYBE 0) = 0
(1 AND 1] = 1

So:
[0 AND 2 OR 1 AND 1] = (1 AND 1] =/= (1 AND 1 MAYBE 0)
Thus:
[0 AND 2 OR (1 AND 1] MAYBE 0) = 0 = UNKNOWN:

(1 OR 1 AND [0 MAYBE 1 AND 2) 0R 2]:
(1 OR 1 AND 0 MAYBE 1 AND 2) = 2
[0 MAYBE 1 AND 2 0R 2] = 2
[0 MAYBE 1 AND 2) = 2

So:
(1 OR 1 AND 0 MAYBE 1 AND 2) = [0 MAYBE 1 AND 2 0R 2] = [0 MAYBE 1 AND 2)
Thus:
(1 OR 1 AND [0 MAYBE 1 AND 2) 0R 2] = 1 = TRUE

(1 OR 2 AND [0 MAYBE 1 OR 2) AND 2]:
(1 OR 2 AND 0 MAYBE 1 OR 2) = 2
[0 MAYBE 1 OR 2 AND 2] = 2
[0 MAYBE 1 OR 2) = 0

So:
(1 OR 2 AND 0 MAYBE 1 OR 2) = [0 MAYBE 1 OR 2 AND 2] =/= [0 MAYBE 1 OR 2)
Thus:
(1 OR 2 AND [0 MAYBE 1 OR 2) AND 2] = 2 = FALSE

This may look messy or unconventional, but it is the most effective way to model arguments made at the risk of "mixing apples with computers". There is another contingency where the respective values of the overlapping bracketed sub functions are not all equal and none are equal to the value of their overlap:

(1 OR 2 OR [0 AND 1 MAYBE 1) AND 2]:
(1 OR 2 OR 0 AND 1 MAYBE 1) = 1
[0 AND 1 MAYBE 1 AND 2] = 2
[0 AND 1 MAYBE 1) = 0

So:
(1 OR 2 OR 0 AND 1 MAYBE 1) =/= [0 AND 1 MAYBE 1 AND 2] =/= [0 AND 1 MAYBE 1)
Thus:
(1 OR 2 OR [0 AND 1 MAYBE 1) AND 2] = ?...

...The value of the overall function in this case is best described by the fourth element found in quaternary logic and will be discussed in more depth in future.

 

Parallel Overlap

When nesting is deeper than one level, overlap between different nests can sometimes span a number of levels. The following expression is an example of this problem:
(1 OR (2 AND 0 OR [0 MAYBE 1) OR 2 MAYBE 1) AND 2]

In this case, a single overlap is divided between two separate levels:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR 2 MAYBE 1) AND 2]

As you can see, we define this overlap as the widest single overlap that exists between both bracket types or classes simultaneously. In such a case, the overlap itself is treated as a comparison of parts. The nature of that comparison is defined by the leftmost outermost operator (in this case OR) - and known as the overlap controller:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR 2 MAYBE 1) AND 2]

For this definition to hold, the inner or deeper portion of the overlap is deduced:
(1 OR ( 2 AND 0 OR [ 0 MAYBE 1 ) OR 2 MAYBE 1) AND 2] Leading one to solve for:
0 MAYBE 1 = 1

Then the outer or higher portion of the overlap is deduced while reserving the use of the overlap controller such that:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR 2 MAYBE 1 ) AND 2] Leading one to solve for:
2 MAYBE 1 = 2

Now we use the overlap controller to derive the value of overlap from the parts calculated above:
1 OR 2 = 1    IE: (0 MAYBE 1) OR (2 MAYBE 1) = 1

Then the calculation continues as normal:
(1 OR ( 2 AND 0 OR 0 MAYBE 1) OR 2 MAYBE 1) = 1
[0 MAYBE 1 OR 2 MAYBE 1 AND 2] = 2
(0 MAYBE 1) OR (2 MAYBE 1) = 1   
{PARALLEL OVERLAP}
So:
(1 OR ( 2 AND 0 OR 0 MAYBE 1) OR 2 MAYBE 1) = (0 MAYBE 1) OR (2 MAYBE 1) =/= [0 MAYBE 1 OR 2 MAYBE 1 AND 2]
Thus:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR 2 MAYBE 1) AND 2] = 0 = UNKNOWN

 

Deep Overlap

The solution of problems involving deep overlap begins with the solution of deepest overlap. Take this problem for example:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] MAYBE 0]

Note the confusing array of overlaps and how the problem as a whole (or as a whole part of a larger problem) ends with a different type or class of bracket from the one it starts with.  When in search of the deepest overlap, there are traps for the inexperienced. Take the next highlights for example:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] MAYBE 0]:

The overlap functions highlighted above are only portions of the total overlap, and combined they do not represent the overlap as a whole. The trick is to draw a Venn Diagram corresponding to the nature of diverse brackets in the problem you are to solve. It is also easier if you start by identifying the controlling overlap which is the largest whole overlap between the highest levels of bracketing. The controlling overlap in this expression is highlighted as follows:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] MAYBE 0]

The deepest sub bracket regions are highlighted respectively as follows:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] MAYBE 0]
(1 OR ( 2 AND 0 OR
[0 MAYBE 1) OR [2) AND 2] MAYBE 0]

The sub overlap where these two sub bracketed regions overlap happens to be the deepest overlap in this problem and it is highlighted as follows:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] MAYBE 0]

The beginning of the overlap calculation starts here at the overlap of the deepest pair of sub bracketed regions. That this is the deepest overlap is confirmed by the fact that both classes of bracket are open and thus ignored in the following calculation. This deep overlap is used in he calculation of the overlap on the next level up as follows:
0 MAYBE 1 OR 2 = 0

Now, taking each class of sub bracketed region of the next level up in the overlap:
(2 AND 0 OR 0 MAYBE 1) OR 2 = 0
0 MAYBE 1 OR [2 AND 2] = 0

So:
( 2 AND 0 OR 0 MAYBE 1) OR 2 = 0 MAYBE 1 OR [2 AND 2] = 0 MAYBE 1 OR 2
Thus:
( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] = 1

Now
1 OR ( 2 AND 0 OR 0 MAYBE 1) OR 2 = 0
0 MAYBE 1 OR [2 AND 2] = 0

So:
1 OR ( 2 AND 0 OR 0 MAYBE 1) OR 2 = 0 MAYBE 1 OR [2 AND 2] =/= ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2]
Thus:
(1 OR ( 2 AND 0 OR [0 MAYBE 1) OR [2) AND 2] MAYBE 0] = 2 = FALSE

Deep overlap has the potential to be substantially more difficult to solve than parallel overlap. However, when combined; both deep and parallel highlight promise to cause even the brightest minds to pause, if only for a moment!

 

Multiple Overlap

Multiple overlap occurs when more than two bracket classes overlap. The following is a simple example of this type of problem:
(2 AND [1 OR {0 MAYBE 2) MAYBE 1 AND 0] OR 1 AND 0}

In this case, things are not so clear. However, if we look for the common ground that is shared by all three, this takes on the role of overlap to the exclusion of all else. Let us consider the contents of each unique bracket region:

(2 AND 1 OR 0 MAYBE 2) = 0
[1 OR 0 MAYBE 2 MAYBE 1 AND 0] = 2
{0 MAYBE 2 MAYBE 1 AND 0 OR 1 AND 0} = 0

Now the common ground held by all three bracket regions is as follows:
(2 AND [1 OR {0 MAYBE 2) MAYBE 1 AND 0] OR 1 AND 0} Leading one to solve for:
0 MAYBE 2 = 0
So:
0 MAYBE 2 = (2 AND 1 OR 0 MAYBE 2) = {0 MAYBE 2 MAYBE 1 AND 0 OR 1 AND 0} =/= [1 OR 0 MAYBE 2 MAYBE 1 AND 0]
Thus:
(2 AND [1 OR {0 MAYBE 2) MAYBE 1 AND 0] OR 1 AND 0} = 0 = UNKNOWN

 

Apparent Depth

(2 AND [1 AND {0 MAYBE 2) MAYBE 1} AND 0]

In this case, we seem to have a deep overlap structure. However, no single bracket class exists inside another. This means that the depth we could interpret would not be true depth, even if in this case, it leads us to the area of mutual overlap. This illusion can be misleading when attempting to prioritise overlap calculations in deeply nested expressions. Therefore, if we are to solve:
(2 AND [1 AND {0 MAYBE 2) MAYBE (1} OR [0) OR 1] AND 0]

The contents of the highlighted brackets are the deepest part of this expression, and must be solved first along with any associated overlap:
(2 AND [1 AND {0 MAYBE 2) MAYBE (1} OR [0) OR 1] AND 0]

There are three bracket classes attached to this part of the expression by overlap but there is no common overlap shared by all bracket classes as well as this this part of the expression. Therefore, we deal with the overlap with this part of the expression firstly and separately from any other part or overlap in the expression. The relevant controlling overlap is highlighted as follows:
(2 AND [1 AND {0 MAYBE 2) MAYBE (1} OR [0) OR 1] AND 0]

Notice how this controlling overlap is itself overlapping with another bracket class. The solution of this controlling overlap will be required for the solution of the next part of this expression, as we work our way along the overlap chain:
(1 OR 0) = 1
[0 OR 1] = 1
0 = 0

So:
(1 OR 0) = [0 OR] 1 =/= 0
Thus:
(1 OR [0) OR 1] = 2

Given that there is another function connected by overlap with this overlap, the next relevant controlling overlap to solve is as highlighted:
(2 AND [1 AND {0 MAYBE 2) MAYBE (1} OR [0) OR 1] AND 0]

Now
(1 OR [0) OR 1] = 2
{0 MAYBE 2 MAYBE 1} = 0
1 = 1

So:
1 =/= {0 MAYBE 2 MAYBE 1} =/= (1 OR [0) OR 1]
Thus:
{0 MAYBE 2) MAYBE (1} OR [0) OR 1] = 3

Here, things take an unexpected turn as one of the bracket classes for the next overlap is:
{0 MAYBE 2) MAYBE (1} OR [0) OR 1]

Instead of:
{0 MAYBE 2 MAYBE 1}

So the final leg in the solution of this problem is:
(2 AND 1 AND 0 MAYBE 2) = 2
[1 AND 0 MAYBE 2 MAYBE 1 AND 0] = 2
{0 MAYBE 2) MAYBE (1} OR [0) OR 1] = 3

So:
(2 AND 1 AND 0 MAYBE 2) = [1 AND 0 MAYBE 2 MAYBE 1 AND 0] =/= {0 MAYBE 2) MAYBE (1} OR [0) OR 1]
Thus:
(2 AND [1 AND {0 MAYBE 2) MAYBE (1} OR [0) OR 1] AND 0] = 2 = FALSE

This is about as complicated as logic gets without considering the multitude of functions, and incredible depths to which nesting can be taken. Even a fairly short problem featuring a combination of multiple and parallel overlaps at depth will probably challenge even the brightest minds...    ...and if not, this class of problem gets exponentially more difficult as the expression gets larger.

As you can see, compound nesting can become incredibly complex very quickly, and generally increases complexity too quickly in progressively compounding problems, to be practical for inclusion in any intelligence testing regime. The fact that this is a vital part of Quaternary logic is why IQ Power does not ship with Quaternary Logic tests.

 

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