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 subfunctions 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.
