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#Post#: 627--------------------------------------------------
Question about Boolean Algebra
DIR By: TromFan
Date: February 14, 2020, 12:24 pm
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I am now reading Insanity Point for about the third or fourth
time now. Each time I read it, the Boolean Algebra makes a
little more sense to me. Between my TROM studies and doing the
actual processing, I have developed an intuitive understanding
of it and what the phenomena is, and could probably explain it
pretty well in plain English, but still, at times it's a
headache for me trying to figure out what the actual symbols are
intended to mean.
I think if I can understand this ONE THING I believe either it
will all make sense or at least I will know what question to
answer next.
Here's the passage, followed by my question:
"Now this postulate “X(1-X)=1” has some very interesting
deductions, very interesting deductions. I'll give them to you.
I won't prove these deductions but they can be, I can assure
you, every one I'm giving to you can be proven very easily in
Boolean algebra.
X(1-X)=1, X is and is not simultaneously
Here we go. We can deduce from “X(1-X)=1” that “X +(1-X)=0.”
X +(1-X)=0, neither X exists nor not X exists.
In other words it's a state of affairs where neither X exists
nor not X exists. Get it?
X + (1-X)=1, either X exists or not X exist or both exist
“X +(1-X)=0” now that's a state of unreason because reason
maintains that “X +(1-X)=1” that's what reason maintains.
But unreason, insanity, the IP, says that “X + (1-X)=0”
X +(1-X)=0, neither X exists nor not X exists.
Now this is a particularly interesting deduction from our point
of view because it tells us that while the person is in the IP
state the reasonable part of the postulate set is reduced to
zero. "
So, what is the difference between X(1-X) and X + (1-X)? More
specifically, what is X(1-X) supposed to mean? I took regular
algebra in school and I know that when you put two values
together like this, it is supposed to mean multiplication, But
I know he does not mean that here. I looked and I know that
with a plus sign it means "and" or "or" but Dennis does not
really explain what it means if there is no symbol between the
two values.
I have a vague understanding that it means both values
simultaneously, but not added, but I am not 100% sure.
I know someone on this forum has to understand this. I'm no
logician. I have other strong points, but that is not one of
them. Any help explaining is greatly appreciated.
#Post#: 702--------------------------------------------------
Re: Question about Boolean Algebra
DIR By: Colleen
Date: July 11, 2021, 3:37 pm
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Sorry for the late reply. At one moment of extra-clarity I
really "GOT IT". But then I figured it was unnecessary to
actually do the first few levels, although I sure appreciate
Dennis' sense of logic helping him to figure out and break down
the the foundational postulates of this universe/mind.
#Post#: 712--------------------------------------------------
Re: Question about Boolean Algebra
DIR By: Jur93n
Date: July 11, 2021, 9:28 pm
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Don't know if you have come closer to understanding the boolean
algebra?
So just my 2 cents here about notation, the "+" symbol should be
read as OR, the multiplication symbol should be read as AND.
Also you may read "1" as being the concept of True (exist) and
"0" the concept of False (not exist) (in computer science we
would write !X for not-X, where ! is the symbol, example: !False
equals True, !True equals False)
(1-X) is the same as Not-X or !X
If we look at the reasonable (sane) equation “X +(1-X)=1” it
says that X exists OR not-X (the absence of X).
For the insanity equation it states “X +(1-X)=0”, so here the
being is believing that "neither X exists nor not X exists"
Through boolean logic you can NOT the whole equation and you
will get "!(X and !X) = 1" or "X and !X = 1", here the being is
stating that both X and !X exist at the same time which is the
state of insanity.
Let me know if you need more clarification, glad to help on
this.
#Post#: 713--------------------------------------------------
Re: Question about Boolean Algebra
DIR By: mjh
Date: July 12, 2021, 10:02 am
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Hi,
(Note: use as a reference page 81 in the 2016 edition of TROM,
also I have utilised (.) to represent the AND in the logic)
I'll give my two pennies worth here - though I will work from
Dennis' two points in reverse, I believe it makes the first
concept easier to understand as it defines an aspect of it.
[attachimg=1]
Here we can see that the Universe (1) is made up of the Concept
[X] (let's say: Dogs) and the absence of the Concept [1-X]
(let's say: No Dogs)
In TROM this would be 'X' (To Know) and '1-X' (To Not Know)
That is the 2nd Axiom DS offers on the subject of logic - As DS
states - "The subject of logic rests upon two fundamental
axioms"
Let's now look at the First Axiom:
1) The common class of a concept and its absence does not exist.
x.(1-x)=0
This is the fundamentals of what is known as something being
mutually exclusive (x and 'not x' [1-x] is not possible)
HTML https://www.youtube.com/watch?v=X6usGgwXFyU
Part of the OP question was about x.(1-x) and x + (1-x)
Well, when put together x.(1-x), they are a single class. And we
can see that from the Axiom 1 that 'x.(1-x)' is not possible -
you cannot have dogs and no dogs in the same class together.
But, you can have a two separate classes as per Axiom 2 - A
class with dogs (x) OR a class with no dogs (1-x)
'x+(1-x)=1'
This is more apparent when we look at the actual goals packages.
x= to know
y= to be known
therefore 1-x = to not know
and 1-y = to not be known
This creates 4 overlap classes (scenarios) in a single Universe
(1):
x.y = the class of 'to know' and 'to be known' (complementary)
(1-x).y = the class of 'to not know' and 'to be known'
(conflicting)
x.(1-y) = the class of 'to know' and 'to not be known'
(conflicting)
(1-x).(1-y) = the class of 'to not know' and 'to not be known'
(complementary)
The above proves: x.y + x.(1-y) + y.(1-x) + (1-x).(1-y) = 1
all possible combinations in the Universe (1) are represented.
Note that the (+) symbol is separating the classes. As Jur93n
states you may read the + as OR (though this is a situation
where it is either those classes or it is both of those
classes), along with . as AND
Once you have this information correct in your mind - move onto
the Insanity Point material.
#Post#: 714--------------------------------------------------
Re: Question about Boolean Algebra
DIR By: Jur93n
Date: July 12, 2021, 8:26 pm
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Awesome write-up Khepri!
#Post#: 849--------------------------------------------------
Re: Question about Boolean Algebra
DIR By: mjh
Date: September 26, 2021, 6:38 pm
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I have recently re-transcribed the IP materials.
There were errors in the original transcript, these could cause
much confusion.
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