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Axioms of Boolean Algebra got from Concept Algebra
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PostPosted: Wed May 24, 2006 10:04 am    Post subject: Axioms of Boolean Algebra got from Concept Algebra Reply with quote

Axioms of Boolean Algebra got from Concept Algebra

By Shilong Wu March 06
All Rights Reserved
http://conceptalgebras.eponym.com/blog

Abstract:

All of the Axioms on Boolean algebra will be proved on Concept
algebra at this article. It is proved that the Boolean algebra is
sub algebra of Concept algebra<5>.

1. Introduction

The first kind of complete algebra<4> was the numerical algebra
based on the four binary operations“Addition Minus Times and
Divide”.
Based on this first kind of complete algebra makes the mankind get into
the material civilization from the savage times. It also could say that
the recent development of technology and science could not be reached
if the first kind of complete algebra did not be found. The more
deeper problem is how to establish the math base on the philosophy
after the material civilization reached. Especially the math based of
the logic thinking must reconsidered first.

Mankind research the law of logic thinking continuously since the
syllogism found by Aristotle 300B.C. A few logic laws had been got
after Aristotle using more than two thousands years. But the logic laws
only used on logic reasoning. As we know the logic reasoning is one
part of the logic thinking. Until the mid of the nineteenth century
an incomplete algebra appeared by G. Boolean called Boolean algebra.
There are only two binary operations on this algebra. It is similar as
figure addition and figure multiplication on the numerical math that
could not constitute a complete numerical algebra, the Boolean algebra
that only includes two binary operations was not a complete algebra
yet.
The only one standard of the complete algebra is that the terms
on this algebra can be moved from one side of the equation to another
by the theorem of this algebra. For example hypothesis A, B and C are
the variable on Boolean algebra, how can we move the variable B from
left of following equation to right?

A * B = C

It is clearly that the variable B at above equation can not be moved.
According to the definition of complete algebra a lot of algebra are
not the complete algebra such as group theory, ring theory and
set theory so on.

Until ninetieth of twenty century I designed a complete algebra
using the “theorem founder”. At this algebra there are four binary
operations, one unary operation and one constant. The axioms was
designed and tested on the “Theorem Founder”. The new algebra was
established called as “Special Wu Algebra” <1>.

The eight compound operations had been found based on the four
basic binary operations after researching this new special Wu algebra.
The “General Wu Algebra” <2> was established. The application of
the General Wu algebra was researched, I found the procedure of logic
thinking could be expressed completely, so that the name of this
algebra has been changed called “Concept Algebra” <3>. The relative

article will be found at web “http://www.conceptalgebra.cn”.

2 Definition of Concept Algebra

The Concept Algebra<3> is the another name of General Wu Algebra<2>.
The Concept Algebra researches the relationship among the concepts.
As we know the concept is the primary element of the logic thinking,
so that the concept algebra also is the math base of logic thinking.
As we know the special Wu algebra<1> is the second kind of the complete
algebra. The Concept Algebra extends from special Wu algebra, so that
the Concept Algebra also is one kind of the complete algebra. There
are four binary operations, eight compound operations, one unary
operation and one constant on concept algebra. The definition of
concept algebra is as follows:

The Concept algebra is an algebra with structure {GW, +, /, *, -,
@, #, 〉, !〉, <,!<, ^, !^, ‘, υ}. The operation +, /, *, -, @,
#,
〉, !〉, <,!<, ^ and !^ are binary operations, ‘ is unary
operation.
υ [ upsilon ] is a constant on this algebra. In the difference
algebra, the operations and the constant have different means.
The axioms of this algebra are as follows:

x + x’ = υ GW1
υ - x = x’ GW2
x / υ = x GW3
x / x = υ GW4
x - y = y’ - x’ GW5
x / y = y’ / x’ GW6
(x / y) / z = (x / z) / y GW7
x / (y + z) = (x / y) * (x / z) GW8
(x / y) * z = (x * z) / (y / z) GW9
(x / y) - z = (x - z) / (y + z) GW10
(x @ y) = ((x / y) * (y / x)) GW11
(x # y) = ((x - y) + (y - x)) GW12
(x 〉 y) = ((x + y) @ x) GW13
(x < y) = ((x * y) @ x) GW14
x !〉 y = (x + y) # x GW15
x !< y = (x * y) # x GW16
x ^ y = (y / x) # x GW17
x !^ y = (y / x) @ x GW18

Because of the binary operations on Boolean algebra are the basic
operations, so that the procedures of proving axioms on Boolean
algebra, the axioms of compound operation on concept algebra are not
used. The axioms from GW1 to GW10 are the same as the axioms on
special Wu algebra, so that the special Wu algebra is the sub algebra
of concept algebra.

3 Axioms of Boolean algebra
All of the axioms on Boolean algebra is shown at follows:

x * x’ = υ’ 4.1
x * υ = x 4.2
x + υ’ = x 4.3
x + x’ = υ 4.4
x + y = y + x 4.5
x * y = y * x 4.6
(x + y) + z = x + (y + z) 4.7
(x * y) * z = x * (y * z) 4.8
x * (y + z) = (x * y) + (x * z) 4.9
x + (y * z) = (x + y) * (x + z) 4.10

The deference between axioms of Boolean and these formulas is the
constant. The two constants on Boolean algebra have been replaced by
one constant and its complement on concept algebra

4 Proving Procedure of Axioms on Boolean Algebra

4.1 x * x’ = υ’
Proof:
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x 4.1.1
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using y’ to replace z, get
x / (y + y’) = (x / y) * (x / y’)
According to Axiom GW1, get
x / υ = (x / y) * (x / y’)
According to Axiom GW3, get
x = (x / y) * (x / y’)
Using υ’ to replace x, get
υ’ = (υ’ / y) * (υ’ / y’)
According to 4.1.1, get
υ’ = (υ’ / y) * y
Using y’ to replace y, get
υ’ = (υ’ / y’) * y’
According to 4.1.1, get
υ’ = y * y’
Proved.

4.2 x * υ = x
Proof:
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
Using υ to replace z, get
(x / y) * υ = (x * υ) / (y / υ)
According to Axiom GW3, get
(x / y) * υ = (x * υ) / y
Using υ’ to replace x, get
(υ’ / y) * υ = (υ’ * υ) / y
According to 4.1, get
(υ’ / y) * υ = υ’ / y
Using y’ to replace y, get
(υ’ / y’) * υ = υ’ / y’
According to 4.1.1, get
y * υ = y
Proved.

4.3 y + υ’ = y
Proof
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
Using υ to replace z, get
(x / y) * υ = (x * υ) / (y / υ)
According to Axiom GW3, get
(x / y) * υ = (x * υ) / y
Using υ’ to replace x, get
(υ’ / y) * υ = (υ’ * υ) / y
According to 4.1, get
(υ’ / y) * υ = υ’ / y
Using y’ to replace y, get
(υ’ / y’) * υ = υ’ / y’ 4.3.1
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x 4.3.2
So that according to 4.3.1 get
y * υ = y
To complement between two sides of this equation, get
(y * υ)’ = y’ 4.3.3
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using υ’ to replace x, Using y’ to replace y and z’ to replace
z, get
υ’ / (y’ + z’) = (υ’ / y’) * (υ’ / z’)
According to 4.3.2, get
(y’ + z’)’ = y * z
Using y’ to replace y and z’ to replace z, get
y + z = (y’ * z’)’
So that 4.3.3could be wrote as
υ’ + y’ = y’
Rewrite it.
Proved.

4.4 x + x’ = υ
This is the same as GW1.

4.5 x + y = y + x
Proof
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x 4.5.1
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using y’ to replace z, get
x / (y + y’) = (x / y) * (x / y’)
According to Axiom GW1, get
x / υ = (x / y) * (x / y’)
According to Axiom GW3, get
x = (x / y) * (x / y’)
Using υ’ to replace x, get
υ’ = (υ’ / y) * (υ’ / y’)
According to 4.5.1, get
υ’ = (υ’ / y) * y
Using y’ to replace y, get
υ’ = (υ’ / y’) * y’
According to 4.5.1, get
υ’ = y * y’ 4.5.2
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
Using z to replace y, get
(x / z) * z = (x * z) / (z / z)
According to Axiom GW4, get
(x / z) * z = (x * z) / υ
According to Axiom GW3, get
(x / z) * z = (x * z)
Using υ’ to replace x, get
(υ’ / z) * z = (υ’ * z)
Using z’ to replace z, get
(υ’ / z’) * z’ = (υ’ * z’)
According to 4.5.1, get
z * z’ = (υ’ * z’)
According to 4.5.2, get
υ’ = (υ’ * z’) 4.5.3
According to Axiom GW2
υ - x = x’
According to Axiom GW5, get
x’ - υ’ = x’
So that
x - υ’ = x 4.5.4
According to Axiom GW5, get
υ’’ - x’’ = x’
Using x to replace x’, get
υ’’ - x’ = x
Using υ to replace x, get
υ’’ - υ’ = υ
According to 4.5.4, get
υ’’ = υ 4.5.5
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x
According to Axiom GW6, get
x’’ / υ’’ = x
According to 4.5.5, get
x’’ / υ = x
According to Axiom GW3, get
x’’ = x 4.5.6
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using υ’to replace x, Using y’to replace y and z’to replace z,
get
υ’ / (y’ + z’) = (υ’ / y’) * (υ’ / z’)
According to 4.5.1, get
(y’ + z’)’ = y * z
Using y’to replace y and z’to replace z, and according to 4..5.6,
get
y + z = (y’ * z’)’ 4.5.7
Quote:
From 4.5.1
υ’ / x’ = x

Using x’ to replace x, get
υ’ / x’’ = x’
According to 4.5.7, get
υ’ / x = x’ 4.5.8
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
Using υ’ to replace x, get
(υ’ / y) * z = (υ’ * z) / (y / z)
According to 4.5.8, get
y’ * z = (υ’ * z) / (y / z)
According to 4.5.3, get
y’ * z = υ’ / (y / z)
According to 4.5.8, get
y’ * z = (y / z)’ 4.5.9
According to 4.5.9
y’ * x = (y / x)’
Rewrite it
(y’ * x)’ = y / x 4.5.10
According to 4.5.7, get
y + x’ = y / x
According to Axiom GW6
x / y = y’ / x’
According to 4.5.10, get
x + y’ = y’ + x
Using y to replace y’, get
x + y = y + x
Proved.

4.6 x * y = y * x
Proof
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x 4.6.1
According to Axiom GW2
υ - x = x’
According to Axiom GW5, get
x’ - υ’ = x’
So that
x - υ’ = x 4.6.2
According to Axiom GW5, get
υ’’ - x’’ = x’
Using x to replace x’, get
υ’’ - x’ = x
Using υ to replace x, get
υ’’ - υ’ = υ
According to 4.6.2, get
υ’’ = υ 4.6.3
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x
According to Axiom GW6, get
x’’ / υ’’ = x
According to 4.6.3, get
x’’ / υ = x
According to Axiom GW3, get
x’’ = x 4.6.4
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using υ’to replace x, Using y’to replace y and z’to replace z,
get
υ’ / (y’ + z’) = (υ’ / y’) * (υ’ / z’)
According to 4.6.1, get
(y’ + z’)’ = y * z
Using y’to replace y and z’to replace z, and according to 4..6.4,
get
y + z = (y’ * z’)’ 4.6.5
Quote:
From 4.4 and 4.6.5 the proof is directly.
Proved.


4.7 (x + y) + z = x + (y + z)
Proof
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x 4.7.1
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using y’ to replace z, get
x / (y + y’) = (x / y) * (x / y’)
According to Axiom GW1, get
x / υ = (x / y) * (x / y’)
According to Axiom GW3, get
x = (x / y) * (x / y’)
Using υ’ to replace x, get
υ’ = (υ’ / y) * (υ’ / y’)
According to 4.7.1, get
υ’ = (υ’ / y) * y
Using y’ to replace y, get
υ’ = (υ’ / y’) * y’
According to 4.7.1, get
υ’ = y * y’ 4.7.2
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
Using z to replace y, get
(x / z) * z = (x * z) / (z / z)
According to Axiom GW4, get
(x / z) * z = (x * z) / υ
According to Axiom GW3, get
(x / z) * z = (x * z)
Using υ’ to replace x, get
(υ’ / z) * z = (υ’ * z)
Using z’ to replace z, get
(υ’ / z’) * z’ = (υ’ * z’)
According to 4.7.1, get
z * z’ = (υ’ * z’)
According to 4.7.2, get
υ’ = (υ’ * z’) 4.7.3
According to Axiom GW2
υ - x = x’
According to Axiom GW5, get
x’ - υ’ = x’
So that
x - υ’ = x 4.7.4
According to Axiom GW5, get
υ’’ - x’’ = x’
Using x to replace x’, get
υ’’ - x’ = x
Using υ to replace x, get
υ’’ - υ’ = υ
According to 4.7.4, get
υ’’ = υ 4.7.5
According to Axiom GW3
x / υ = x
According to Axiom GW6, get
υ’ / x’ = x
According to Axiom GW6, get
x’’ / υ’’ = x
According to 4.7.5, get
x’’ / υ = x
According to Axiom GW3, get
x’’ = x 4.7.6
According to Axiom GW8
x / (y + z) = (x / y) * (x / z)
Using υ’to replace x, Using y’to replace y and z’to replace z,
get
υ’ / (y’ + z’) = (υ’ / y’) * (υ’ / z’)
According to 4.7.1, get
(y’ + z’)’ = y * z
Using y’to replace y and z’to replace z, and according to 4..7.6,
get
y + z = (y’ * z’)’ 4.7.7
Quote:
From 4.7.1
υ’ / x’ = x

Using x’ to replace x, get
υ’ / x’’ = x’
According to 4.7.7, get
υ’ / x = x’ 4.7.8
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
Using υ’to replace x, get
(υ’ / y) * z = (υ’ * z) / (y / z)
According to 4.7.8, get
y’ * z = (υ’ * z) / (y / z)
According to 4.7.3, get
y’ * z = υ’ / (y / z)
According to 4.7.8, get
y’ * z = (y / z)’ 4.7.9
According to 4.7.9
y’ * x = (y / x)’
Rewrite it
(y’ * x)’ = y / x
According to 4.7.7, get
y + x’ = y / x 4.7.10
According to Axiom GW7
(x / y) / z = (x / z) / y
According to 4.7.10, get
(x + y’) + z’ = (x + z’) + y’
Rewrite it
(x + y) + z = (x + z) + y
Proved.

4.8 (x * y) * z = x * (y * z)
Proof
From4.7.6
x + (y + z) = (x + y) + z
Get
(x + (y + z))’ = ((x + y) + z)’
According to 4.7.7, get
x’ * (y’ * z’) = (x’ * y’) * z’
Rewrite it
x * (y * z) = (x * y) * z
Proved.

4.9 x * (y + z) = (x * y) + (x * z)
Proof
According to Axiom GW9
(x / y) * z = (x * z) / (y / z)
According to 4.7.10, get
(x + y’) * z = (x * z) + (y / z)’
According to 4.7.9, get
(x + y’) * z = (x * z) + (y’ * z)
Rewrite it
(x + y) * z = (x * z) + (y * z)
Proved.

4.10 x + (y * z) = (x + y) * (x + z)
Proof
Quote:
From 4.7.9, get
y’ * x = (y / x)’

= (x’ / y’)’ GW6
= (x * y’) 4.7.9
So that following equation is hold
x * y = y * x 4.10.1
According to Axiom GW10
(x / y) - z = (x - z) / (y + z)
Using z’ to replace y, get
(x / z’) - z = (x - z) / (z’ + z)
According to Axiom GW1, get
(x / z’) - z = (x - z) / υ
According to Axiom GW3, get
(x / z’) - z = (x - z)
Rewrite it
(x / z) - z’ = (x - z’)
Using z to replace x, get
(z / z) - z’ = (z - z’)
According to Axiom GW4, get
υ - z’ = (z - z’)
According to Axiom GW2, get
z = (z - z’) 4.10.2
According to Axiom GW10
((z / y) - x) = ((z - x) / (y + x))
According to Axiom GW5 and GW6, get
(x’ - (z / y)' ) = ((y + x)' / (z - x)')
Using x’ to replace x, get
(x’’ - (z / y)' ) = ((y + x’)' / (z - x’)')
Using z to replace x, get
(z - (z / y)' ) = ((y + z’)' / (z - z’)')
According to 4.10.2, get
(z - (z / y)' ) = ((y + z')' / z')
According to 4.7.7 and GW6, get
(z - (y’ / z’)' ) = ((y’ * z) / z')
Using υ’ to replace z, get
(υ’ - (y’ / υ’’)' ) = ((y’ * υ’) / υ’')
According to Axiom GW3, get
(υ’ - y) = (y’ * υ’)
According to 4.10.1, get
(υ’ - y) = (υ’ * y’)
According to 4.7.3, get
(υ’ - y) = υ’ 4.10.3
According to Axiom GW10
((z / y) - x) = ((z - x) / (y + x))
According to Axiom GW5 and GW6, get
((y’ / z’) - x) = ((x’ - z’) / (y + x))
Using υ’ to replace z, get
((y’ / υ) - x) = ((x’ - υ) / (y + x))
According to Axiom GW3, get
(y’ - x) = ((x’ - υ) / (y + x))
According to Axiom GW5, get
(y’ - x) = ((υ’ - x) / (y + x))
According to 4.10.3, get
(y’ - x) = (υ’ / (y + x))
According to 4.7.1, get
(y’ - x) = (y + x)’
Using y’ to replace y, get
(y - x) = (y’ + x)’ 4.10.4
According to Axiom GW10
(x / y) - z = (x - z) / (y + z)
Quote:
From 4.10.4
y - x = (y’ + x)’

According to 4.7.7, get
y - x = y * x’ 4.10.5
According to 4.10.5, get
(x / y) * z’ = (x * z’) / (y + z)
According to 4.7.10, get
(x + y’) * z’ = (x * z’) + (y + z)’
According to 4.7.7, get
(x + y’) * z’ = (x * z’) + (y’ * z’)
Rewrite it
(x + y) * z = (x * z) + (y * z)
Proved.

5. Conclusion
All of the Axioms on Boolean algebra have been proved on Concept
algebra at this article. It is proved that the Boolean algebra is
sub algebra of Concept algebra.


Reference:
<1> Shilong Wu “Special Wu Algebra”
<2> Shilong Wu “Generalized Wu Algebra(GWA)”
<3> Shilong Wu “Concept Algebra”
<4> Shilong Wu “Discussion of Two Kinds of Complete Algebra”
<5> Shilong Wu “Concept Algebra”
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