Axioms of Boolean Algebra got from Concept Algebra
Conbra
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
>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
>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
>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
>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)
>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”
William Elliot
> Axioms of Boolean Algebra got from Concept Algebra
>
> By Shilong Wu March 06
> All Rights Reserved
> http://conceptalgebras.eponym.com/blog
>
Your web site is unreadable, full of cyptic letters without explaination.
> 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 operationsâ@]. Addition Minus Times and
> Divideâ@].
Why frequent usage of those weird characters like â@] and others?
It makes reading your post annoying and bothersome while seemingly adding
nothing to the discourse.
William Elliot
> Axioms of Boolean Algebra got from Concept Algebra
>
> By Shilong Wu March 06
> All Rights Reserved
> http://conceptalgebras.eponym.com/blog
>
> Abstract:
>
Learn about cross posting.
Jan Burse
> 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.
Not a very intelligent thing to do.
I mean what is gained by that?
(Big Question Mark here,
gained in terms of substantial
novellity, i.e. some open questions
closed)
I can create millions of algebras that
*contain* the boolean algebra.
Just observe that any set theory
leads to a boolean algebra, by forming
the following boolean ring:
Starting point:
Sets, i.e. U, intersection (^), union (u), complement (~)
A \ B := A ^ ~B.
End point:
Boolean algebra, i.e. U, add (+), mul (*), neg (-)
Define it as follows:
A * B := A ^ B.
A + B := (A \ B) v (B \ A) (The symmetric difference)
- A := ~ A.
Now we have:
A*(B + C) = A*B + A*C (in Sets A^(BvC)= A^B v A^C )
But not:
A+B*C = (A+B)*(A+C) not in general
(in Sets Av(B^C) = (AvB) ^ (AvC) holds)
(Simple counter example:
A={1,2}, B={1}, C={2}
A+B*C = {1,2}, (A+B)*(A+C) = {} )
Etc..
Jan Burse
A v B := A + B + A*B
A ^ B := A * B
~ A := - A.
See also:
http://citeseer.ist.psu.edu/143687.html
Quote:
The isomorphic relationship between Boolean algebras
and Boolean rings was found in 1936 by Stone and could
probably date back to 1927 by Zhegalkin.
Maybe one can go even further, back to Boole?
George Boole (1815-1864), also don't forget
Claude Shannon thesis (1938).
http://de.wikipedia.org/wiki/George_Boole
http://www.staugustine.net/mathematicalanalysisoflogic.html
Jan Burse
That - - A = A, is also not so
exciting. It would be more exciting
if not - - A = A in general.
george
Jan Burse wrote:
> Quote:
> The isomorphic relationship between Boolean algebras
> and Boolean rings was found in 1936 by Stone and could
> probably date back to 1927 by Zhegalkin.
>
> Maybe one can go even further, back to Boole?
> George Boole (1815-1864), also don't forget
> Claude Shannon thesis (1938).
Then the usual link:
http://en.wikipedia.org/wiki/George_Boole
But the wikipedia article on Boolean Algebra,
after listing all the following topics as related to it,
does NOT list one from above:
# Algebra of sets
# Ampheck
# And-inverter graph
# Boole, George
# Boolean algebra
# Boolean domain
# Boolean function
# Boolean logic
# Boolean implicant
# Boolean prime ideal theorem
# Boolean-valued function
# Boolean-valued model
# Boolean satisfiability problem
# Boole's syllogistic
# Canonical form (Boolean algebra)
# Characteristic function
# Compactness theorem
# Complete Boolean algebra
# De Morgan, Augustus
# De Morgan's laws
# Duality (order theory)
# Entitative graph
# Existential graph
# First-order logic
# Formal system
# Free Boolean algebra
# Heyting algebra
# Indicator function
# Interior algebra
# Jevons, William Stanley
# Johnston diagram
# Karnaugh map
# Laws of Form
# Lindenbaum-Tarski algebra
# Logic gate
# Logical connective
# Logical graph
# Logical matrix
# Logical value
# Monadic Boolean algebra
# Peirce, Charles Sanders
# Peirce's law
# Propositional calculus
# Sole sufficient operator
# Stone, Marshall Harvey
# Stone duality
# Stone's representation theorem for Boolean algebras
# Stone space
# Topological Boolean algebra
# Truth table
# Two-element Boolean algebra
# Venn, John
# Venn diagram
# Zeroth-order logic
It does not list Boolean Ring.
Worse, it does not list Boolean Lattice, although
that might be forgiveable since the Boolean Lattice,
being so close to the Boolean Algebra to begin with,
is also discussed in the SAME article.
Somebody really needs to start with the axioms for a lattice
and then show what you need to add to make it a Boolean Lattice,
and then what you need to add to THAT to make it a Boolean Ring.
Jan Burse
> Somebody really needs to start with the axioms for a lattice
> and then show what you need to add to make it a Boolean Lattice,
> and then what you need to add to THAT to make it a Boolean Ring.
>
Hi you don't need to add something
to a boolean lattice to obtain a boolean
ring. They a different representations
of the same thing.
Hm, repeating myself:
A * B := A ^ B.
A + B := (A \ B) v (B \ A)
- A := ~ A.
Makes a boolean ring from a boolean
lattice.
Hm, repeating myself:
A v B := A + B + A*B
A ^ B := A * B
~ A := - A.
Makes a boolean lattice from a boolean
ring.
The "makes" are adding additional
function symbols, not via axioms
between the function symbols, but
by hierarchical definitions.
(i.e Theory T' = Theory T und
hierarchical definition of new
symbols S)
So going from a lattice to a boolean
lattice is not the same as going from
a boolean lattice to a boolean ring.
When you go from a lattice (i.e. a
complete ordering, correct?) to boolean
lattice (with de Morgan law etc.) then
the set of models becomes smaller, because
you really add restricting axioms.
But if you go from a boolean lattice
to a boolean ring, the set of models
does not really become smaller.
But actually I don't know what happens
if you have infinite sums/unions. But
for the finite operations, it should
be quite simple to show that indeed
A v B = A + B + B ^ C
where X+Y := (X \ Y) v (Y \ X)
(Tipp use truth table of xor, or
a circle diagramm)
Jan Burse
Jan Burse wrote:
> A v B = A + B + B ^ C
> where X+Y := (X \ Y) v (Y \ X)
Oops, tipo:
A v B = A + B + A^B
Jan Burse
Jan Burse
Jan Burse wrote:
> This is really really nice:
> http://www.ux1.eiu.edu/~cfdmb/idempotents/boolean.pdf
See slide 7 and 8 about the
relationship between boolean
algebra and boolean ring.
Bye
george
Jan Burse wrote:
> you don't need to add something
> to a boolean lattice to obtain a boolean
> ring. They a different representations
> of the same thing.
The fact that two things are isomorphic
does not make them different representations of
the same thing. What it means for axiom-
systems over different langauges to be "isomorphic"
is problematic to put it charitably.
The best isomorphism between them would map
one symbol in each language to one in the other.
> Hm, repeating myself:
>
> A * B := A ^ B.
That is redundant.
Whether something is a boolean ring or a boolean
lattice does NOT depend on how you SPELL any
of the operators.
William Elliot
> Hm, repeating myself:
>
> A * B := A ^ B.
> A + B := (A \ B) v (B \ A)
> - A := ~ A.
>
> Makes a boolean ring from a boolean
> lattice.
>
> Hm, repeating myself:
>
> A v B := A + B + A*B
> A ^ B := A * B
> ~ A := - A.
>
> Makes a boolean lattice from a boolean ring.
>
No, ~0 = 1, -0 = 0; ~A = 1 - A
or ~A = 1 + A, as A + A = 0.
In addition, the ring has to have an identity.
Nam Nguyen
george wrote:
> The fact that two things are isomorphic
> does not make them different representations of
> the same thing. What it means for axiom-
> systems over different langauges to be "isomorphic"
> is problematic to put it charitably.
If we *first* define 2 isomorphic languages (which we can),
then we can define 2 isomorphic axiom-systems over 2
different (but isomorphic) languages, without a problem.
> The best isomorphism between them would map
> one symbol in each language to one in the other.
Again, the mapping between languages/formulae would
come first, before we could care to talk about the utility
of an isomorphism between 2 theories.
I think what you'd like to say here is that such utility is
almost null: there doesn't seem to be much of a benefit.
I think that's true, but only for finitely axiomatizable theories.
For recursively axiomatizable ones, in general, that 2
theories are isomorphic (in the sense above) don't necessarily
mean the provability or (un)decidability of a particular formula
F in one theory is that in the other theory. And that's because
the (meta) description of a system T, with infinite numbers of axioms,
in general is not precise enough to spell out all the axioms.
As such, when we use to isomorphic languages to over such meta
description, some of the n-ary symbols are bound to be overridden,
from one language to the other. An example would be defining a theory
T to be an extension of ZF: would T then prove AC? Obviously it's
undecidable (in the meta level) that T would prove AC. Therefore
in this case a principle of conservatism has to be applied: since
we don't know that whether or not T would prove AC, neither could
we say an isomorphic T' (viz-a-viz T) would be "the same" as T; T
might prove AC while T' might AC' (which is an isomorphic version
of AC).
Something like that, imho.
--
----------------------------------------------------
Time passes, there is no way we can hold it back.
Why then do thoughts linger, long after everything
else is gone?
Ryokan
----------------------------------------------------
Nam Nguyen
Nam Nguyen wrote:
I meant "As such, when we use two isomorphic languages over such meta"
Nam Nguyen
Nam Nguyen wrote:
I really hate typo, but it should have been "while T' might ~AC'"
>> of AC).
"of ~AC", of course.
Bertie Reed
> Jan Burse wrote:
>> you don't need to add something
>> to a boolean lattice to obtain a boolean
>> ring. They a different representations
>> of the same thing.
>
> The fact that two things are isomorphic
> does not make them different representations of
> the same thing. What it means for axiom-
> systems over different langauges to be "isomorphic"
> is problematic to put it charitably.
> The best isomorphism between them would map
> one symbol in each language to one in the other.
>
There are various standard ways of expressing the equivalence, here is one:
Any model of the theory of boolean rings interprets a boolean algebra (on
the whole of its domain) and visa versa.
If the theorys were complete, then one would write that they are
biinterpretable. (infact definable
Conbra
> A={1,2}, B={1}, C={2}
> A+B*C = {1,2}, (A+B)*(A+C) = {} )
(A+B) = {1,2}
(A+C) = {1,2}
so that (A+B)*(A+C) = {1,2}
A+B*C = (A+B)*(A+C) in general
Conbra
Conbra
text formation.
Conbra
> (Simple counter example:
> A={1,2}, B={1}, C={2}
> A+B*C = {1,2}, (A+B)*(A+C) = {} )
(A+B) = {1,2}
(A+C) = {1,2}
so that (A+B)*(A+C) = {1,2}
because A+B*C = {1,2}
so that
Jan Burse
Sorry, I left out the 0 and 1.
In the boolean algebra we:
1_ba = glb = U
0_ba = lub = {}
In the boolean ring that is derived
from the boolean algebra by my definitions
as above:
1_br := 1_ba
0_br := 0_ba
It is easy to see that 0_br is the neutral
element of ring addition, i.e.:
0_br + A = (A \ {}) v ({} \ A) = A.
It is also easy to see that 1_br is the
neutral element of ring multiplication, i.e.:
1_br * A = U ^ A = A.
It is easy to see that - is the inverse
for addition, i.e.:
A + (- A) = (A \ ~A) v (~A \ A) = {} = 0_br
As the question is about a ring, one has
not to show that there is an inverse for
multiplication.
Bye
Jan Burse
Conbra wrote:
Nope, + is symmetric difference, hence:
A+B = (A \ B) v (B \ A) = {2} v {} = {2}
A+C = (A \ C) v (C \ A) = {1} v {} = {1}
(A+B)*(A+C) = {2} ^ {1} = {}
But:
A + B*C = ({1,2} \ {}) v ({} \ {1,2}) = {1,2}
So we have a COUNTER EXAMPLE to:
A + B*C = (A+B)*(A+C)
Actually one expects a proper boolean ring to behave
like that (i.e. only one distributivity),
because if it would not behave like that
it were a boolean lattice (i.e. two distributivities).
Bye
Jan Burse
Conbra wrote:
> your suppose is
> so that
> A+B*C = (A+B)*(A+C) in general
Also, it is a little bit dangerous if
you could eliminate *one* counter example
to claim then that something holds *in general*.
I can produce much more counter examples,
and also examples where A+B*C = (A+B)*(A+C).
One counter example is enough to show that
A+B*C = (A+B)*(A+C) does not hold in general.
But to show that A+B*C=(A+B)*(A+C) holds,
would necessiciate to show that there are
no couter examples at all.
Bye
William Elliot
> Sorry, I am new comer, I will not be cross posting later
>
You may find useful
http://oakroadsystems.com/genl/unice.htm
Use cross posting instead of posting same question two newsgroups.
Please include context as you see others do.
-- To Google and MathForum users:
Reply only if adequate context is included _within_ the reply.
Otherwise all contexts are removed from my view,
the flow of thought disrupted and chaos reigns.
In particular for Google users:
Instead of simply hitting the prominent "Reply" link, which doesn't
include a copy of the post to which one is replying, click the "Show
Options" link (toward the top of an item in the thread), which causes
a shaded area of links to appear next to the top of the item, including
"Reply" (first) that does introduce a copy of the previous text (offset
by > signs in the usual fashion).
----
george
Jan Burse wrote:
> Just observe that any set theory
> leads to a boolean algebra, by forming
> the following boolean ring:
>
> Starting point:
> Sets, i.e. U, intersection (^), union (u), complement (~)
Well, OK, if you insist,
but the point is, we do NOT start "with any set theory".
Set theory IS LONG AND COMPLICATED.
It has a dozen axioms. Or infinitely many if you
blow up the schemata. What we DO start with is AXIOMS,
in a first-order language. These are going to be DIFFERENT
for a ring vs. for a lattice. And sets simply DO NOT NEED to
be involved AT ALL, for a ring OR for a lattice.
> A \ B := A ^ ~B.
> End point:
> Boolean algebra, i.e. U, add (+), mul (*), neg (-)
I thought you said this was going to be a boolean ring!
Obviously, given any universal set, its powerset is[the domain of]a
boolean algebra. That was the whole initial purpose. That is sort
of the original isomorphism between logic and set theory. HOWEVER,
1) it ONLY works in set theories WITH a universal set, which, since
the canonical set theory DOESN'T have a universal set, makes doing
ANY of this IN TERMS OF sets JUST STUPID
> Define it as follows:
>
> A * B := A ^ B.
> A + B := (A \ B) v (B \ A) (The symmetric difference)
> - A := ~ A.
>
> Now we have:
>
> A*(B + C) = A*B + A*C (in Sets A^(BvC)= A^B v A^C )
>
> But not:
>
> A+B*C = (A+B)*(A+C) not in general
> (in Sets Av(B^C) = (AvB) ^ (AvC) holds)
You still haven't clarified here whether you mean the ring or the
lattice.
In any case, I repeat, sets are irrelevant. What you START with is NOT
ANY
sets BUT RATHER one set of axioms for the algebra and another for the
ring.
William Elliot
> William Elliot wrote:
> > On Wed, 24 May 2006, Jan Burse wrote:
> >> A * B := A ^ B.
> >> A + B := (A \ B) v (B \ A)
> >> - A := ~ A.
> >>Makes a boolean ring from a boolean lattice.
> >> A v B := A + B + A*B
> >> A ^ B := A * B
> >> ~ A := - A.
> >>Makes a boolean lattice from a boolean ring.
> > No, ~0 = 1, -0 = 0; ~A = 1 - A
> > or ~A = 1 + A, as A + A = 0.
> > In addition, the ring has to have an identity.
>
> Sorry, I left out the 0 and 1.
> In the boolean algebra we:
>
> 1_ba = glb = U
> 0_ba = lub = {}
>
Huh? Missing argument for glb lub. What's U?
> In the boolean ring that is derived from the boolean algebra by my
> definitions as above:
>
> 1_br := 1_ba
> 0_br := 0_ba
>
> It is easy to see that 0_br is the neutral
> element of ring addition, i.e.:
>
> 0_br + A = (A \ {}) v ({} \ A) = A.
>
> It is also easy to see that 1_br is the
> neutral element of ring multiplication, i.e.:
>
> 1_br * A = U ^ A = A.
>
> It is easy to see that - is the inverse
> for addition, i.e.:
>
> A + (- A) = (A \ ~A) v (~A \ A) = {} = 0_br
>
That's not correct. A\~A = A, ~A\A = ~A. A v ~A = 1
Again, -A /= ~A; ~A = 1 + A
-0 = 0, ~0 = 1. Thus -0 /= ~0
~0 = 1 + 0 = 1
What's correct is
A + A = A\A v A\A = 0 v 0 = 0.
Hence -A = A
So again if ~A = -A, then A = -A = ~A which is absurd.
~A does not equal -A.
Exercise: ~A = 1 + A = 1 - A
> As the question is about a ring, one has not to show that there is an
> inverse for multiplication.
>
Exercise: A ring is a Boolean ring iff for all x, xx = x.
Ie, prove x + x = 0 and xy = yx from for all x, xx = x.
Exericse: If R is a ring without the axiom x + y = y + x
but instead the axiom xx = x, show x + y = y + x is theorem.
Ie, for Boolean rings, commutativity of + is redundant.
Jan Burse
> Huh? Missing argument for glb lub. What's U?
glb = greates lower bound
U = universe
> That's not correct. A\~A = A, ~A\A = ~A. A v ~A = 1
Oops, youre right. Excuse my sloppyness.
So going from boolean algebra to boolean ring,
the following should work:
-A := A
And going from boolean ring to boolean algebra,
the following should work as well:
~A := 1_br + A
The last def also explains why its better to
carry along the 1 of the ring.
Bye
Jan Burse
> In any case, I repeat, sets are irrelevant.
It can be shown that every finite Boolean algebra
is isomorphic to the Boolean algebra of all subsets
of a finite set. Therefore, the number of elements
of every finite Boolean algebra is a power of two.
Stone's celebrated representation theorem for Boolean
algebras states that every Boolean algebra A is isomorphic
to the Boolean algebra of all closed-open sets in some
(compact totally disconnected Hausdorff) topological
space.
Hence for any boolean algebra A I can find a set W,
such that in set theory, the following definitions:
X u_W Y := { x in W | x in X or x in Y }
X ^_W Y := { x in W | x in X and x in Y }
~_W X := { x in W | not x in X }
1_W := W
0_W := {}
Leads to boolean algebra B=<W,u_W,^_W,~_W,1_W,0_W>
which is isomorphic to A.
The question is only whether we can formulate the
construction of W as well in the set theory.
Bye
Jan Burse
Jan Burse wrote:
> Hence for any boolean algebra A I can find a set W,
> such that in set theory, the following definitions:
>
> X u_W Y := { x in W | x in X or x in Y }
> X ^_W Y := { x in W | x in X and x in Y }
> ~_W X := { x in W | not x in X }
> 1_W := W
> 0_W := {}
>
> Leads to boolean algebra B=<W,u_W,^_W,~_W,1_W,0_W>
> which is isomorphic to A.
>
> The question is only whether we can formulate the
> construction of W as well in the set theory.
Oops, we need also a set V, V subset P(W),
B=<V,u_W,^_W,~_W,1_W,0_W>.
Bye
Jan Burse
The Stone representation theorem cannot be proven within the
Zermelo-Fraenkel axioms. It is equivalent to the Boolean prime ideal
theorem which states that every Boolean algebra has a prime ideal. Both
can be proven using the axiom of choice. But the Stone representation
theorem is strictly weaker than the axiom of choice.
george
Jan Burse wrote:
> george wrote:
> > In any case, I repeat, sets are irrelevant.
>
> It can be shown that every finite Boolean algebra
> is isomorphic to the Boolean algebra of all subsets
> of a finite set.
*I* *already* said that.
*I* said, "Obviously, given any universal set, its powerset is[the
domain of]a
boolean algebra. That was the whole initial purpose. That is sort
of the original isomorphism between logic and set theory."
> Therefore, the number of elements
> of every finite Boolean algebra is a power of two.
Of course, but who gives a shit about finitude??
Finitude is not even first-order definable!
First-order axiom-sets as rich as PA *still* have NO
CLUE about finitude! Boolean rings are a lot less rich
than that, so finitude is even less relevant for them.
What IS relevant, if you are going to do it this way,
is a universal set. And, I REPEAT, there is NO universal
set in the usual set theory. So sets are, I REPEAT, IRRELEVANT.
> Stone's celebrated representation theorem for Boolean
> algebras states that every Boolean algebra A is isomorphic
> to the Boolean algebra of all closed-open sets in some
> (compact totally disconnected Hausdorff) topological
> space.
Shit. How Utterly CIRCULARLY irrelevant.
GOOD LUCK even DEFINING compact, total, disconnected,
Hausdorff, OR topological WITHOUT A PRIOR notion of a
boolean algebra. JEEzus.
>
> Hence for any boolean algebra A I can find a set W,
> such that in set theory,
*NOBODY* GIVES A SHIT.
Set theory is MUCH LONGER AND MORE COMPLICATED
than boolean algebra. Anybody trying to understand boolean
lattices and boolean rings is trying to understand two concepts
FAR SIMPLER THAN set theory! There is a nice little isomorphism
in the finite case with a universal set, but that is a degenerate case.
It is NOT important. If you are doing a typical first-order set
theory
then you 1) have no idea what "finite" even means in the first place,
and 2) have no universal set to begin with, so all these things
depending
on complement are just irrelevant.
Conbra
Jan Burse wrote
>Nope, + is symmetric difference, hence:
>A+B = (A \ B) v (B \ A) = {2} v {} = {2}
>A+C = (A \ C) v (C \ A) = {1} v {} = {1}
> (A+B)*(A+C) = {2} ^ {1} = {}
>But:
>A + B*C = ({1,2} \ {}) v ({} \ {1,2}) = {1,2}
>So we have a COUNTER EXAMPLE to:
> A + B*C = (A+B)*(A+C)
>Actually one expects a proper boolean ring to behave
>like that (i.e. only one distributivity),
>because if it would not behave like that
>it were a boolean lattice (i.e. two distributivities).
A+B = (A \ B) v (B \ A)
operation "\" is subtraction between two sets, so that the meaning
of anther operation "+" must be explained as "complement equation
of". This is a compound operation, similar as the operation "#"
on concept algebra. This is a limitation to take the value of A and B
using this formula. At distribution law the operation "+"
definition is a simple meaning. It is only to present addition between
two sets. I'll explain the operation meanings among concept algebra,
set algebra and logic algebra at anther article on sci.logic.
Bye
george
Jan Burse wrote:
> Hence for any boolean algebra A I can find a set W,
What utter bullshit.
If ALL you have is a boolean algebra A then
YOU HAVE NO SETS, PERIOD.
To GET some sets, you need SOME MORE AXIOMS,
far more than the ones defining a boolean lattice
or a boolean ring.
> such that in set theory,
Except that WE'RE NOT IN set theory.
We're in the first-order theory of boolean lattices,
wondering whether this will or won't get us to the
first-order theory of boolean rings.
Jan Burse
Conbra wrote:
> First the meaning of your rule
> A+B = (A \ B) v (B \ A)
I wouldn't call it a rule, and I also used another
notation. I didn't use the equation sign (=). I
used the asignment sign (:=) first. Hence it was not
a rule, but a definition. Initially I wrote:
A + B := (A \ B) v (B \ A). (2)
Later when I write:
A=...
B=...
A + B = (A \ B) v (B \ A) =... (3)
Then (3) is only an expansion by (2). A
definitional expansion. Also in my initional
post I defined that v ist the set union,
\ the set difference, and hence + the symmetric
differention. So everything should have been
cristal clear.
The difference between the equation sign (=)
and the assignment sign (:=) is similar to the
difference between the biconditional sign (<=>)
and the logical definition sign (:<=>).
The clue by using definitional signs is that
they signal at the same time the name that is
defined and the way the name is defined. So
if I have a logic with v and ~. I can simply
say:
A -> B :<=> A v ~B.
This will in the same time introduce a new
symbol, in this case the conditional (->),
and and also the content of this symbol,
namely A -> B <=> A v ~B.
Etc..
Jan Burse
Hi
george wrote:
> then you 1) have no idea what "finite" even means in the first place,
> and 2) have no universal set to begin with, so all these things
> depending on complement are just irrelevant.
I didn't use an universal set. I wrote W and not U.
W is dependent on the boolean algebra, one wants
to isomorphically map into a set.
> Shit. How Utterly CIRCULARLY irrelevant.
> GOOD LUCK even DEFINING compact, total,
> disconnected, Hausdorff, OR topological WITHOUT
> A PRIOR notion of a boolean algebra. JEEzus.
One must distinguish between genera and specifica.
A notion such as boolean algebra is a genera.
In a stone proof there a two specifica of a boolean
algebra, namely the boolean algebra A and the
resulting algebra B.
So the proof does not work so much with the general
notion of boolean algebra. Lets make an analog
example, take the following sentence:
A boy becomes a man.
Although we have a boy is a human, a man is a human,
the sentence is not circular. A prior notion of man
is not necessary to understand the sentence.
Jan Burse
> If you are doing a typical first-order set
> theory then you 1) have no idea what "finite"
> even means in the first place, and 2) have no
> universal set to begin with, so all these things
> depending on complement are just irrelevant.
Its relatively simple to define a predicate FIN
in ZF. Here it goes:
FIN(S) :<=> ~INFIN(S)
INFIN(S) :<=> exists f, U(U subset S & U<>S & f:U->S & f bijectiv)
An infinite set, has a proper subset with can be
mapped to the set itself.
Also I didn't assume an universal set when
I mentioned W. It depends on the initial
boolean algebra. So its not universal.
Jan Burse
george wrote:
> What utter bullshit.
> If ALL you have is a boolean algebra A then
> YOU HAVE NO SETS, PERIOD.
> To GET some sets, you need SOME MORE AXIOMS,
> far more than the ones defining a boolean lattice
> or a boolean ring.
What are then the axioms of a boolean algebra
talking about? The structure of a boolean
algebra is:
A set A, <<<<<<<<<<<<<< A SET, HUCH!!!!!!!
A function ^: AxA -> A
A function u: AxA -> A
A function ~: A -> A
A constant 0: A
A constant 1: A
If you have an axiom:
x ^ ~x = 0
Then this means:
forall x in A(^(x,~(x))=0())
Ok, you can ignore this reading of an
axiom, and completely work with the
"algebra", i.e. the shuffling of symbols.
Historically I don't know just now where
this pure view point belongs to. And
whether it is in contradication to my
set interpretation of an axiom.
Maybe you can tell where these two views
result in a collision.
george
Jan Burse wrote:
> What are then the axioms of a boolean algebra
> talking about?
No axioms are ever talking about anything
in particular. ALL axioms are SIMULTANEOUSLY
"about" ALL their models! All the axioms hold
in ALL of the models of the axioms, completely IRrespective
of whether the objects in the domains of those models are sets,
numbers, letters, or your toenails.
>The structure of a boolean
> algebra is:
>
> A set A, <<<<<<<<<<<<<< A SET, HUCH!!!!!!
What UTTER bullshit.
There is NO SUCH THING as "the" structure of a boolean
algebra. A boolean algebra IS ANY structure SATISFYING
THE RELEVANT AXIOMS. That is THE ONLY criterion.
Jan Burse
george wrote:
> What UTTER bullshit.
> There is NO SUCH THING as "the" structure of a boolean
> algebra. A boolean algebra IS ANY structure SATISFYING
> THE RELEVANT AXIOMS. That is THE ONLY criterion.
I am saying the same thing. What I said
was the structur of a boolean algebra is:
bla bla
And then I went on, what it means for such
a structure to satisfy the axioms.
Why do you think I said something else
than you are saying? You should have noticed
that I said "the structur of >>A<< boolean
algebra".
I didn't mean intend or what ever that
there is only one boolean algebra around.
This has been wrongly interpreted into my
saying.
In all of my posting I never said that
there is >>one<< boolean algebra. I always
went on in my saying >>a<< boolean algebra,
>>a<< set theory etc..
Bye
Bye
Jan Burse
george wrote:
> No axioms are ever talking about anything
> in particular. ALL axioms are SIMULTANEOUSLY
> "about" ALL their models! All the axioms hold
> in ALL of the models of the axioms, completely IRrespective
> of whether the objects in the domains of those models are sets,
> numbers, letters, or your toenails.
Here you got something wrong. When I say
A set S,
Than this is the domain of the boolean
algebra. Whether the elements of A are sets,
numbers, letters or toenails is open, yes.
But saying a set S is the domain, is not
the same as saying a set P(W) is the domain.
If I say a set S is the domain, then the
domain can contain elements which are either
sets, or something other or a combination
of both.
When I would say a set P(W) is the domain,
then the elements in the domain are indeed subsets
of the set W. And when I did say V subset P(W)
is the domain, then the domain is indeed some
of the subsets of the set W.
But this is actually the passage of the stone
theorem. A boolean algebra over S is isomorphic
to some boolean algebra over V subset P(W). And
this is the content of my postings that should
have been conveyed to you.
The content of my postings is not that every
boolean algebra has a domain V subset P(W)
a priori. Hope this clears everything.
Bye
Jan Burse
george wrote:
>>The structure of a boolean
>>algebra is:
>>
>> A set A, <<<<<<<<<<<<<< A SET, HUCH!!!!!!
> What UTTER bullshit.
What is your point? Should I say
A class S instead of a A set S?
How would you define a boolean algebra?
(Oops I am using the letter S now
to avoid confusion with the english
article a/an)
Jan Burse
Nam Nguyen wrote:
> One would, as GG pointed out, define it in a normal way using axioms,
> and then construct a model where the theorems are true. A model of
> this boolean algebra system does *not* have to be a set: it could be
> numbers, etc...
I didn't say the elements of a boolean algebra have to
be sets. I said the elements of a boolean algebra form
a set together. For example if your elements of the boolean
algebra are numbers, they still form a set.
For example if your boolean algebra is based on the elements
0, 1, 2 and 3. Then the domain is {1,2,3,4} which is a set.
And all I was saying, in the post that g was refering to,
was that we have such a set. I didn't say that the set must
contain elements which are sets again.
> The point I think GG is making is that you keep referring to
> "a boolean algebra", without mentioning axioms, as if it could
Why should I repeat here the 32 or more axioms, that
are usually refered by the name "boolean algebra".
> stand on its own. It doesn't. I mean even if you could come up
> (as you did) with some binary operations on say a ZF set that
> one could call it a boolean algebra, what about other examples
> in which ZF sets are not the individuals of the domains of
> the operations? To consistently refer to them as "boolean
> algebra's", in cross domains as such, we must have axioms.
I didn't refer to the set based boolean algebra as "the"
boolean algebra. I just said that it makes up >>a<<
boolean algebra. Maybe I should have called it a set lattice,
to avoid confusion about certain readers.
Bye
Jan Burse
Nam Nguyen wrote:
> Jan Burse wrote:
>> How would you define a boolean algebra?
> One would, as GG pointed out, define it in a normal way using axioms,
> and then construct a model where the theorems are true.
So you admit that models can be constructed. Why not
construct a model first. And then talk about axioms.
What prevents you from doing that.
You can construct a model, talk about axioms and
than see other models. Why am I forced at first hand
to see all models.
What school prescribes a certain way of working with
axioms and models? Clearly I know the axiomatic method.
But is the axiomatic method enough nowadays?
Isn't the axiomatic method doomed to be unnecessary
restrictive as we are able to construct models nowaday.
Isn't for example the essence of showing independence
of axioms, in constructing two models, with contrary
valuation?
Are algebraists just bad logicians? (P.S.: My answer
is algebraists are normally excellent logicians)
Bye
Barb Knox
Jan Burse <janb...@fastmail.fm> wrote:
> george wrote:
>
> > If you are doing a typical first-order set
> > theory then you 1) have no idea what "finite"
> > even means in the first place, and 2) have no
> > universal set to begin with, so all these things
> > depending on complement are just irrelevant.
>
> Its relatively simple to define a predicate FIN
> in ZF. Here it goes:
>
> FIN(S) :<=> ~INFIN(S)
> INFIN(S) :<=> exists f, U(U subset S & U<>S & f:U->S & f bijectiv)
>
> An infinite set, has a proper subset with can be
> mapped to the set itself.
Yes, that will ensure that INFIN(S) implies S is infinite, but it does
NOT ensure that FIN(S) implies S is finite. INFIN does not capture all
possible infinite sets, because there are non-standard models of ZF in
which some of the ~INFIN sets are in actual fact infinite (as seen from
outside the model), but the model lacks the bijections which would show
that fact, inside the model.
This is similar to non-standard PA, which contains infinite numbers.
And unlike PA, there is no agreement about what should be "the" standard
model for ZF.
Jan Burse
Barb Knox wrote:
> In article <44763923...@fastmail.fm>,
> Jan Burse <janb...@fastmail.fm> wrote:
>> FIN(S) :<=> ~INFIN(S)
>> INFIN(S) :<=> exists f, U(U subset S & U<>S & f:U->S & f bijectiv)
> Yes, that will ensure that INFIN(S) implies S is infinite, but it does
> NOT ensure that FIN(S) implies S is finite. INFIN does not capture all
> possible infinite sets, because there are non-standard models of ZF in
> which some of the ~INFIN sets are in actual fact infinite (as seen from
> outside the model), but the model lacks the bijections which would show
> that fact, inside the model.
But if you have non-standard models,
than it is also not provable the INFIN(S),
and hence the FIN(S) is also not provable.
So its only a problem of the completness
of the definition, not of the correctness
of the definition.
Or will some inconsistencies arise? Can
you show such an inconsistency?
Is the lack of such a function f provable
somehow, for your acclaimed non standard model?
Thus, do you have a theory X such that
X |= ZF(C) & forall f, U (U subset S & U<>S & f:U->S -> ~f bijectiv)
Whereby some construction of X we know S
is infinite. How is the construction of X
done?
Bye
george
> > A boolean algebra IS ANY structure SATISFYING
> > THE RELEVANT AXIOMS. That is THE ONLY criterion.
Jan Burse wrote:
> I am saying the same thing.
NO, you are not. YOU are talking about sets.
THE AXIOMS FOR BOOLEAN ALGEBRA do NOT talk about sets.
> What I said
> was the structur of a boolean algebra is:
And the axioms for a boolean algebra DO NOT say that.
NO axioms say ANYthing about any PARTICULAR structure,
unless they are categorical. Otherwise, there are a GREAT MANY
DIFFERENT structures satisfying the axioms, and the axioms are
simultaneously ABOUT ALL of them IN GENERAL and NOT about
any PARTICULAR one!
> bla bla
> And then I went on, what it means for such
> a structure to satisfy the axioms.
IF that's what you did, then you did something completely
impermissible. NOBODY CARES about whether "such a " structure
satisfies anything! The whole point is that it does NOT need to be
"such a" structure! It can be ANYthing! It in particular does NOT
need
to be "a structure described via set theory".
NOTHING NEEDS to be said, EVER, about "what it means for such
a structure to satisfy the axioms". ALL that NEEDS to be said ARE THE
AXIOMS *themselves*. What structures satisfy them simply FOLLOWS
from them.
> Why do you think I said something else
> than you are saying? You should have noticed
> that I said "the structur of >>A<< boolean
> algebra".
There is no such thing as THE structure of A boolean algebra.
There is ONE set of axioms defining boolean algebra.
There are a GREAT MANY DIFFERENT structures satisfying it.
ALL of them are boolean algebras and they all have DIFFERENT
structures.
> I didn't mean intend or what ever that
> there is only one boolean algebra around.
> This has been wrongly interpreted into my
> saying.
There is, however, up to logical equivalence (and therfore
isomorphism), only ONE SET OF AXIOMS DEFINING what a boolean
algebra is. The question is whether the set of axioms defining what
a boolean ring is is or isn't equivalent to it.
Jan Burse
> There is ONE set of axioms defining boolean algebra.
> There are a GREAT MANY DIFFERENT structures satisfying it.
> ALL of them are boolean algebras and they all have DIFFERENT
> structures.
I didn't say that the structures are the same.
But they have the same signature:
A set S:
^: S x S -> S
v: S x S -> S
~: S -> S
1: S
0: S
That is to each boolean algebra belongs a set S,
and the above functions, whether you want it or not.
The set S hasn't to belong to a particular set theory.
If you don't have this structure you have only symbols,
which is also fine. But if you say that a boolean algebra
has an operation ^, v, ~ and constants 1, 0. Then you
are using first order model structure terminology.
I don't know how you describe your axioms. Maybe you say
my axioms are simply concatenations of symbols, i.e. strings
of symbols. Than that's also fine with me. But this is not
my position. And we can disagree on that, and name it.
GG View: Pure symbolic algebra.
JB View: Structure based algebra.
So what. There are two views. Nobody has asked you to advocated
one view. I was only trying to convey an idea depending on
a certain view. Destroying this view doesn't help conveing the
idea behind my view. It does neither bring you any further nor me.
Bye
Jan Burse
george wrote:
> There is, however, up to logical equivalence (and therfore
> isomorphism), only ONE SET OF AXIOMS DEFINING what a boolean
> algebra is. The question is whether the set of axioms defining what
> a boolean ring is is or isn't equivalent to it.
This question has been settled (BA~~>BR,
BR~~>BA). At least for a structure based
algebraist as me. I don't know what problems
persist to other views. Tell me please.
Bye
Jan Burse
george wrote:
>>> A boolean algebra IS ANY structure SATISFYING
>>>THE RELEVANT AXIOMS. That is THE ONLY criterion.
> There is, however, up to logical equivalence (and therfore
> isomorphism), only ONE SET OF AXIOMS DEFINING what a boolean
> algebra is. The question is whether the set of axioms defining what
> a boolean ring is is or isn't equivalent to it.
BTW: Equational varieties are relatively simple
beasts. And the equations for a boolean algebra
and the axioms for a boolean ring are purely
equational.
Because it is only a boolean ring, you don't have
negation, connectives or quantifiers in
the axioms. If it were a "körper" one would have
at least a little bit more exciting axiom:
x!=0 -> x*(x^-1)=1
Which has a disjunction in it. Although equational
varienties are very simple. The equivalence and
other things is not decidable. Just imagine a
Post word problem. Or some such.
See also:
http://www.math.uwaterloo.ca/~snburris/htdocs/WWW/PDF/boolean.pdf
george
Jan Burse wrote:
> > There is no such thing as THE structure of A boolean algebra.
> > There is ONE set of axioms defining boolean algebra.
> > There are a GREAT MANY DIFFERENT structures satisfying it.
> > ALL of them are boolean algebras and they all have DIFFERENT
> > structures.
>
> I didn't say that the structures are the same.
> But they have the same signature:
THEY *DO* *NOT*, you *IGNORANT* dipshit!
There are HUNDREDS OF DIFFERENT *equivalent* axiomatizations
of boolean algebra AND THEREFORE hundreds of DIFFERENT signatures
for it!
george
Jan Burse wrote:
> > There is no such thing as THE structure of A boolean algebra.
> > There is ONE set of axioms defining boolean algebra.
> > There are a GREAT MANY DIFFERENT structures satisfying it.
> > ALL of them are boolean algebras and they all have DIFFERENT
> > structures.
>
> I didn't say that the structures are the same.
> But they have the same signature:
>
> A set S:
Idiot, please: NO set is part of ANY signature.
A signature is a collection of FUNCTION NAMES
with associated ARITIES.
george
Jan Burse wrote:
> I didn't say that the structures are the same.
> But they have the same signature:
Nobody cares what signature a STRUCTURE has.
The ISSUE is what signature the LANGUAGE has.
> A set S:
> ^: S x S -> S
> v: S x S -> S
> ~: S -> S
> 1: S
> 0: S
The structure has to MATCH the language in the sense that
the function-names occurring in the language have to be interpreted
as functions occurring in the structure.
And this signature is bloated and redundant.
Of the six things you have specified here, THE ONLY TWO
you actually NEED are
^ and ~ . Under FOL *with* equality,
the following 3 axioms define boolean algebra:
(x^y)^z = x^(y^z)
xy = yx
~(~(x^y) ^ ~(x^~y)) = x
THAT IS ALL. THAT is boolean algebra.
EVERYthing else you might want to call boolean
algebra, REGARDLESS of what signature you spell it with,
IS boolean algebra BECAUSE it is equivalent TO THAT.
You do NOT need to put 0 or 1 into the signature BECAUSE
you can PROVE THEIR EXISTENCE AS THEOREMS from these axioms.
0 is just an abbreviation for x^~x. 1 is just an abbreviation for ~0.
Boolean alegbra therefore canNOT be defined by any PARTICULAR
signature because a great many DIFFERENT signatures get the
job done. IN PARTICULAR, THE ONE FOR BOOLEAN RINGS *ALSO*
gets it done (though of course there is more than 1 for boolean rings
just as there is more than 1 for boolean algebra).
NO STRUCTURES OR SETS have anything to do with this,
REGARDLESS of what signatures they may have. Alleging the
relevance of any particular structure WITH any particular signature
is just obscuring the fact that that particular signature is NOT
relevant.
george
> >>algebra is:
> >>
> >> A set A, <<<<<<<<<<<<<< A SET, HUCH!!!!!!
I replied,
> > What UTTER bullshit.
Jan B. replied
> What is your point? Should I say
> A class S instead of a A set S?
> How would you define a boolean algebra?
I apologize. I have been unduly abusive.
I never expected you to ask me to define it.
I thought you were already confident you knew what it was.
I would define it THE SAME WAY I DEFINE EVERYTHING ELSE,
of course, VIA A RECURSIVE FIRST-ORDER AXIOM-SET.
The fascinating thing about this particular case (boolean algebra)
is that many different axiomatizations, with different signatures,
are all possible, and they all mean the same thing, EVEN though
they use completely different languages! There is sort of an
underlying
fact that there are only 16 possible boolean functions of 2 (or fewer)
boolean arguments. A boolean algebra is just a realm in which all
16 of them are definable. But you can vary greatly in which you take
as basic and which you take as defined. And from the standpoint
of whether this is or isn't a boolean algebra, THAT (which signature,
which basis) does NOT matter.
In particular I would not define it as being dependent on some
universal set and the subsets thereof. That is one possible
interpretation
but it is not special. A boolean algebra is ANYthing that you can
satisfy
THE AXIOMS with. INCLUDING a boolean ring.
I posted a preferred axiomatization with a preferred signature earlier.
The traditional model-theoretic approach is overly limiting.
It presupposes that only sets can be models and it only wants
to know whether some structure-of-sets is a model or not.
That is not the issue. The issue IS, in what sense are a boolean
algebra and a boolean ring the same thing, and in what sense
are they different things?
Conbra
William Elliot wrote:
> On Wed, 24 May 2006, Conbra wrote:
>
> > Axioms of Boolean Algebra got from Concept Algebra
> >
> > By Shilong Wu March 06
> > All Rights Reserved
> > http://conceptalgebras.eponym.com/blog
> >
> Your web site is unreadable, full of cyptic letters without explaination.
>
> > 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".
>
> Why frequent usage of those weird characters like " and others?
>
> It makes reading your post annoying and bothersome while seemingly adding
> nothing to the discourse.
I send this artice again in the form of ms_word format to replace the
text format that was sent early. And the sign errors could be overcome.
My web is
Jan Burse
george wrote:
> I would define it THE SAME WAY I DEFINE EVERYTHING ELSE,
> of course, VIA A RECURSIVE FIRST-ORDER AXIOM-SET.
and then:
> The traditional model-theoretic approach is overly limiting.
> It presupposes that only sets can be models and it only wants
What is the difference between first order axioms
and first order set theoretic models.
According to the completness theorem this is the
same. That is |= == |-.
Do you have a case/proof that the completness theorem
of FOL fails?
Bye
Jan Burse
george wrote:
> There are HUNDREDS OF DIFFERENT *equivalent* axiomatizations
> of boolean algebra AND THEREFORE hundreds of DIFFERENT signatures
> for it!
May be signatures "for it". But a boolean algebra
has a definite signatures. If you have a notion
such as:
boolean algebra
boolean ring
boolean lattice
etc..
Each of these notions comes with a signature
and some axioms. Irrelevant of whethere they
can be converted to each other.
The idea of the names boolean algebra etc..
is to give a name to a signature and a set
of axioms.
So strictly speaking, if you come up with
another signature and other axioms you have
to invent a new name for that.
But there are ways in language to do it
simpler. For example you can say a bla bla
algebra with bla bla axioms. So overriding
the axiom of a given algebra.
So in fact if you have another axiomatization
at hand you already have another algebra at
hand, also if you reach at an equivalence
proof.
Another language trick to shape this manifold
of axiom systems and names for them is to
use the authors name to denote an axiomatization.
So one uses phrases like the axioms system of
ford, the axioms of mckinsey, etc.. And the
algebra to override follows from the context.
Bye
Bertie Reed
> Jan Burse wrote:
>> george wrote:
>> > In any case, I repeat, sets are irrelevant.
>>
>> It can be shown that every finite Boolean algebra
>> is isomorphic to the Boolean algebra of all subsets
>> of a finite set.
>
> *I* *already* said that.
> *I* said, "Obviously, given any universal set, its powerset is[the
> domain of]a
> boolean algebra. That was the whole initial purpose. That is sort
> of the original isomorphism between logic and set theory."
This is not the same thing at all.
Generally, 1 and 2 below are different.
1. If A has Property 1, P(A) has Property 2
2. If B has Property 2, then it is P(A) for some A with Property 1.
Regards
Bertie
Bertie Reed
Try reading what Jan wrote. You will note the disparity between what Jan
wrote, and what you argued with.
Bertie Reed
> Nam Nguyen wrote:
>
>> One would, as GG pointed out, define it in a normal way using axioms,
>> and then construct a model where the theorems are true. A model of
>> this boolean algebra system does *not* have to be a set: it could be
>> numbers, etc...
This is gramatically incorrect:
A model (singluar) could numbers (plural).
What is this supposed to mean? "A model is a number" is a statement in
English. It's not clear what one would mean by it though.
A model could be a set of numbers (together with interpretations), however.
If you don't mean that, then you will have to explain what you de mean.
Bertie Reed
>>>>The structure of a boolean
>>>>algebra is:
>>>>
>>>> A set A, <<<<<<<<<<<<<< A SET, HUCH!!!!!!
>
> I replied,
>>> What UTTER bullshit.
>
> Jan B. replied
>> What is your point? Should I say
>> A class S instead of a A set S?
>> How would you define a boolean algebra?
>
> I apologize. I have been unduly abusive.
> I never expected you to ask me to define it.
> I thought you were already confident you knew what it was.
> I would define it THE SAME WAY I DEFINE EVERYTHING ELSE,
> of course, VIA A RECURSIVE FIRST-ORDER AXIOM-SET.
So you start with a _set_ of axioms, or is an axiom-set not a set of
axioms?
> The fascinating thing about this particular case (boolean algebra)
> is that many different axiomatizations, with different signatures,
> are all possible, and they all mean the same thing, EVEN though
> they use completely different languages! There is sort of an
> underlying
> fact that there are only 16 possible boolean functions of 2 (or fewer)
> boolean arguments. A boolean algebra is just a realm in which all
> 16 of them are definable.
What is a realm?
Note that if I have a boolean algebra, then I have, in particular "a realm"
and 16 different sets of axioms.
However, I quote
"If ALL you have is a boolean algebra A then
YOU HAVE NO SETS, PERIOD.
To GET some sets, you need SOME MORE AXIOMS,
far more than the ones defining a boolean lattice
or a boolean ring."
Hmm. This sounds like a contradiction to me.
> But you can vary greatly in which you take
> as basic and which you take as defined. And from the standpoint
> of whether this is or isn't a boolean algebra, THAT (which signature,
> which basis) does NOT matter.
>
> In particular I would not define it as being dependent on some
> universal set and the subsets thereof. That is one possible
> interpretation
> but it is not special. A boolean algebra is ANYthing that you can
> satisfy
> THE AXIOMS with.
How can you tell if a realm satisfies an axiom?
> INCLUDING a boolean ring.
>
> I posted a preferred axiomatization with a preferred signature earlier.
>
> The traditional model-theoretic approach is overly limiting.
> It presupposes that only sets can be models and it only wants
> to know whether some structure-of-sets is a model or not.
Whereas you allow "realms" to be models.
This is indeed much more flexible, but merely because you haven't said what
a realm is.
Bertie Reed
> Jan Burse wrote:
>> I didn't say that the structures are the same.
>> But they have the same signature:
>
> Nobody cares what signature a STRUCTURE has.
> The ISSUE is what signature the LANGUAGE has.
>
>> A set S:
>> ^: S x S -> S
>> v: S x S -> S
>> ~: S -> S
>> 1: S
>> 0: S
>
> The structure has to MATCH the language in the sense that
> the function-names occurring in the language have to be interpreted
> as functions occurring in the structure.
>
> And this signature is bloated and redundant.
> Of the six things you have specified here, THE ONLY TWO
> you actually NEED are
> ^ and ~ . Under FOL *with* equality,
> the following 3 axioms define boolean algebra:
>
> (x^y)^z = x^(y^z)
> xy = yx
Hmm what is xy?
george
> > Jan Burse wrote:
> >> I didn't say that the structures are the same.
> >> But they have the same signature:
> >> ^: S x S -> S
> >> v: S x S -> S
> >> ~: S -> S
> >> 1: S
> >> 0: S
I replied
> > And this signature is bloated and redundant.
> > Of the six things you have specified here, THE ONLY TWO
> > you actually NEED are
> > ^ and ~ . Under FOL *with* equality,
> > the following 3 axioms define boolean algebra:
> >
> > (x^y)^z = x^(y^z)
> > xy = yx
> > ~( ~(x^y) ^ ~(x^~y) ) = x
Bertie Reed asked,
> Hmm what is xy?
xy was supposed to be x^y, and yx was supposed to be y^x.
Conjunction is like multiplication, which is often represented
by juxtaposition.
Your axioms, and therefore your signature,
don't need v, O, or I , because they are all definable.
This signature is also good because you can do the boolean
ring axioms in it as well; you just have to use a different
definition for the ring's + from the one you use for the algebra's v.
But this axiomatization shows that the "underlying" signature and
axioms are in fact not merely isomorphic BUT IDENTICAL.
Of course, the hard part is figuring out whether you can or cannot
repeat this feat in FOL withOUT identity.
george
Bertie Reed wrote:
> So you start with a _set_ of axioms, or is an axiom-set not a set of
> axioms?
No, you DON'T start with a SET of axioms.
Any more than you start with a SET of strings.
If we already knew what sets were then we wouldn't
need axioms for set theory.
The prior question is whether we do or don't know what
the usual finite natural numbers are.
We do this whole thing in a paradigm of "a first order language"
in which every string in the language has to have a length
that is a natural number, and every predicate and functor in the
language has to have a number-of-arguments that is a natural
number, and yet, When The Smoke Clears, thanks to Godel's
1st Incompleteness Theorem, we canNOT even SAY what a finite
natural number IS (for that, you have to use 2nd-order logic).
My point is that we presume the naturals in defining the framework.
Similarly, if we are going to have axioms, we presume that we can
present and list them withOUT knowing how lists (or sets) are FORMALLY
defined. It is unaddressed and unimportant. You have to presume
SOMEthing or you could never start. And we are NOT addressing
what-we-have-to-presume-in-order-to-coherently-use-character-
strings-and-sentences-as-a-mode-of-presentation. We're just
presuming we know how to do that. The things that we ARE addressing,
we're addressing with axioms written in that presumed paradigm.
> > The fascinating thing about this particular case (boolean algebra)
> > is that many different axiomatizations, with different signatures,
> > are all possible, and they all mean the same thing, EVEN though
> > they use completely different languages! There is sort of an
> > underlying
> > fact that there are only 16 possible boolean functions of 2 (or fewer)
> > boolean arguments. A boolean algebra is just a realm in which all
> > 16 of them are definable.
>
> What is a realm?
Well, whatever it is, it is not a model and it is not a language.
If we invoked enough category theory I suppose we could call it
a variety. None of that is a point. The point is that YOU KNOW that
there are 16 binary boolean functions. The question of what to
call the collection of them is fundamentally conventional and
arbitrary.
Any particular choice of framework for presenting them is going to
insert implementation detail that is not germane to the full generality
of the concept/realm/variety/sort/whatEVER-the-fuck.
> Note that if I have a boolean algebra, then I have, in particular "a realm"
> and 16 different sets of axioms.
Who is noting this? You or me?
Even if you DON'T have a boolean algebra, even if you INSTEAD
just have a notion of a function on 2 arguments of the same type
producing a result of the same type, and the authority to decree that
the type only has 2 elements, THEN you have 16 functions.
Boolean algebra doesn't necessarily have anything to do with this.
> However, I quote
>
> "If ALL you have is a boolean algebra A then
> YOU HAVE NO SETS, PERIOD.
> To GET some sets, you need SOME MORE AXIOMS,
> far more than the ones defining a boolean lattice
> or a boolean ring."
>
> Hmm. This sounds like a contradiction to me.
Well, it isn't.
The boolean algebra in question cannot define "3" but that
does not prohibit us from choosing to define it using THREE axioms
and THREE symbols.
> A boolean algebra is ANYthing that you can
> > satisfy
> > THE AXIOMS with.
>
> How can you tell if a realm satisfies an axiom?
In that context, it's not a realm, it's a structure or an
interpretation. I was talking about a collection of
functions, most of which aren't directly mentioned in
the axioms. The "realm" does not even come into existence
until AFTER The Confirmation of the Existence of SOMEthing
that satisfies the axioms.
> > The traditional model-theoretic approach is overly limiting.
> > It presupposes that only sets can be models and it only wants
> > to know whether some structure-of-sets is a model or not.
>
> Whereas you allow "realms" to be models.
No, really, I don't. The collection of 16 functions that are definable
under every boolean algebra is not itself the relevant model of
a boolean algebra. This collection of 16 functions is in some sense
MORE important than ANY model because while there are MANY models,
there is only ONE collection of these 16 functions.
> This is indeed much more flexible, but merely
> because you haven't said what a realm is.
That's entirely intentional.
If I tell you that there are 16 of some things then it does not
NECESSARILY EVEN MATTER what the collection is called,
or if the collection even exists. Maybe they will just always be
16 separate things and NEVER become 1 collective thing with 16 parts.
Nam Nguyen
Jan Burse wrote:
> Hi
>
> Nam Nguyen wrote:
>
>> Jan Burse wrote:
>>
>>> How would you define a boolean algebra?
>>
>> One would, as GG pointed out, define it in a normal way using axioms,
>> and then construct a model where the theorems are true.
>
>
> So you admit that models can be constructed. Why not
> construct a model first. And then talk about axioms.
> What prevents you from doing that.
Technically speaking, what prevents us from doing that is
the models' dependency on axioms. A model is more than just
a set of certain individuals: among other things, it has to
basically involve a mapping between the language's n-ary symbols
to the actual n-aries of the "structure"; and then we have to
interprete all the theorem-formulae (axioms included) to be true,
before we could call that a model of the *axiom*-system. And this
is the short answer.
The longer answer is that axioms (and formulae in general) is supposed
to be invariant, while models *involve* *interpretation*, which is
variant, from one reasoner to another. The consequence of this is
that you could cite something M as a model of an un-axiomatized "theory"
and I might agree to it; but how do we both formalize that agreement?
how do you know for sure what I interpret is the same as what you'd
interpret on the structure M, notwithstanding that I verbally agree
with you? Let me give an example. Let M be a set of all the points
on a square (of a 2-dimensional Euclidean space). Suppose we both have
chosen an appropriate language L that has vocabularies such as "point"
, "line-segment", "line-segment-length", "parrallel-line-segment",...
Suppose further that p1, p2 are 2 distinct points on the left side
of M, and p3, p4 are 2 parallel counterparts on the right side. Now
we both could agree to claim both segments (p1,p3) and (p2,p4) of the
model M are parallel. But could we claim the 2 segments are of
different length? The long and the short of it is that anybody could
"see" (or interprete) the 2 line-segments to be of different length, or
of the same length, for that matter! (Hint: think of the intersections
of concentric circles on the square!). In other words, without axioms,
eventually there won't be any guarantee that our reasoning conversation
would always be consistent!
>
> You can construct a model, talk about axioms and
> than see other models. Why am I forced at first hand
> to see all models.
Given that a formulae F is a theorem iff it's true in *all* models of
a theory, you're not forced to see all model (or any model at all!).
You just have to exercise the option to produce a proof of F,
naturally.
> What school prescribes a certain way of working with
> axioms and models? Clearly I know the axiomatic method.
I suppose the school's name is something like "FOL theory
axiomatization".
> But is the axiomatic method enough nowadays?
What does "enough" mean in this context of axiom/model debate?
>
> Isn't the axiomatic method doomed to be unnecessary
> restrictive as we are able to construct models nowaday.
I thought the other way around is true. Given that a theorem
is true in all models of a theory, before we could even
come up with an existence of any model, in what sense should the
axiomatic method be accused of as being "restrictive"?
> Isn't for example the essence of showing independence
> of axioms, in constructing two models, with contrary
> valuation?
That seems to be a limit of FOL reasoning. And your posting
seems to have a merit in that we might typically come up
with an intuition of a model of some "vague" theory which
has not been completed formalized yet; and this is OK.
But what you seem to have also advocated is to abolish axioms,
which I think is very wrong.
>
> Are algebraists just bad logicians? (P.S.: My answer
> is algebraists are normally excellent logicians)
>
> Bye
>
--
----------------------------------------------------
Time passes, there is no way we can hold it back.
Why then do thoughts linger, long after everything
else is gone?
Ryokan
----------------------------------------------------
Nam Nguyen
Bertie Reed wrote:
> On Sat, 27 May 2006 21:49:24 +0200, Jan Burse wrote:
>
>
>>Nam Nguyen wrote:
>>
>>
>>>One would, as GG pointed out, define it in a normal way using axioms,
>>>and then construct a model where the theorems are true. A model of
>>>this boolean algebra system does *not* have to be a set: it could be
>>>numbers, etc...
>
>
> This is gramatically incorrect:
>
> A model (singluar) could numbers (plural).
>
> What is this supposed to mean? "A model is a number" is a statement in
> English. It's not clear what one would mean by it though.
>
> A model could be a set of numbers (together with interpretations), however.
> If you don't mean that, then you will have to explain what you de mean.
As you suspected, it was just an informal/quick talk. What I meant is
the individuals of a model of the boolean-algebra theory don't have
to be ZF(C) sets: they could be natural numbers, for example.
george
Jan Burse wrote:
> You are contradicting yourself:
No, really, I'm not.
>
> george wrote:
> > I would define it THE SAME WAY I DEFINE EVERYTHING ELSE,
> > of course, VIA A RECURSIVE FIRST-ORDER AXIOM-SET.
Right.
> and then:
> > The traditional model-theoretic approach is overly limiting.
> > It presupposes that only sets can be models and it only wants
>
> What is the difference between first order axioms
> and first order set theoretic models.
OK, I have NOT been overly abusive.
You really do deserve to be abused if you can't
tell the difference between an axiom and a structure.
Set theory is not absolutely necessary to either of these
but it is traditional (first-order ZFC is traditional) for describing
structures (that ist he traditional model-theoretic approach).
Axioms are strings (well, trees, actually) and neither graph theory
nor string theory is a special case of set theory.
george
Jan Burse wrote:
> What is the difference between first order axioms
> and first order set theoretic models.
>
> According to the completness theorem this is the
> same. That is |= == |-.
You're being ridiculous.
|- is about proofs.
|= is about structures.
Structures don't have to be described in first-order ZFC.
That was not the framework in which the completeness
theorem was proved. ANYthing can be a structure as long
as it looks appropriately "functional" when "corresponded to"
the signature of the language.
It is traditional to describe structures in terms of first-order ZFC
but that does not mean that axioms are the same as structures.
And the completeness theorem certainly doesn't say that.
What it says is that AFTER you have described your structures,
if ALL of them that make the axioms true ALSO make the theorem true,
then there is a (syntactic) proof of the theorem from the axioms.
This result is not limited to
structures-that-are-described-in-first-order-ZFC,
and it certainly does not equate any such structure with any axioms.
Jan Burse
> A model (singluar) could be numbers (plural).
I meant:
A model could be [a set of] numbers
(together with interpretations), however.
I just left out the tiny phrase [a set of].
Whether it is grammatically disallowed to
mix singular with plural by the verb being,
is not obvious for me. There are nouns which
have no plural form.
For example: Water is H2O molecules.
Makes sense for me.
Jan Burse
> We do this whole thing in a paradigm of "a first order language"
> in which every string in the language has to have a length
> that is a natural number, and every predicate and functor in the
> language has to have a number-of-arguments that is a natural
This means for me that you are advocating doing mathematics
in a concrete domain and refrain from doing logic.
Because in logic these objects, i.e. the language with all
its finite stuff in it, becomes the object of study.
Usually one makes the distinction between the:
object level: sets, functions, relations, etc..
meta object level: formulas, proofs, etc..
Logicians and the field of mathematical logic, studies
the meta object level with same tools that object levels
are studied by ordinary mathematicians.
The distinction between object level and meta object level
is insofar relevant to the whole discussion, because
I think here is a collision between people working in
the object level and people working in the meta object level.
The collision is such that the meta object level is in
bad reputation, due to all the brable with ZFC and gödel.
But it should be pointed out, that one should work normally
with the meta object level as one would work with the
object level.
So if for example you are a mathematician that is inclined
to make indirect proofs, why not do it with the meta object
level etc.. Or if you are used to work with sets, so please
walk as you talk as well in meta object level.
Don't worry, be happy.
Bye
Jan Burse
george wrote:
> Jan Burse wrote:
> You're being ridiculous.
> |- is about proofs.
> |= is about structures.
Nevertheless they are the same.
It is effectively for FOL |- == |-.
See for example:
http://xn--gdels-completeness-theorem-cvc.area51.ipupdater.com/
Not all logics have completness results.
Jan Burse
george wrote:
> Jan Burse wrote:
>>According to the completness theorem this is the
>>same. That is |= == |-.
> Structures don't have to be described in first-order ZFC.
> That was not the framework in which the completeness
> theorem was proved. ANYthing can be a structure as long
> as it looks appropriately "functional" when "corresponded to"
> the signature of the language.
In case you have axioms T for your structure.
And these axioms are FOL.
Then the completness/soundness theorem tells you,
that in the positive case:
T |- A.
At least the consequence is also valid for
all set structures. That is:
forall M M[T]=1 -> M[A]=1
If there were an ANY-Structure X of yours with:
X[T]=1 & X[A]=0
Then either your proof system is also blind for
this ANY-Structure X, because your proof systems
says T |- A.
Or if your proof system is not blind for this ANY-
Structure X, then it must be incorrect. Because
your proof systems says T |- A, and semantically
this does not hold.
You see there is a consequence back to the axioms,
and your proof system, if you allow ANY-Structures.
In the negative case, if:
T |/- A.
Then the completness result says, that we can
find a set theoretic structure Y, with:
Y[T]=1 & Y[A]=0
I personally don't mind, that the theorem doesn't
say, that there is an ANY-Structures X, with the
above property. I am happy enough to get sets.
> It is traditional to describe structures in terms of first-order ZFC
> but that does not mean that axioms are the same as structures.
> And the completeness theorem certainly doesn't say that.
I didn't talk about axioms. |- is syntactic inference, and
|= is semantic inference. In both cases we have axioms T.
> What it says is that AFTER you have described your structures,
> if ALL of them that make the axioms true ALSO make the theorem true,
> then there is a (syntactic) proof of the theorem from the axioms.
> This result is not limited to structures-that-are-described-in-
> first-order-ZFC
Yes, but please see above. If you really want to do something
with ANY-Structures you have to care about both sides. I.e.
completness and soundness. I am happy to learn about such
approaches. Please tell me more about your ANY-Structures and
the merit of your approaches.
> and it certainly does not equate any
> such structure with any axioms.
Yes, the approach is still axiomatic. That means one does
not assume one particular structure. The work is done with
axioms, and consequences.
Bye
Jan Burse
have even been formalized later on.
See for example partial algebras.
Is that what you are heading on, georg?
george
> > We do this whole thing in a paradigm of "a first order language"
> > in which every string in the language has to have a length
> > that is a natural number, and every predicate and functor in the
> > language has to have a number-of-arguments that is a natural
Jan Burse wrote:
> This means for me that you are advocating doing mathematics
> in a concrete domain and refrain from doing logic.
You are lying as usual.
EVERY logician KNOWS what propositional logic is.
Guess what: the propositions in propositional logic
HAVE to be FINITE. But 0th-order logic is NOT capable
of DEFINING what a finite natural number is.
Most logicians know that first-order-logic-as-we-know-it
is conducted in something called A FIRST ORDER LANGUAGE.
As I said before, the very DEFINITION of this language requires
every individual sentence in the language to be finite, and requires
every function or predicate to accept a finite number of arguments.
BUT THIS LOGIC IS ALSO *NOT* powerful enough to define "finite".
So I am NOT ANYwhere advocating that anyone "refrain" from doing
logic. I am just pointing out that you can't require that EVERYthing
be
defined in advance because that leads to infinite regress. You
eventually
have to STOP demanding prior definitions for SOME things.
Finite collections are the canonical example of such things.
> Because in logic these objects, i.e. the language with all
> its finite stuff in it, becomes the object of study.
You can study it all you want; that doesn't mean that you
are in prior doubt as to what it is. You can make ANYthing an
object of study. The language is PREsupposed DESPITE the
fact that you CAN'T define it "logically" (in first-order logic,
anyway).
> Usually one makes the
SHUT UP.
YOU do NOT know BETTER than *I* do just
WHAT is USUAL!
> distinction between the:
> object level: sets, functions, relations, etc..
You're being ridiculous. ANYthing can be the object level.
If I want to have a first-order theory of cars or genetics, or
graphs, or categories, or strings, OR SENTENCES IN A LANGUAGE
OR NUMBERS, then I *can* have one. The object level IS NOT
sets, functions, or relations, UNLESS the theory in question IS
a set theory, or a relation theory, or a function theory. FUNCTIONS
are PRESUPPOSED *throughout* the paradigm!
> meta object level: formulas, proofs, etc..
That is not a level, meta, object, or otherwise.
Sentences in a language are just strings, period.
You can have a theory in which the object level is about
strings, if you want to have a string theory.
> Logicians and the field of mathematical logic, studies
> the meta object level with same tools that object levels
> are studied by ordinary mathematicians.
Do NOT PRESUME to DEFINE "logic" to me.
As for "mathematical" logic, it is not inherently different
from any other kind. What YOU are WRONGLY *calling*
"the field of mathematical logic" might as well be called
"first-order logic with the canonical model-theory of first-
order ZFC". That is the primary thing that ZFC is important
for, in the context of mathematical logic (constructing models
and interpretations). But that is not the only way and it is by
NO means definitive of the field.
> The distinction between object level and meta object level
> is insofar relevant to the whole discussion, because
> I think here is a collision between people working in
> the object level and people working in the meta object level.
Well, you're just wrong.
> The collision is such that the meta object level is in
> bad reputation, due to all the brable with ZFC and gödel.
NObody CALLS *this* "the meta-object level" except you!
Models are the OBJECT level in any case!
And Godel's theorem is not "brable"; all it means is that
rich axiom-sets have multiple models. This is important!
> But it should be pointed out, that one should work normally
> with the meta object level as one would work with the
> object level.
No, really, one shouldn't, and one certainly shouldn't be fool enough
to point it out in public.
Can we please go back to talking about boolean algebras and boolean
rings? There is more than one way to present boolean algebra.
NONE of the ways REQUIRE any mention of sets. That was SUPPOSED
to be what we were talking about. Do you have anything more to say
about boolean algebra or boolean rings? I gave you a set of axioms
for boolean algebra. Are you willing to come up with a translation for
boolean rings from it?
Jan Burse
george wrote:
> > meta object level: formulas, proofs, etc..
> That is not a level, meta, object, or otherwise.
> Sentences in a language are just strings, period.
> You can have a theory in which the object level is about
> strings, if you want to have a string theory.
It is usual practice to make the distinction
between meta object level and object level. But this
distinction does not exclude that both levels can
contain the same objects.
See for example:
Kleene, S.C: Introduction to Meta-Mathematics,
§15: "To Hilbert is due now, first, the emphasis
that strict formalization of a theory involves
the total abstraction from meaning, the result
being called a formal system or formalism; and
second, his method of making the formal system
as a whole the object of a mathematical strudy
called meta-mathematics or proof theory.
Metamathematics includes the description of
definition of formal systems as well as the
investigation of properties of formal systems.
In dealiung with a particular formal system,
we may call the system the object theory, and
the metamathematics relating to it its metatheory."
[..]
(Some authors use "meta-" to identify a language
or theory in which another language or theory is made
the object of a study not restricted to finitary
methods. Also "syntax language" vs "object langauge" is
used in this connection. Cf. Carnap 1934; also cf §37.
In this book, we only use "meta-" when the methods are
finitary.)
Actually most of the confusion and collision for
people working in the object level, is mostly caused
by the fact that the meta object level and the object
level can be the same structures. This is a beginners
problem.
But as a mathematician you can work as informally as
you want in both levels. Working 100% formal in both
levels gives tedious long proofs and is eventually
counter productive. Because the listener might understand
the sender also when the sender is informal.
What godel and tarski did, was a reflection of the
meta object level in the object level. Thus codings
of formulas A in the object level as "A" and
predicates such as PROOF("A") or TRUE("A") are
reflections.
Again what godel and tarksi did, should prevent one
from doing mathematical logic or mathematics. The
world is not falling appart. Only some limitations
of FOL showed up.
Bye
Jan Burse
george wrote:
> I gave you a set of axioms
> for boolean algebra. Are
> you willing to come up with
> a translation for boolean
> rings from it?
What you did was not writing down the axioms of a
boolean algebra. You wrote down the DUAL of the
axioms from Herbert Robbins (1933), namely:
(x^y)^z = x^(y^z)
xy = yx
~(~(x^y) ^ ~(x^~y)) = x
A path would probably be to adapt the original proof
of Robbins, so that it works with the ^ instead of
v, and show that your axiom system is equivalent to
a boolean algebra. And then pick some proof from
the literature for the passage from a boolean algebra
to a boolean ring.
Bye
george
Jan Burse wrote:
> So you admit that models can be constructed.
But not that they ever SHOULD be.
> Why not construct a model first.
Occam's razor: Why construct a model AT ALL?
The whole import of the completeness theorem is that
you DON'T NEED models! ANY theorem you can confirm using models
(i.e., that something follows from something, or that
something doesn't have any models because it's inconsistent)
YOU CAN *ALSO* show via proof! THAT is the point of the completeness
theorem! EITHER of syntax (proof) or semantics (models) suffices
BY ITSELF. You DON'T NEED one if you have the other, because the
other IS EQUIVALENT to the one.
Of course, one kind of question is more easily answered by constructing
a proof, while the other is more easily answered by constructing a
model.
If you want to confirm that something DOES follow from something else
(or, in the inconsistent case, that a contradiction follows from
something
else), then you'd be better off constructing a proof than constructing
a
model, because you'd have to examine All models in order to confirm
that. Conversely, if you want to show that
something is NOT provable, then you're probably better off constructing
a model wherein it comes up false, than trying to do a generalized
meta-
proof showing that ALL proofs fail to prove the unprovable conclusion.
THAT is the reason why people retain both toolsets/viewpoints.
But it simply is not the case that either is prior to or better than
the other,
and if that tie were going to be broken, it would certainly be broken
in
favor of proofs AND NOT of models.
> And then talk about axioms.
> What prevents you from doing that.
Common sense.
In real life, we don't NEED to "construct" the model.
In real life, we already have some field that we are trying to
investigate, and the intended model is already factually present.
The problem is, if you are trying to PROVE things, then again,
because of the completeness theorem, everything that's
provABLE will hold in EVERY model, so the particular model
that you are investigating and that inspired you to research
this problem IS NOT any longer IMPORTANT!
> You can construct a model, talk about axioms and
> than see other models. Why am I forced at first hand
> to see all models.
You've got it backwards. I am asking you (this is just
my syntactic/proof-theoretic perspective/preference;
your preference is apparently semantic/model-theoretic,
i.e., the opposite) to see NO models, EVER, because they
DON'T MATTER, because EVERYthing that's PROVABLE (and,
therefore, everything that DOES matter) is INdependent of what
happens in any particular model.
> Isn't for example the essence of showing independence
> of axioms, in constructing two models, with contrary
> valuation?
Of course, but that is something you do in a context where
you DON'T know what axioms you need. That is something
you do when your grasp of things has gotten sufficiently murky
that you suspect you may need to find new axioms. That is not
the general state of affairs. It is especially not the state of
affairs
in the context of defining boolean algebras or boolean rings.
george
Jan Burse wrote:
> It is usual practice to make the distinction
> between meta object level and object level. But this
> distinction does not exclude that both levels can
> contain the same objects.
Obviously.
Anybody's meta-level can be turned into anybody
else's object level simply by deciding to. I could
have a first-order formal theory of natural language
if I wanted to.
>
> See for example:
> Kleene, S.C: Introduction to Meta-Mathematics,
> §15: "To Hilbert is due now, first, the emphasis
> that strict formalization of a theory involves
> the total abstraction from meaning, the result
> being called a formal system or formalism; and
> second, his method of making the formal system
> as a whole the object of a mathematical strudy
> called meta-mathematics or proof theory.
Actually, nowadays, that's generally called model
theory. The proof-theoretic side gets talked about
a lot less, which is unfortunate. This has turned out
in practice to be the case because the sexy indpendence
results were easier to obtain by constructing disagreeing
models than by proving meta-proofs about unprovability
in proof theory.
> Metamathematics includes the description of
> definition of formal systems as well as the
> investigation of properties of formal systems.
> In dealing with a particular formal system,
> we may call the system the object theory, and
> the metamathematics relating to it its metatheory."
Just because somebody famous said this does NOT
make it important. "Meta-" already had a meaning BEFORE
this quote came along. The point is that you can have a meta-
take IN NATURAL LANGUAGE, on ANY object-area-of-study,
JUST by SAYING that you are choosing that viewpoint.
How much of either level you do or don't bother to formalize
is a personal question. What makes this field math is that
the object levels are formal. At the meta levels, everybody has
a completely free hand, and they can wave it all they want.
Which may not be much; they may want to make the meta-
levels equally formal. But that is not only unnecessary, it is,
EVENTUALLY, IMPOSSIBLE. You have to STOP formalizing every
next level, at SOME point. Otherwise you are in an infinite regress
of prior formal definitions.
> [..]
> (Some authors use "meta-" to identify a language
> or theory in which another language or theory is made
> the object of a study not restricted to finitary
> methods.
Exactly. And "finitary" is not the only thing that is relaxable
in this context.
> Also "syntax language" vs "object langauge" is
> used in this connection. Cf. Carnap 1934; also cf §37.
> In this book, we only use "meta-" when the methods are
> finitary.)
That makes this book wrong.
Again, this is a case of somebody famous doing something UNusual
precisely because his fame enables HIM to get AWAY with it.
What he actually MEANS by "we use meta- when the methods are
finitary" is "other people's resorting to non-finitary means in THEIR
meta-methods is DEPRECATED by ME because I'M so smart that
I can achieve great results in META-mathematics with One Hand
Tied Behind My Back" (i.e. by restricting myself to methods that
more strongly resemble what we're restricted to at the object level).
> Actually most of the confusion and collision for
> people working in the object level, is mostly caused
> by the fact that the meta object level and the object
> level can be the same structures. This is a beginners
> problem.
Oh, SHUT *YOUR* *BEGINNER* ASS UP!
People who are NOT beginners UNDERSTAND the IMPORTANCE
of NOT phrasing things in such a way as to cause beginners to
trip over what OUGHT to be SIMPLE distinctions.
> But as a mathematician you can work as informally as
> you want in both levels. Working 100% formal in both
> levels gives tedious long proofs and is eventually
> counter productive. Because the listener might understand
> the sender also when the sender is informal.
Excuse me? I thought that was MY side of the debate!
If we agree on THAT then WHAT are we arguing over??
> What godel and tarski did, was a reflection of the
> meta object level in the object level. Thus codings
> of formulas A in the object level as "A" and
> predicates such as PROOF("A") or TRUE("A") are
> reflections.
They didn't just "do" this. Godel-numberings "are there",
as an abstract possibility, whether Godel ever gets around
to pointing them out or not. Even though no particular Godel-
numbering is sacred, some statements do wind up being
INescapably "about" the meta-theory. The correspondence
is inescapably possible, and THAT has implications.
Every predicate of primitive recursive arithmetic is
definable and usable in PA *even* if you *don't * add (as PRA
does) a symbol for it to the language.
> Again what godel and tarksi did, should not prevent one
> from doing mathematical logic or mathematics. The
> world is not falling appart. Only some limitations
> of FOL showed up.
Well, of course, we agree on all that.
I am starting to need clarification about what
we disagreed on.