What logical conclusions will be drawn from “Set A belongs to Set B” and “Set C belongs to Set D” ?
Conbra
and “Set C belongs to Set D” ?
These two propositions are independent of each other. It is
clearly that no any logic law could be used to reason new conclusions
from these two propositions. One wants to solve this problem, he must
study the new theory base of discrete mathematics, name concept
algebra at following web:
http://blog.tom.com/concept_algebra/article/3360.html
After studying concept algebra, this problem will be solved by
reader simply.
I'd like to solve this problem at following post.
Conbra
Wonderer
>What logical conclusions will be drawn from “Set A belongs to Set B”
>and “Set C belongs to Set D” ?
>
> These two propositions are independent of each other. It is
>clearly that no any logic law could be used to reason new conclusions
>from these two propositions. One wants to solve this problem,
What problem?
Conbra
>
> - Show quoted text -
The proplem is how to draw logical conclusions from these two
propositions?
Conbra
Conbra
The propositions “Set A belongs to Set B” and “Set C belongs to Set D”
are the relation of the four sets. The set A and set B in one
proposition, and the set C and set D in another proposition. How to
connect the relation among these four sets?
According to concept algebra the “belongs to” is one compound
operation, “<”. So that these two proposition could be written as
“Set A < Set B” and “Set C < Set D”
using the basic operation “*” to replace “and” on concept algebra, get
“Set A < Set B” * “Set C < Set D”
so that the following concept equation will be got
X / (Set A < Set B) * (Set C < Set D) = Dao
X is unknown variable; Dao is only one constant on concept algebra.
After solving this equation there are a lot of logical conclusion will
be drawn. The relation among these four sets is clearly.
Conbra
Wonderer
> Wonderer <i...@here.now> wrote:
>
>> >What logical conclusions will be drawn from “Set A belongs to Set B”
>> >and “Set C belongs to Set D” ?
>> >
>> >One wants to solve this problem [...]
>>
>> What problem?
>
>The proplem is how to draw logical conclusions from these two
>propositions?
There's no problem as it can already be done in second-order logic.
Given that a set of sets is a second-order predicate, "set A belongs
to B" is expressed B(A), and so "set C belongs to D" is expressed
D(C). Now we can for example derive this logical conclusion:
1. | B(A) assume
2. | | D(C) assume
3. | | B(A) & D(C) Intro & 1,2
4. | D(C) -> (B(A) & D(C)) Intro -> 2-3
5. B(A) -> (D(C) -> (B(A) & D(C))) Intro -> 1-4
The logical conclusion in step 5 is a tautology. ~Wonderer
Conbra
Wonderer wrote
> 5. B(A) -> (D(C) -> (B(A) & D(C))) Intro -> 1-4
>
> The logical conclusion in step 5 is a tautology. ~Wonderer
There is no new relation appeared at the conclusion.
Conbra
Conbra
>
> - 显示引用的文字 -
X / (Set A < Set B) * (Set C < Set D) = Dao
First solution of this equation is
X = (Set C < Set B) / (Set D < Set A)
The explanation of this solution in words is
X = If Set D belong to Set A, then Set C also belong to Set B.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if Set
D belong to Set A, then Set C also belong to Set B".
Above law is logical.
Conbra
Wonderer
>On Jul 19, 8:32 am, Wonderer <i...@here.now> wrote:
>> Conbra <s...@sh163.net> wrote:
>> > Wonderer <i...@here.now> wrote:
>>
>> >> >What logical conclusions will be drawn from “Set A belongs to Set B”
>> >> >and “Set C belongs to Set D” ?
>>
>> >> >One wants to solve this problem [...]
>>
>> >> What problem?
>>
>> >The proplem is how to draw logical conclusions from these two
>> >propositions?
>>
>> There's no problem as it can already be done in second-order logic.
>> Given that a set of sets is a second-order predicate, "set A belongs
>> to B" is expressed B(A), and so "set C belongs to D" is expressed
>> D(C). Now we can for example derive this logical conclusion:
>>
>> 1. | B(A) assume
>> 2. | | D(C) assume
>> 3. | | B(A) & D(C) Intro & 1,2
>> 4. | D(C) -> (B(A) & D(C)) Intro -> 2-3
>> 5. B(A) -> (D(C) -> (B(A) & D(C))) Intro -> 1-4
>>
>> The logical conclusion in step 5 is a tautology. ~Wonderer
>
>There is no new relation appeared at the conclusion.
Uh, you said the problem was "how to draw logical conclusions from
these two propositions?" Well, I proved an answer: step 5 (above) is a
logical conclusion drawn for the two propositions. But now you've
switched the problem to apparently how to draw a "new relation" from
the propositions. Well, I also did that, but it seems you don't know.
The logical connectives are functions (and all functions are
relations). So we can drop the usual infix notation and for example
instead of writing P & Q we can write &(P,Q) to make the functional
(and thus relational) nature of the statement clear. So several
relations were drawn from the two propositions. ~Wonderer
Conbra
>
X / (Set A < Set B) * (Set C < Set D) = Dao
Second solution of this equation is
X = (B > C) / (A > D)
The explanation of this solution in words is
X = If Set A includes Set D, then Set B also includes Set C.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if Set
A includes Set D, then Set B also includes Set C".
Conbra
Third solution of this equation is
X = (A < D) / (B < C)
The explanation of this solution in words is
X = If Set B belong to Set C, then Set A also belong to Set D.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if Set
B belong to Set C, then Set A also belong to Set D".
Jan Burse
> X / (Set A < Set B) * (Set C < Set D) = Dao
>
> Third solution of this equation is
> X = (A < D) / (B < C)
> The explanation of this solution in words is
> X = If Set B belong to Set C, then Set A also belong to Set D.
>
> That is to say, this one conclusion could be explained as:
>
> If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if Set
> B belong to Set C, then Set A also belong to Set D".
>
> Above law is logical.
>
> Conbra
You talking to yourself now?
To raise some interest in your work, you need to
show some translation results. Something like:
- My calculus is as powerful as the well know calculus X
- My calculus is less powerful than the well know calculus X
- My calculus is more powerful than the well know calculus X
For example for a calculus L define c(L) = { A | |-_L A }, i.e.
all tautologies that can be derived in the calculus. Let
denote:
LK = some classical logic calculus
CB = conbras calculus
Questions:
- c(LK) subset c(CB)?
- c(CB) subset c(LK)?
- None of the two above?
There some obstacles in the quest. You actually have to
carry out some proofs to answer the above questions. And
there are plenty of methods around:
- Bisimulation
- Lindenbaum Algebra
- Logic Matrices
- etc..
Work, work, work, ... dont sleep.
Bye
Conbra
I am showing the results from concept algebra to the web, any one
could comment these results. If there is any classical method can draw
the same results, tell me please.
The practice is only one standard for the theory.
Conbra
Conbra
Fourth solution of this equation is
X = (D > A) / (C > B)
The explanation of this solution in words is
X = If Set C includes Set B, then Set D also includes Set A.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if Set
C includes Set B, then Set D also includes Set A".
Jan Burse
> I am showing the results from concept algebra to the web, any one
> could comment these results. If there is any classical method can draw
> the same results, tell me please.
So you are expecting the community to do the
work for you? Shame on you.
You have to do the research first, and then give
the community enough hints that it can follow you.
And after all that the community can relate
to existing things.
So give us some logical translation results please?
Best Regards
Conbra
Jan Burse wrote
>So you are expecting the community to do the
>work for you? Shame on you.
No any meaning for expecting the community to do the work for me. Any
one who does not understand the concept algebra, he could not know the
concept expression. The community only could comment that can the
results got from concept algebra be got from other methods.
>So give us some logical translation results please?
If you have studied on web “http://blog.tom.com/concept_algebra/
article/3360.html”, you could understand the concept expressions of
the results, the logical translation is clearly.
Conbra
Conbra
Fifth solution of this equation is
X = (~D ^ ~B) / (~C ^ ~A)
The explanation of this solution in words is
X = If Complement of Set C does not relate to Complement of Set A,
then Complement of Set D does not relate to Complement of Set B,
either.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set C does not relate to Complement of Set A, then
Complement of Set D does not relate to Complement of Set B, either.".
Jan Burse
>> So give us some logical translation results please?
> If you have studied on web “http://blog.tom.com/concept_algebra/
> article/3360.html”, you could understand the concept expressions of
> the results, the logical translation is clearly.
>
> Conbra
Logical translation results should mean:
* Literally taking a logic book from somebody else.
* Picking the semantics and/or calculus there.
* And then make a comparative study.
I dont mean a logical translation for understanding the
calculus in itself. I mean logical translations to
existing approaches.
Bye
Jan Burse
>
> Logical translation results should mean:
> * Literally taking a logic book from somebody else.
For a start you could pick the paper by Gentzen. It
contains 3 calculi in it:
- Natural deduction
- Multi Sequent deduction
- Hilbert style
Pick one of the 3. And then work, work, work, ...
compare the picked calculus with yours.
I am sure such a venture will be very fruitful.
Conbra
The results posted at this web were calculated by concept calculator.
This calculator has the properties: when the conditions “Set A belong
to Set B” and “Set C belong to< Set D” input to this calculator, the
logical results will be output to the user. For example
If Set D belong to Set A, then Set C also belong to Set B
If Set A includes Set D, then Set B also includes Set C
……
Conbra
Conbra
Sixth solution of this equation is
X = (~D ^ ~B) / (~D ^ ~C)
The explanation of this solution in words is
X = If Complement of Set D does not relate to Complement of Set C,
then Complement of Set D does not relate to Complement of Set B,
either.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set D does not relate to Complement of Set C, then
Complement of Set D does not relate to Complement of Set B, either".
Conbra
Seventh solution of this equation is
X = (A < D) / (~D ^ ~C)
The explanation of this solution in words is
X = If Complement of Set D does not relate to Complement of Set C,
then Set A belong to Set D.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set D does not relate to Complement of Set C, then Set A
belong to Set D".
Above law is logical.
(All sign at this post are the operations on concept algebra:, see
following web:
http://blog.tom.com/concept_algebra/article/3360.html )
Conbra
Conbra
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Eighth solution of this equation is
X = (D > A) / (~D ^ ~C)
The explanation of this solution in words is
X = If Complement of Set D does not relate to Complement of Set C,
then Set D includes Set A
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set D does not relate to Complement of Set C, then Set D
includes Set A".
Above law is logical. All words in the result were produced by concept
calculator automatically.
This example is proved that the logic thinking of mankind can be
automated by machine. Any questions touch me freely.
Conbra
Tenth solution of this equation is
X = (C < B) / (~B ^ ~A)
The explanation of this solution in words is
X = If Complement of Set B does not relate to Complement of Set A,
then Set C belong to Set B
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set B does not relate to Complement of Set A, then Set C
belong to Set B".
Above law is logical. All words in the result were produced by concept
Conbra
Conbra
>
> 阅读更多 >>- 隐藏被引用文字 -
>
> - 显示引用的文字 -
What logical conclusions will be drawn from "Set A belongs to Set B"
and "Set C belongs to Set D" ?
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Eleventh solution of this equation is
X = (B > C) / (~B ^ ~A)
The explanation of this solution in words is
X = If Complement of Set B does not relate to Complement of Set A,
then Set B includes Set C
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set B does not relate to Complement of Set A, then Set B
includes Set C".
Above law is logical. All words in the result were produced by concept
Conbra
>
> read more >>- Hide quoted text -
>
> - Show quoted text -
What logical conclusions will be drawn from "Set A belongs to Set B"
and "Set C belongs to Set D" ?
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Twelfth solution of this equation is
Conbra
>
> 阅读更多 >>- 隐藏被引用文字 -
>
> - 显示引用的文字 -
What logical conclusions will be drawn from "Set A belongs to Set B"
and "Set C belongs to Set D" ?
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Thirteenth solution of this equation is
X = (~B ^ ~D) / (~B ^ ~A)
The explanation of this solution in words is
X = If Complement of Set B does not relate to Complement of Set A,
then Complement of Set B does not relate to Complement of Set D,
either
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if
Complement of Set B does not relate to Complement of Set A, then
Complement of Set B does not relate to Complement of Set D, either".
Above law is logical. All words in the result were produced by concept
calculator automatically.
This example is proved that the logic thinking of mankind can be
automated by machine. Any questions touch me freely.
(All sign at this post are the operations on concept algebra:, see
following web:
http://blog.tom.com/concept_algebra/article/3360.html
Conbra
Conbra
>
> read more >>- Hide quoted text -
>
> - Show quoted text -
What logical conclusions will be drawn from "Set A belongs to Set B"
and "Set C belongs to Set D" ?
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Fourteenth solution of this equation is
X = ~C > ~D
The explanation of this solution in words is
X = Complement of Set C includes Complement of Set D
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then
Complement of Set C includes Complement of Set D.
Conbra
>
> read more >>- Hide quoted text -
>
> - Show quoted text -
What logical conclusions will be drawn from "Set A belongs to Set B"
and "Set C belongs to Set D" ?
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Fifteenth solution of this equation is
X = ~A > ~B
The explanation of this solution in words is
X = Complement of Set A includes Complement of Set B.
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then
Complement of Set A includes Complement of Set B.
Conbra
>
> read more >>- Hide quoted text -
>
> - Show quoted text -
What logical conclusions will be drawn from "Set A belongs to Set B"
and "Set C belongs to Set D" ?
When this condition inputs to the concept calculator, the following
concept equation was established.
X / (Set A < Set B) * (Set C < Set D) = Dao
This concept equation was solved on the concept calculator and got a
lot of solutions.
Sixteenth solution of this equation is
X = (C ^ A) / (B ^ D)
The explanation of this solution in words is
X = If Set B does not relate to Set D, then Set C does not relate to
Set A, either
That is to say, this one conclusion could be explained as:
If "Set A belongs to Set B" and "Set C belongs to Set D"; then "if Set
B does not relate to Set D, then Set C does not relate to Set A,
either".
Above law is logical. All words in the result were produced by concept