Quantification

B

bahen

New member
Joined
Dec 25, 2014
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15
Hi all,

Could someone please help me quantify the following rules?

(1) (A ⋀ B) → C
(2) (A ⋀ ¬B) → C
(3) (¬A ⋀ B) → C
(4) (¬A ⋀ ¬B) → ¬C

Ultimately I would like to be able to represent the above in set theory notation such that they would read in the following way (respectively):

(1) If A and B coincide, then C (or so too do they coincide with C)
(2) If A coincides with not-B, then C
(3) If not-A and B coincide, then C
(4) If not-A and not-B coincide, then not-C.

Many apologies for the rudimentary query. I'm not versed in formal notation but need to make use of it nonetheless.

With many thanks,

bahen
 
bahen said:
Could someone please help me quantify the following rules?
(1) (A ⋀ B) → C
(2) (A ⋀ ¬B) → C
Ultimately I would like to be able to represent the above in set theory notation such that they would read in the following way (respectively):
(1) If A and B coincide, then C (or so too do they coincide with C)
(2) If A coincides with not-B, then C
Click to expand...
I taught logic at all levels for over forty years. I have never seen the term coincide used in this context.
What does it mean?

The usual English translation for 1) is If A and B then C.
That is equivalent to If not C then not (A and B); Not A or not B or C; C is necessary for (A and B); (A and B) is sufficient for C; & (A and B) only if C).
 
Hello,

Thanks very much for your reply. I realise 'coincide' is non-technical. I should clarify. What I mean is that I want to be able to translate the transitivity of (1) into the notation of set theory. Thus, 'coincide' just means intersect (hope that link works).

This means the following for the above:

(1) If A and B intersect, then C.
(2) If A and not-B intersect, then C.
(3) If not-A and B intersect, then C.
(4) If not-A and not-B intersect, then not-C.

Again, I am trying to quantify the above in set theory notation.

Thanks again and please let me know what you think.

Best wishes,

bahen


pka said:
I taught logic at all levels for over forty years. I have never seen the term coincide used in this context.
What does it mean?

The usual English translation for 1) is If A and B then C.
That is equivalent to If not C then not (A and B); Not A or not B or C; C is necessary for (A and B); (A and B) is sufficient for C; & (A and B) only if C).
Click to expand...
 
bahen said:
Hello,

Thanks very much for your reply. I realise 'coincide' is non-technical. I should clarify. What I mean is that I want to be able to translate the transitivity of (1) into the notation of set theory. Thus, 'coincide' just means intersect (hope that link works).

This means the following for the above:

(1) If A and B intersect, then C.
(2) If A and not-B intersect, then C.
(3) If not-A and B intersect, then C.
(4) If not-A and not-B intersect, then not-C.

Again, I am trying to quantify the above in set theory notation.

Thanks again and please let me know what you think.

Best wishes,

bahen
Click to expand...



More specifically, I am interested in a quantified expression of rule number 3. It describes the set in which not-A and B obtain together, and therefore imply C.
 
bahen said:
I realise 'coincide' is non-technical. I should clarify. What I mean is that I want to be able to translate the transitivity of (1) into the notation of set theory. Thus, 'coincide' just means intersect (hope that link works).
This means the following for the above:
(1) If A and B intersect, then C.
(2) If A and not-B intersect, then C.
(3) If not-A and B intersect, then C.
(4) If not-A and not-B intersect, then not-C.
Click to expand...

I would really like to know your native language. You are using a word with a very well agreed upon meaning in a way of your own choosing. One can do that as long as you take the time and effort to redefine the word. That you appear reluctant to do.

The phrase "not P" in logic is \(\displaystyle \mathcal{P^c}\) which is read "complement of P"

"All dogs are animals" means that "the set of dogs is a subset of the set of animals": \(\displaystyle D\subseteq A\).

So (2) "If A and not-B intersect, then C" would be \(\displaystyle (A\cap B^c)\subseteq C\)

(4) If not-A and not-B intersect, then not-C. would be \(\displaystyle (A^c\cap B^c)\subseteq C^c\)

We are not here to give tutorial lessons. So get yourself a basic set theory textbook and teach yourself.
 
PKA, I would appreciate it if you moved on and allowed others to help. Your message is both unconstructive and patronising. I will not be responding further to your posts.

To any one else who might read this, I would be hugely grateful if you could help with the above queries!

Many thanks,

bahen
 
bahen said:
PKA, I would appreciate it if you moved on and allowed others to help. Your message is both unconstructive and patronising. I will not be responding further to your posts.

To any one else who might read this, I would be hugely grateful if you could help with the above queries!

Many thanks,

bahen
Click to expand...
I'm sorry, but nobody else is going to be able to divine your unique definitions, nor is anybody else going to attempt to provide here a course in Boolean logic.

Our best wishes to you. ;)
 
stapel said:
I'm sorry, but nobody else is going to be able to divine your unique definitions, nor is anybody else going to attempt to provide here a course in Boolean logic.

Our best wishes to you. ;)
Click to expand...

Hi Stapel,

Thanks for your message. I mentioned that I'm new to this and am aware that I am very likely using the wrong language. I'm a lecturer in my field and we don't usually use formal notation in our work. Except that I think there is a case or two where set theory may be especially instructive. Hence my amateurish forays and hence also my request for help. Which unique definitions do you have in mind? By 'coincide' I just meant that the objects obtain in the same set. Does that make sense?

Best wishes,

bahen
 
bahen said:
By 'coincide' I just meant that the objects obtain in the same set. Does that make sense?
Click to expand...
NO!
obtain ?? objects?? "the objects obtain in the same set"?? Each of which makes all more confusing!

stapel said:
...nor is anybody else going to attempt to provide here a course in Boolean logic.
Click to expand...
But there is hope for you to help yourself.
Keith Devlin was born, raised, and educated in the UK.
Using his book you can learn the vocabulary of set theory.
 
bahen said:
Hi Stapel,

Thanks for your message. I mentioned that I'm new to this and am aware that I am very likely using the wrong language. I'm a lecturer in my field and we don't usually use formal notation in our work. Except that I think there is a case or two where set theory may be especially instructive. Hence my amateurish forays and hence also my request for help. Which unique definitions do you have in mind? By 'coincide' I just meant that the objects obtain in the same set. Does that make sense?

Best wishes,

bahen
Click to expand...
It might help to know in what field you are a lecturer.
 
HallsofIvy said:
It might help to know in what field you are a lecturer.
Click to expand...

Hi HallsofIvy,

Thanks for your message. I lecture philosophy at Cambridge. The language I have been using is standard in my field. 'Obtain' = exists; 'object' = something that exists. I'll try to rewrite what I meant in English and then perhaps you can instruct how I can quantify the relations?

(1) There is at least one case wherein A, B, and C intersect. [This is another way of saying that there is at least one case where A, B, and C are a set.]
(2) There is at least one case wherein A, B', and C intersect.
(3) There is at least one case wherein A', B, and C intersect.
(4) There is at least one case wherein A', B', and C' intersect.

Am I making sense?

With thanks,

bahen
 
bahen said:
Hi HallsofIvy,

Thanks for your message. I lecture philosophy at Cambridge. The language I have been using is standard in my field. 'Obtain' = exists; 'object' = something that exists. I'll try to rewrite what I meant in English and then perhaps you can instruct how I can quantify the relations?

(1) There is at least one case wherein A, B, and C intersect. [This is another way of saying that there is at least one case where A, B, and C are a set.]
(2) There is at least one case wherein A, B', and C intersect.
(3) There is at least one case wherein A', B, and C intersect.
(4) There is at least one case wherein A', B', and C' intersect.

Am I making sense?

With thanks,

bahen
Click to expand...

If you wish to make yourself understood here (and probably any one else expecting a logic/set theory/mathematical type discussion), you will either have to learn the logic/set theory/matematical vocabulary or explain in more detail as you have done here. For example, what I would say for (1) is 'the intersection of A, B, and, C is not empty' and write it as
\(\displaystyle A \cap B \cap C \ne \phi\)
or possibly 'there exists an x such that x belongs to A, x belongs to B, and x belongs to C'
∃ x \(\displaystyle \mid\space x \epsilon A,\space x \epsilon B,\space x \epsilon C\)
or some other equivalent statement.
 
Ishuda said:
If you wish to make yourself understood here (and probably any one else expecting a logic/set theory/mathematical type discussion), you will either have to learn the logic/set theory/matematical vocabulary or explain in more detail as you have done here. For example, what I would say for (1) is 'the intersection of A, B, and, C is not empty' and write it as
\(\displaystyle A \cap B \cap C \ne \phi\)
or possibly 'there exists an x such that x belongs to A, x belongs to B, and x belongs to C'
∃ x \(\displaystyle \mid\space x \epsilon A,\space x \epsilon B,\space x \epsilon C\)
or some other equivalent statement.
Click to expand...

Excellent, many thanks for your help. I am learning the ropes, that much is clear. Would it be possible to combine the four rules into a single rule? (I mean this in the non-obvious sense, i.e. without just insert conjunctions between the rules.)
 
bahen said:
Excellent, many thanks for your help. I am learning the ropes, that much is clear. Would it be possible to combine the four rules into a single rule? (I mean this in the non-obvious sense, i.e. without just insert conjunctions between the rules.)
Click to expand...

I don't believe they can be combined in a non-trivial way. For example, compare (1) and (2):
(1) ∃ x [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]C[/FONT]
(2) ∃ x [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]B'[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]C[/FONT]
where, I would assume, B' was the complement of B, that is all elements not in B. Were one to try to combine statements (1) & (2) in a non-trivial way, for example, then it appears to me that one would be saying there exists an x belonging to neither B, since, by (2), it belongs to B', nor B', since, by (1) it belongs to B.

One could, as you mentioned, always combine them trivially as '(1) and (2)' where the x's are, this time, not the same elements.
 
Ishuda said:
I don't believe they can be combined in a non-trivial way. For example, compare (1) and (2):
(1) ∃ x [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]C[/FONT]
(2) ∃ x [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]B'[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]ϵ[/FONT][FONT=MathJax_Math]C[/FONT]
where, I would assume, B' was the complement of B, that is all elements not in B. Were one to try to combine statements (1) & (2) in a non-trivial way, for example, then it appears to me that one would be saying there exists an x belonging to neither B, since, by (2), it belongs to B', nor B', since, by (1) it belongs to B.

One could, as you mentioned, always combine them trivially as '(1) and (2)' where the x's are, this time, not the same elements.
Click to expand...

Thank you.
 
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