How do I express this in FOL?

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Reply Sun 13 Jul, 2008 08:24 pm
I'm trying to express this in FOL:

Every word has at least one concept associated w/ it, and every concept has at least one word associated w/ it. Would it be:

For all (x)(Cx<->Wx)

or

(for all x(w(x) -> there exists y(c(y) & r(x,y)))) & (for all y(c(y) -> there exists x(w(x) & r(x,y)))), where r(x,y) denotes the concept associate w/ the word x. Anyway to make that clearer? What would this actually look like on paper?

Thanks!
 
Theaetetus
 
Reply Sun 13 Jul, 2008 08:27 pm
@Protoman2050,
Protoman2050 wrote:
I'm trying to express this in FOL:

Every word has at least one concept associated w/ it, and every concept has at least one word associated w/ it. Would it be:

For all (x)(Cx<->Wx)

or

(for all x(w(x) -> there exists y(c(y) & r(x,y)))) & (for all y(c(y) -> there exists x(w(x) & r(x,y)))), where r(x,y) denotes the concept associate w/ the word x. Anyway to make that clearer? What would this actually look like on paper?

Thanks!


Even though I took a logic course and feel I am logical what the hell in FOL?
 
Protoman2050
 
Reply Sun 13 Jul, 2008 09:03 pm
@Theaetetus,
First order logic; propostional logic w/ existential quantifiers; First-order logic - Wikipedia, the free encyclopedia
 
Ron C de Weijze
 
Reply Mon 14 Jul, 2008 03:38 am
@Protoman2050,
Why do you want to express this in FOL? BTW why not UML? Isn't the expression you want to arrive at an app?
 
Protoman2050
 
Reply Mon 14 Jul, 2008 10:45 am
@Ron C de Weijze,
Ron C. de Weijze wrote:
Why do you want to express this in FOL? BTW why not UML? Isn't the expression you want to arrive at an app?


No, it's not for anything compsci related...just to condense my writing of a proof for God: http://www.philosophyforum.com/forum/philosophy-religion/1655-proof-god-s-self-evidence.html
 
VideCorSpoon
 
Reply Mon 14 Jul, 2008 05:47 pm
@Protoman2050,
In quantificational logic, (x) (Cx<->Wx) translates as; For any (x), C(x) if and only if W(x). Your elaboration paints a different picture.

For FOL...you would need to assert something (i.e. universal or existential quantifier) to establish a categorical syntactical structure (unless you are assuming it).

Also, you would need to elaborate on the sentence constants (i.e. C and W) to show what exactly you are trying to prove.

Also, you may want to reconsider your notation with the bi-conditional and quantificational logic.
 
Protoman2050
 
Reply Mon 14 Jul, 2008 05:58 pm
@VideCorSpoon,
VideCorSpoon wrote:
In quantificational logic, (x) (Cx<->Wx) translates as; For any (x), C(x) if and only if W(x). Your elaboration paints a different picture.

You would need to assert something (i.e. universal or existential quantifier) to establish a categorical syntactical structure (unless you are assuming it).

Also, you would need to elaborate on the sentence constants (i.e. C and W) to show what exactly you are trying to prove.

Also, you may want to reconsider your notation with the bi-conditional and quantificational logic.


So it would work better as --where C(x) means concept referred to by x, and W(x) means word linked to the concept of x; R(x,y) means the concept y associated w/ x--:
for all x(W(x) -> there exists y(C(y) & R(x,y))) & for all x(C(x) -> there exists y(W(y) & R(x,y)))

Can you show me how that would be read and what that would look like as symbols?

Thanks!
 
VideCorSpoon
 
Reply Mon 14 Jul, 2008 06:15 pm
@Protoman2050,
You may be over complicating the process of quantificational logic. First and foremost, what is the literal (x) (Cx --> Wx) (note: I substituted the bi-conditional for the conditional because of the lingual assumption.)

That formula translates as follows (in the universal affirmative); Every x such that x is C such that x is W.

Also, there are universal affirmatives here, but in your elaboration you use existential quantifications. This violates wff. You may want to revise that portion.
 
Protoman2050
 
Reply Mon 14 Jul, 2008 06:18 pm
@VideCorSpoon,
I am trying to convey the idea that, for every word, there are one or more ideas associated w/ it, and for every idea, there are one or more words associated w/ it.
 
VideCorSpoon
 
Reply Mon 14 Jul, 2008 06:44 pm
@Protoman2050,
Oh. You don't need to go into quantificational logic for that. You can stay within the realm of propositional logic and still accomplish your goal without predicative inferences.

All you need is a standard bi-conditional and a truth table to prove it.
 
Protoman2050
 
Reply Mon 14 Jul, 2008 07:35 pm
@VideCorSpoon,
VideCorSpoon wrote:
Oh. You don't need to go into quantificational logic for that. You can stay within the realm of propositional logic and still accomplish your goal without predicative inferences.

All you need is a standard bi-conditional and a truth table to prove it.


But I'm trying to say for ALL words, and ALL concepts...there are NO meaningless words, and NO unnamed concepts.
 
 

 
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