Help with Converting Sentences to FOL

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  • Thread starter hanzla
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  • #1
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Hi, I need to convert below senteces into FOL, but I have difficulty doing it. Could someone peale help?

Lara ate exactly two apples.

Every city is either smaller than London or polluted.
 

Answers and Replies

  • #2
Start by determining the relations used in these statements and giving names to those relations.
 
  • #3
Hi, I need to convert below senteces into FOL, but I have difficulty doing it. Could someone peale help?

Lara ate exactly two apples.

Every city is either smaller than London or polluted.
The 2nd sentence can be converted in a way more accessible to pridicate calculus so we have :

if x is a city then x is smaller than London or x is polluted

Your pridicates here are: is a city , is smaller than, is polluted 1st and 3rd are one pridicates and the 2nd a two place pridicate:

Now denote:
is a city by capital C
is polluted by P
Is smaller than by S
London by L

Can you carry on from here
 
  • #4
We symbolize like this C(x): x is a city
P(x): x ia polluted
S(x): x is smaller than y
l for London

Does it have sa quatifier, cose its true for very city? Does the symbolization look like this?
∀x C(x) → (S(x,l) V P(x))
 
  • #5
S(x): x is smaller than y
You forgot the argument y of S.
∀x C(x) → (S(x,l) V P(x))
Correct, but it is better to enclose everything after ∀x in parentheses.
 
  • #6
I also have doubt about sentence `London is not a polluted city.` Is it correct like this. C(x): x is a city; P(x): x is polluted; l for London
¬P(l) ∧ C(x) does it need a identifying quantifier too?
 
  • #7
The formula is $C(l)\land\neg P(l)$. I am not sure what the phrase "identifying quantifier" means.
 
  • #8
Its existantial quantifier ∃ and it says “for some”, “there exists”, “there is a”, or “for at least one”.
∃x C(l)∧¬P(l)
 
  • #9
Its existantial quantifier ∃ and it says “for some”, “there exists”, “there is a”, or “for at least one”.
∃x C(l)∧¬P(l)
The above ∃x C(l)∧¬P(l) is not correct .The correct formula is

$\exists x(C(x)\wedge\neg P(x))$ or simply $C(l)\wedge\neg P(l)$

Only when you have a variable x,y,z...you can use a quantifier
 
Last edited:

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