Help with Converting Sentences to FOL

  • MHB
  • Thread starter hanzla
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In summary: In this sentence the quantifier is ∃ and it says “for some”, “there exists”, “there is a”, or “for at least one”. $C(l)\wedge\neg P(l)$The sentence is correct.The correct formula is $\exists x(C(x)\wedge\neg P(x))$.
  • #1
hanzla
4
0
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.
 
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  • #2
Start by determining the relations used in these statements and giving names to those relations.
 
  • #3
hanzla said:
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
hanzla said:
S(x): x is smaller than y
You forgot the argument y of S.
hanzla said:
∀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
hanzla said:
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:

FAQ: Help with Converting Sentences to FOL

What is FOL?

FOL stands for First-Order Logic, which is a formal language used in mathematics, computer science, and philosophy to express statements and relationships in a precise and unambiguous manner.

Why is it important to convert sentences to FOL?

Converting sentences to FOL allows us to analyze and reason about complex statements and relationships using formal rules and logical deductions. It also helps to eliminate ambiguity and ensure consistency in our reasoning.

How do you convert a sentence to FOL?

To convert a sentence to FOL, you need to identify the key concepts and relationships in the sentence and represent them using logical symbols and quantifiers. This process requires a good understanding of the language and the underlying logic.

What are the common symbols used in FOL?

The most commonly used symbols in FOL include logical operators such as conjunction (AND), disjunction (OR), negation (NOT), and implication (IF-THEN). Quantifiers such as "for all" (∀) and "there exists" (∃) are also frequently used.

What are some tips for converting sentences to FOL?

Some tips for converting sentences to FOL include breaking down complex sentences into smaller parts, using precise and unambiguous language, and being familiar with the rules and conventions of FOL. It also helps to check your work for consistency and validity using logical equivalences and truth tables.

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