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Which of the following first-order logic sentence matches closest with the sentence "All students are not equal"?

  1. $\forall x \exists y[\operatorname{student}(x) \wedge \operatorname{student}(y)] \Rightarrow \neg \operatorname{Equal}(x, y)$
  2. $\forall x \forall y[\operatorname{student}(x) \wedge \operatorname{student}(y)] \Rightarrow \neg \operatorname{Equal}(x, y)$
  3. $\forall x \exists y[\operatorname{student}(x) \wedge \operatorname{student}(y) \wedge \neg \operatorname{Equal}(x, y)]$
  4. $\forall x \forall y[\operatorname{student}(x) \wedge \operatorname{student}(y) \wedge \neg \operatorname{Equal}(x, y)]$

     

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The sentence "All students are not equal" can be translated into First-Order Logic (FOL) in different ways, but only one option accurately reflects the meaning. Let's analyze each option:

A. βˆ€π‘₯ βˆƒπ‘¦[𝑠𝑑𝑒𝑑𝑒𝑛𝑑(π‘₯) ∧ 𝑠𝑑𝑒𝑑𝑒𝑛𝑑(𝑦)] β‡’ Β¬πΈπ‘žπ‘’π‘Žπ‘™(π‘₯, 𝑦)

  • This translates to "For every student x, there exists another student y such that x is not equal to y."
  • This option is incorrect because it doesn't say that all students are different, just that for any student, there exists another who is different.

B. βˆ€π‘₯ βˆ€π‘¦[𝑠𝑑𝑒𝑑𝑒𝑛𝑑(π‘₯) ∧ 𝑠𝑑𝑒𝑑𝑒𝑛𝑑(𝑦)] β‡’ Β¬πΈπ‘žπ‘’π‘Žπ‘™(π‘₯, 𝑦)

  • This translates to "For every student x and every student y, x is not equal to y."
  • This option is too strong because it implies that no student can be equal to themselves, which is not the intended meaning of the original sentence.

C. βˆ€π‘₯ βˆƒπ‘¦[𝑠𝑑𝑒დ𝑒𝑛𝑑(π‘₯) ∧ 𝑠𝑑𝑒𝑑𝑒𝑛𝑑(𝑦) ∧ Β¬πΈπ‘žπ‘’π‘Žπ‘™(π‘₯, 𝑦)]

  • This translates to "For every student x, there exists another student y such that x is student and y is student and they are not equal."
  • This option is correct. It captures the essence of the original sentence by stating that for any student, there exists another student who is different.

**D. βˆ€π‘₯ βˆ€π‘¦[𝑠𝑑𝑒𝑑𝑒𝑛𝑑(π‘₯) ∧ 𝑠𝑑𝑒𝑑𝑒𝑛𝑑(𝑦) ∧ Β¬πΈπ‘žπ‘’π‘Žπ‘™(π‘₯, 𝑦)] **

  • This is the same as option B and is therefore too strong and incorrect.

Therefore, the First-Order Logic sentence that closest matches the original sentence "All students are not equal" is C. βˆ€π‘₯ βˆƒπ‘¦[π‘ π‘‘π‘’π‘‘π‘’αž“π‘‘(π‘₯) ∧ 𝑠𝑑𝑒𝑑𝑒𝑛𝑑(𝑦) ∧ Β¬πΈπ‘žπ‘’π‘Žπ‘™(π‘₯, 𝑦)].

1 comment

Option C is correct but your reasoning for why A is incorrect is wrong. You have basically stated option A is the same as option C.

Interpretation of option A would be "For every x there exists a Y such that if X is a student and Y is a student then it implies they are not equal." So it would mean no 2 students are equal which is not followed from the given statement.
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4

For all students X &  Y ,if X is a student  & Y is student, then  X  is not equal to Y

Same as no two students  are equal

So all students are not equal

Student (x): true if & only if X is true

Equal(X,y): predicte is true if & only if X any Y are equal

Conclusion: equal (X,y) are not equal

4 Comments

Bro can you tell which subject question is this and from which book to study it.
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Propositional logic.
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I don't agree @shivakundank.

"All students are not equal" is not equivalent to saying "No two students  are equal".

"All students are not equal" is more close to the statement - "Some students are equal but not all" or "Some student are equal and some are not." or "All the students taken together are not equal"

I that case option C makes more sense.

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@6rivu Depends on the domain of variable x and y, say the domain is everyone at school (including faculty members) then the expression in C and D will become false even if in reality there exists a non equal pair of student as student(some faculty member) is false.

We need a way to ignore all non students members, B should be correct IMO

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