A person who is radical \((R)\) is electable \((E)\) if he/she is conservative…

2020

A person who is radical \((R)\) is electable \((E)\) if he/she is conservative \((C)\), but otherwise not electable.

Few probable logical assertions of the above sentence are given below.

(A) \((R \wedge E) \Leftrightarrow C\)

(B) \(R \rightarrow (E \leftrightarrow C)\)

(C) \(R \Rightarrow ((C \Rightarrow E) \vee \neg E)\)

(D) \((\neg R \vee \neg E \vee C) \wedge (\neg R \vee \neg C \vee E)\)

Which of the above logical assertions are true?

Choose the correct answer from the options given below:

  1. A.

    (B) Only 

  2. B.

    (C) Only 

  3. C.

    (A) and (C) Only 

  4. D.

    (B) and (D) Only 

Attempted by 107 students.

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Correct answer: D

Interpretation: A radical person (R) is electable (E) if and only if they are conservative (C); if they are not conservative then they are not electable. In logical terms this means: if R and C then E, and if R and not C then not E.

  • Formalize the two clauses: (R ∧ C) → E and (R ∧ ¬C) → ¬E.

  • Combine them to get: R → ((C → E) ∧ (¬C → ¬E)). This is equivalent to R → (E ↔ C).

Now check each given assertion against this interpretation:

  • Assertion (R ∧ E) ↔ C: This is too strong because it relates C to R ∧ E even when R is false. For example, if R is false, C true and E false, the original sentence imposes no constraint but this formula would be false. Hence it is not equivalent to the intended meaning.

  • Assertion R → (E ↔ C): This exactly matches the combined interpretation above, so it is correct.

  • Assertion R → ((C → E) ∨ ¬E): This is too weak. Counterexample: let R = true, C = false, E = true. The assertion becomes true, but the original sentence requires that a radical non-conservative person not be electable (E false), so this formula does not capture the original meaning.

  • Assertion (¬R ∨ ¬E ∨ C) ∧ (¬R ∨ ¬C ∨ E): This CNF is equivalent to the pair of implications (R ∧ E) → C and (R ∧ C) → E. For R true these force E ↔ C, so this formula is logically equivalent to R → (E ↔ C) and is therefore correct.

Conclusion: The correct assertions are the implication R → (E ↔ C) and its equivalent conjunctive normal form (¬R ∨ ¬E ∨ C) ∧ (¬R ∨ ¬C ∨ E).

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