Consider the complexity class \(๐ถ๐‘‚โˆ’๐‘๐‘ƒ\) as the set of languages \(L\) suchโ€ฆ

2019

Consider the complexity classย \(๐ถ๐‘‚โˆ’๐‘๐‘ƒ\)ย as the set of languagesย \(L\) such thatย \(\overline{L} \in NP\), and the following two statements:

\(S_1: \: P \subseteq CO-NP\)

\(S_2: \: \text{ If } NP \neq CO-NP, \text{ then } P \neq NP\)

Which of the following is/are correct?

  1. A.

    Onlyย \(๐‘†_1\)

  2. B.

    ย Onlyย \(๐‘†_2\)

  3. C.

    ย Bothย \(๐‘†_1\) andย \(๐‘†_2\)

  4. D.

    ย Neitherย \(๐‘†_1\) norย \(๐‘†_2\)

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Correct answer: C

Answer: Both statements are true.

Key facts: CO-NP is the class of languages whose complements lie in NP. The class P is closed under complement (if a language is in P then its complement is also in P), and P โІ NP.

  • Proof that the first statement is true: For any language L in P, the complement of L is also in P because P is closed under complement. Since P โІ NP, the complement of L is in NP. By definition of CO-NP, this means L โˆˆ CO-NP. Therefore P โІ CO-NP.

  • Proof that the second statement is true: Prove the contrapositive. If P = NP, then NP = P is closed under complement, so NP is closed under complement and thus NP = CO-NP. Taking the contrapositive of this implication gives: if NP โ‰  CO-NP then P โ‰  NP.

Conclusion: Both statements hold, so the correct choice is that both are true.

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