Intuitively if A can reduce to B, and B is NP-Hard, A might be NP Hard but maybe not. If there is a way to solve A that does not involve reducing to B, it might be faster.

How do I formally disprove this statement using a counter example or some well known algorithms, or using some other technique?


1 Answer 1


Let A be the empty language. It is trivially reducible to, say 3SAT: transform any input into the formula $a \land \lnot a$. But only the empty language is reducible to the empty language, thus A is not NP-hard.

If P = NP, then any nontrivial language is NP-hard, thus it will be really hard to find counterexamples where both A and its complement are nonempty.

  • $\begingroup$ can you clarify what empty language is? Is this a SAT where all clauses contain only 𝑎 and ¬𝑎 ? $\endgroup$
    – ChaiTea
    Apr 18, 2020 at 5:57
  • $\begingroup$ In decision problems we are asked to determine if the input string is in the language. An empty language is the language containing no strings. $\endgroup$ Apr 18, 2020 at 6:33

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