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This question is somewhat of a converse to a previous question on sets formed from set operations on NP-complete sets:

If the set resulting from the union, intersection, or Cartesian product of two decidable sets $L_1$ and $L_2$ is NP-complete, is at least one of $L_1, L_2$ necessarily NP-hard? I know that they cannot both be in P (assuming P != NP) since P is closed under these set operations. I also know that the conditions of "decidable" and "NP-hard" are necessary since if we consider any NP-complete set $L$ and another set $B$ outside of NP (whether just NP-hard or undecidable) then we can form two new NP-hard sets not in NP whose intersection is NP-complete. For example: $L_1:= 01L \cup 11B$, and $L_2:= 01L \cup 00B$. However, I don't know how to proceed after that.

I'm thinking that the case of union might not be true since we can take a NP-complete set $A$ and perform the construction in Ladner's Theorem to get a set $B \in$ NPI which is a subset of $A$. Then $B \cup (A \setminus B) = A$ is the original NP-complete set. However, I don't know if $A \setminus B$ is still in NPI or NP-hard. I don't even know where to start for the case of intersection and Cartesian product.

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Ari
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1 Answers1

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The intersection of two non-NP-hard languages can be NP-hard. Example: The solutions of any 3SAT instance are the set intersection of the solutions of a HORN-3SAT instance and an ANTIHORN-3SAT instance. This is because a 3CNF clause must be either a Horn or anti-Horn clause and a 3SAT instance is the conjunction of such clauses. 3SAT is of course NP-complete; HORN-3SAT and ANTIHORN-3SAT are both in P.

Kyle Jones
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