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I have a language $L$ which is NP-hard and I have another language $L_1$, s.t. if I take an instance $q$ of the decision problem corresponding to $L$, and if one of polynomially many instances, $f_1(q),\ldots,f_{p(|q|)}(q)$ is in $L_1$, then $q\in L$.

Is this enough to say that $L_1$ is NP-hard?

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Do you have a polynomial-time algorithm to construct those polynomially instances? If you can construct those polynomially many instances in polynomial time, then you have a Cook reduction (also known as a polynomial-time Turing reduction), not a many-one reduction. So your question then becomes: if we have a Cook reduction from $L$ to $L_1$, and if $L$ is NP-hard, then does it follow that $L_1$ is NP-hard?

The answer depends on the definition of NP-hard that you use, which is a little tricky, so let me start with a simpler question:

If we have a Cook reduction from $L$ to $L_1$, and if $L$ is NP-complete, then does it follow that $L_1$ is NP-complete? Answer: No. But it is strong evidence, if it is also known that $L_1$ is in NP. NP-completeness is defined in terms of many-one reductions, not Cook reductions. The existence of a Cook reduction is not currently known to imply the existence of a many-one reduction (it is a famous open problem, and it might be true - we don't know of any counterexamples - but we don't have a proof of such an implication). See Graph problem known to be $NP$-complete only under Cook reduction, Can we construct a Karp reduction from a Cook reduction between NP problems?.

Back to the question about NP-hardness. That is harder to answer, because there is not universal agreement about what precisely NP-hard means. There are multiple reasonable ways to define NP-hard. One definition is in terms of many-one reductions ($H$ is NP-hard if for every $G$ in NP, there is a many-one reduction from $G$ to $H$), and another definition is in terms of Cook reductions ($H$ is NP-hard if for every $G$ in NP, there is a Cook reduction from $G$ to $H$). The answer to the question is "no" for the first, "yes" for the second.

In particular, if we have a Cook reduction from $L$ to $L_1$, and if $L$ is NP-hard under Cook reductions, then it does follow that $L_1$ is NP-hard under Cook reductions. If we have a Cook reduction from $L$ to $L_1$, and if $L$ is NP-hard under many-one reductions, then it is not known whether this implies $L_1$ is NP-hard under many-one reductions (in particular consider $L$ to be SAT and $L_1$ to be the complement of SAT; then if NP $\ne$ co-NP, as is widely conjectured, then $L$ is NP-hard under many-one reductions but the complement of SAT is not NP-hard under many-one reductions). See also If a problem is Cook-NP hard, and this problem is in NP, does it prove that the problem is Karp-NP-complete?.

Finally, if you don't have a polynomial-time algorithm to construct those polynomially instances, then you cannot conclude anything at all. For instance, consider the case where $L_1$ is a language in P, and $L$ is 3SAT; if you have exponential time, you can determine whether $q$ is in $L$, and then let $f_1(q)$ be a member of $L_1$ if $q$ is in $L$ or a non-member of $L_1$ if $q$ is not in $L$.

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  • $\begingroup$ Thank you for the elaborate explanation. Each of my polynomially many instances can be constructed polynomially. It seems like I do have a Cook reduction in my case, and the language is in NP. If I understood the other sources you linked to correctly, I cannot conclude that this is NPC unless I have some special properties of the language I’m dealing with? Also, if this problem is in P, it still proves that P=NP. So perhaps at least that is something useful I can say. $\endgroup$
    – NL1992
    Feb 10, 2023 at 20:17
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    $\begingroup$ @NL1992, yes, that sounds right to me. As a heuristic, if you have found a Cook reduction, then my guess is that with more work it will likely be possible to find a many-one reduction. $\endgroup$
    – D.W.
    Feb 11, 2023 at 7:20

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