$L=\{$$($$m$,$w$,$n$$)$| $m$ is an encoding of a non-deterministic Turing machine, $w$ is any word/string in the closure of alphabet, i.e. $w\in\Sigma^*$, $n$ is any positive integer, i.e. $n\in\Bbb{Z}$ and $n>0$, and $m$ accepts $w$ for a maximum of $n$ moves/steps/operations $\}$
Alphabet is $\Sigma=\{0,1\}$
If this language is NP Hard then what reduction proves it?
My effort to answer this question myself is:
If $L$ is NP Hard then every NP language reduces to $L$ in polynomial time deterministically.
More formally $L\in\Bbb{NPH}\implies\forall L'\in\Bbb{NP}:\;L'\le_PL$
Let $L'$ be any arbitrary NP language.
According to the assumption that $L'\in\Bbb{NP}$ by definition of the complexity class $\Bbb{NP}$ exists non-deterministic Turing machine $m$ that decides $L'$ in polynomial time.
This means that for every word/string $w\in\Sigma^*$:
If $w\in L'$ then $m$ accepts $w$ for a maximum of $c\cdot N^k$ moves/steps/operations
AND
If $w\notin L'$ then $m$ rejects $w$ for a maximum of $c\cdot N^k$ moves/steps/operations
Where $c$ and $k$ are constant non-negative integers and $N$ is the length of $w$.
If $L'\le_PL$ then exists deterministic Turing machine $m'$ that computes function $f:\Sigma^*\rightarrow\Sigma^*$ in polynomial time so that for every word/string $w\in\Sigma^*$:
$w\in L'\iff f(w)\in L$
Let $m'$ be deterministic Turing machine that in the beginning $m'$ inserts the encoding of $m$ before input word/string $w$, which takes constant time $\mathcal{O}(1)$, because the encoding of $m$ is constant word/string and thus the length of the encoding of $m$ is constant non-negative integer, then $m'$ computes the length of the input word/string $w$ by counting all symbols of $w$, which suppose to take linear time $\mathcal{O}(N)$ where $N$ is the length of the input word/string $w$, then compute the power of the computed length by the constant non-negative integer $k$, which takes polynomial time, then multiply the result by the constant non-negative integer $c$, which also takes polynomial time, and at last append the result after $mw$ and the content of the tape of $m'$ will be $(m,w,c\cdot N^k)$ where $n=c\cdot N^k$
After doing all these actions $m'$ halts.
$m'$ is indeed deterministic Turing machine and runs in polynomial time and $w\in L'\iff (m,w,c\cdot N^k)\in L$ followed by the definitions of complexity class $\Bbb{NP}$ and non-deterministic Turing machines.
Hence by definition $L'\le_PL$ because of $m'$
Conclusion: For arbitrary language $L'$ I showed that $L'\le_PL$ holds.
Therefore every NP language reduces to $L$ in polynomial time deterministically and by definition $L$ is NP Hard.
What I wrote above suppose to be a proof that $L$ is NP Hard, but the proof doesn't reduce already proven NP Hard language to $L$, so I think that I did some mistake.
