= Solution
{c}
We first prove that any two nonzero principal <left ideals> $Ra$ and $Rb$ intersect nontrivially. Suppose instead that $Ra\cap Rb=0$. The nonzero <left ideals>
$$
Ra,\quad Rab,\quad Rab^2,\quad\ldots
$$
then form a <direct sum>. Indeed, take a finite relation
$$
\sum_{i=0}^m u_iab^i=0.
$$
It implies
$$
u_0a=-\left(\sum_{i=1}^m u_iab^{i-1}\right)b\in Ra\cap Rb=0.
$$
As $R$ is a <noncommutative domain> and $a\ne0$, this gives $u_0=0$. Right cancellation of the nonzero element $b$ reduces the remaining relation to one of the same form with one fewer term. Repetition gives $u_1=\cdots=u_m=0$, proving the asserted <direct sum>.
Consequently its finite partial sums give a strictly ascending chain of <left ideals>, contradicting the <ascending chain condition> in a <left Noetherian ring>. Thus $Ra\cap Rb\ne0$ whenever $a,b\ne0$. This is the <left Noetherian domain Ore lemma>.
Now take $s\in S=R\setminus\{0\}$ and $r\in R$. If $r=0$, choose $s'=1$ and $r'=0$. Otherwise choose
$$
0\ne z\in Rr\cap Rs,\qquad z=s'r=r's.
$$
Since $z\ne0$, the coefficient $s'$ is nonzero and hence belongs to $S$. This proves the <left Ore set> condition in all cases.
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