 Right Ore condition

A multiplicative subset $S\subseteq A$ with $1\in S$ and $0\notin S$ of a ring $A$ satisfies the right Ore condition if for every $a\in A$ and $s\in S$ there are $u\in S$ and $b\in A$ such that $au=sb$. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a ring of right fractions $a{s}^{-1}$; every nonzero fraction is invertible. The usual additional denominator reversibility condition is automatic when the elements of $S$ are nonzero in a noncommutative domain.