I am trying to understand the proof of Theorem 7.1.1 of `Noncommutative Rings' I. N. Herstein.
Definition: An element in a ring is said to be regular if it is neither a left nor a right zero divisor in $R$.
Let $Q(R)$ be the left quotient ring of $R$.
Theorem 7.1.1: A necessary and sufficient condition that $R$ have a left quotient ring is: given $a,b \in R$ with $b$ regular then $\exists a_1, b_1 \in R$ with $b_1$ regular such that $b_1 a = a_1 b$.
The proof $Q(R) \Rightarrow$ Ore condition is straight forward. However given the Ore condition, WTS there exists $Q(R)$, I have been stuck on a detail on the proof which says: Let $\mathcal{M} := \{(a,b) | a,b \in R, b\text{ regular} \}$ and we define the following relation $(a,b) \sim (c,d)$ if $d_1a = b_1c$ where $b_1d = d_1b$, $b_1$ regular. Hence it also follows that $d_1$ is regular.
I can figure out why $d_1$ is not a right zero divisor from the equation $b_1d = d_1b$ where $b_1,d$ and $b$ are regular elements. How do we show that $d_1$ is also NOT a left zero divisor?
This is covered in, for example, II.3 of Michael Artin’s notes https://math.mit.edu/~etingof/artinnotes.pdf
We have $t=d_1b$ is regular (since it equals $b_1d$ which is regular). We next use the Ore condition to write $xt=yb$ with $y$ regular. Now $$ xd_1b = xt = yb $$ so $$ (xd_1-y)b=0. $$ Since $b$ is regular we get $xd_1=y$, which is regular, and hence $d_1$ is not a left zero divisor.