In a regular category, an exact sequence is a diagram which is both a coequalizer and a kernel pair: $$R\overset r{\underset s\rightrightarrows} X\to Y$$
Wiki says that in the abelian case, the above sequence is exact in the regular sense iff the following sequence is exact in the abelian sense: $$0\to R\xrightarrow{r-s}X\to Y\to 0$$
This seems strange to me. First of all, $r,s$ are jointly monic but I don't see why $r-s$ should be a mono. Second, if it is, then $P$ would have to be both the domain of kernel of $X\rightarrow Y$ and of its kernel pair, but that seems to contradict the answer to this question.
Is wiki wrong?
As Zhen Lin says in the comments, wiki is wrong.
The exactness of the sequence $R\overset r{\underset s\rightrightarrows} X\xrightarrow f Y$ is equivalent to exactness - in the abelian sense - of the sequence $0\to R\xrightarrow{\begin{pmatrix}r\\s\end{pmatrix}}X\oplus X\xrightarrow{(f,-f)} Y\to 0$.
You can prove $(P,r,s)$ is the kernel pair of $f$ iff $\begin{pmatrix}r\\s\end{pmatrix}$ is the kernel of $(f,-f)$. This is consistent with the linked question - the kernel pair of $f$ is bigger than the kernel of $f$.