#374 abaabababa=babaabaa

Original presentation

\( \left\langle a, b \;\middle|\; a b a^{2} b a b a b a = b a b a^{2} b a^{2} \right\rangle \)

\( \left\langle a, b, c, d \;\middle|\; a c a c^{3} = c^{2} a c a,\; c = d,\; d = b a \right\rangle \)

Using reversed recursive path order (RevRPOCmp) with \(d < a < b < c\):

\( \begin{aligned} c &\rightarrow d \\ b a &\rightarrow d \\ d^{2} a d a &\rightarrow a d a d^{3} \end{aligned} \)