#462 abbaaababa=baabbaaa

Original presentation

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

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

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

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