#468 baabbbbbaa=aba

Original presentation

\( \left\langle a, b \;\middle|\; b a^{2} b^{5} a^{2} = a b a \right\rangle \)

Modified presentation

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

Complete rewriting system

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

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