#432 baaabbbaaa=aba

Original presentation

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

Modified presentation

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

Complete rewriting system

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

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