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#173 baabbabbba=a

Original presentation

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

Modified presentation

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

Complete rewriting system

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

\( \begin{aligned} c b &\rightarrow d \\ b a &\rightarrow c a d c \\ d c a d c &\rightarrow c^{2} \\ c a c^{2} &\rightarrow a \\ b c a d c &\rightarrow c \\ d a &\rightarrow c^{2} a d c \\ b c a d^{2} &\rightarrow d \\ c a c a &\rightarrow a^{2} c^{2} \\ d c a d^{2} &\rightarrow c d \\ a b &\rightarrow c a c d \\ b c a c d &\rightarrow c a d^{2} \\ d c a c d &\rightarrow c^{2} a d^{2} \\ c a^{2} &\rightarrow a^{2} c^{4} \end{aligned} \)