#075 baaababba=aba

Original presentation

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

Modified presentation

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

Complete rewriting system

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

\( \begin{aligned} a c &\rightarrow c d b c \\ b a &\rightarrow c \\ a d &\rightarrow d^{2} b c \\ c d b d &\rightarrow d^{2} b c \\ b c d b c &\rightarrow c^{2} \\ c d c^{2} &\rightarrow d \\ c d c d &\rightarrow d^{2} c^{2} \\ c^{2} d b c &\rightarrow b d \\ b d^{2} b c &\rightarrow c d \\ a b d &\rightarrow c d b^{2} d \\ c d^{2} &\rightarrow d^{2} c^{4} \\ c d c b d &\rightarrow d c d b c \\ b d^{2} c d &\rightarrow d^{2} c^{4} b d \\ b d^{2} b d &\rightarrow d^{2} c^{6} \\ b d^{2} c^{2} &\rightarrow d^{2} c^{4} b c \\ c^{2} d b^{2} d &\rightarrow b d c d b c \\ b d^{2} b^{2} d &\rightarrow d^{2} c^{2} b c \\ b c d b^{2} d &\rightarrow c b d \\ b d^{2} c b d &\rightarrow d^{2} c^{4} b^{2} d \\ b d^{3} &\rightarrow d^{2} c^{4} b d c^{2} \end{aligned} \)