Skip to content

#133 bbbaabbbaa=abbba

Original presentation

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

Modified presentation

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

Complete rewriting system

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

\( \begin{aligned} b^{2} &\rightarrow c \\ a c b a &\rightarrow d \\ b c &\rightarrow c b \\ a d &\rightarrow d^{2} a \\ b d &\rightarrow c^{2} d^{2} a^{2} \\ c^{3} d^{18} a^{2} &\rightarrow d c^{3} d^{2} a^{2} \\ c^{3} d^{6} a^{4} &\rightarrow d \\ a c^{3} d^{2} a^{2} &\rightarrow c^{3} d^{14} a^{3} \\ d c b a &\rightarrow c^{3} d^{14} a^{3} \\ a c^{3} d^{4} a &\rightarrow d c^{3} d^{2} a^{2} \\ c^{3} d^{20} a &\rightarrow d c^{3} d^{4} a \\ c^{3} d^{21} &\rightarrow d c^{3} d^{5} \\ a c^{3} d^{5} &\rightarrow d c^{3} d^{4} a \\ a c^{3} d^{4} c^{3} d^{14} a^{3} &\rightarrow d c^{3} d^{5} \\ c^{3} d^{20} c^{3} d^{14} a^{3} &\rightarrow d c^{3} d^{4} c^{3} d^{14} a^{3} \end{aligned} \)