Skip to content

#130 baaaaabbaa=aaa

Original presentation

\( \left\langle a, b \;\middle|\; b a^{5} b^{2} a^{2} = a^{3} \right\rangle \)

Modified presentation

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

Complete rewriting system

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

\( \begin{aligned} c &\rightarrow a^{2} \\ a^{2} b^{2} &\rightarrow d \\ b a^{3} d a^{2} &\rightarrow a^{3} \\ a^{2} b a^{3} &\rightarrow d a^{3} d a^{2} \\ b a^{3} d^{2} &\rightarrow a d \\ b a^{3} d a d &\rightarrow a^{2} d \\ a^{2} b a d &\rightarrow d a^{3} d^{2} \\ d a^{3} d a^{2} d^{2} &\rightarrow a^{3} d \\ d a^{3} d a^{2} d a d &\rightarrow a^{4} d \\ d a^{3} d a^{2} d a^{2} &\rightarrow a^{5} \\ a^{2} b a^{2} d &\rightarrow d a^{3} d a d \\ b a^{6} d &\rightarrow a^{4} d a^{2} d^{2} \\ b a^{7} d &\rightarrow a^{4} d a^{2} d a d \\ b a^{8} &\rightarrow a^{4} d a^{2} d a^{2} \\ d a^{3} d a^{5} d &\rightarrow a^{6} d a^{2} d^{2} \\ d a^{3} d a^{6} d &\rightarrow a^{6} d a^{2} d a d \\ d a^{3} d a^{7} &\rightarrow a^{6} d a^{2} d a^{2} \end{aligned} \)