/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 }, it remains to prove termination of the 2-rule system { 0 ⟶ 1 , 1 0 1 2 ⟶ 1 2 2 0 0 0 } Loop of length 20 starting with a string of length 11 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .a.abcabccbcc rule a-> b at position 0 .b.abcabccbcc rule babc-> bccaaa at position 0 .bccaaa.abccbcc rule a-> b at position 3 .bccbaa.abccbcc rule a-> b at position 5 .bccbab.abccbcc rule babc-> bccaaa at position 5 .bccbabccaaa.cbcc rule babc-> bccaaa at position 3 .bccbccaaacaaa.cbcc rule a-> b at position 6 .bccbccbaacaaa.cbcc rule a-> b at position 8 .bccbccbabcaaa.cbcc rule babc-> bccaaa at position 6 .bccbccbccaaaaaa.cbcc rule a-> b at position 10 .bccbccbccabaaaa.cbcc rule a-> b at position 12 .bccbccbccababaa.cbcc rule a-> b at position 14 .bccbccbccababab.cbcc rule babc-> bccaaa at position 12 .bccbccbccababccaaa.bcc rule babc-> bccaaa at position 10 .bccbccbccabccaaacaaa.bcc rule a-> b at position 15 .bccbccbccabccaabcaaa.bcc rule a-> b at position 18 .bccbccbccabccaabcaba.bcc rule babc-> bccaaa at position 18 .bccbccbccabccaabcabccaaa.c rule a-> b at position 21 .bccbccbccabccaabcabccbaa.c rule a-> b at position 23 .bccbccbccabccaabcabccbab.c rule babc-> bccaaa at position 21 .bccbccbccabccaabcabccbccaaa.