/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO After renaming modulo { a->0, b->1, c->2 }, it remains to prove termination of the 3-rule system { 0 -> , 0 -> 1 2 , 2 1 1 1 -> 1 0 1 0 } Loop of length 15 starting with a string of length 10 using right expansion and the encoding { 0->a, 1->b, ... }: .a.bbbbbbbbb rule a-> bc at position 0 .bc.bbbbbbbbb rule cbbb-> baba at position 1 .bbaba.bbbbbb rule a-> bc at position 4 .bbabbc.bbbbbb rule cbbb-> baba at position 5 .bbabbbaba.bbb rule a-> bc at position 6 .bbabbbbcba.bbb rule a-> bc at position 9 .bbabbbbcbbc.bbb rule cbbb-> baba at position 10 .bbabbbbcbbbaba. rule cbbb-> baba at position 7 .bbabbbbbabaaba. rule a-> bc at position 8 .bbabbbbbbcbaaba. rule a-> at position 11 .bbabbbbbbcbaba. rule a-> at position 11 .bbabbbbbbcbba. rule a-> bc at position 12 .bbabbbbbbcbbbc. rule cbbb-> baba at position 9 .bbabbbbbbbabac. rule a-> at position 10 .bbabbbbbbbbac. rule a-> bc at position 11 .bbabbbbbbbbbcc.