/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 4-rule system { 0 -> , 0 1 -> 1 2 1 0 , 1 -> 0 , 2 2 -> } Loop of length 20 starting with a string of length 4 using right expansion and the encoding { 0->a, 1->b, ... }: .ab.bb rule ab-> bcba at position 0 .bcba.bb rule b-> a at position 0 .acba.bb rule b-> a at position 2 .acaa.bb rule ab-> bcba at position 3 .acabcba.b rule ab-> bcba at position 2 .acbcbacba.b rule b-> a at position 2 .acacbacba.b rule a-> at position 2 .accbacba.b rule cc-> at position 1 .abacba.b rule b-> a at position 4 .abacaa.b rule ab-> bcba at position 5 .abacabcba. rule ab-> bcba at position 4 .abacbcbacba. rule b-> a at position 4 .abacacbacba. rule a-> at position 4 .abaccbacba. rule cc-> at position 3 .ababacba. rule ab-> bcba at position 2 .abbcbaacba. rule b-> a at position 4 .abbcaaacba. rule a-> at position 4 .abbcaacba. rule a-> at position 4 .abbcacba. rule a-> at position 4 .abbccba. rule cc-> at position 3 .abbba.