/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 -> 1 2 0 0 0 , 0 2 2 -> 1 } The system was reversed. After renaming modulo { 0->0, 1->1, 2->2 }, it remains to prove termination of the 3-rule system { 0 -> , 1 0 -> 0 0 0 2 1 , 2 2 0 -> 1 } Loop of length 14 starting with a string of length 6 using right expansion and the encoding { 0->a, 1->b, ... }: .ba.aaaa rule ba-> aaacb at position 0 .aaacb.aaaa rule ba-> aaacb at position 4 .aaacaaacb.aaa rule a-> at position 4 .aaacaacb.aaa rule a-> at position 4 .aaacacb.aaa rule a-> at position 4 .aaaccb.aaa rule ba-> aaacb at position 5 .aaaccaaacb.aa rule cca-> b at position 3 .aaabaacb.aa rule ba-> aaacb at position 7 .aaabaacaaacb.a rule a-> at position 7 .aaabaacaacb.a rule a-> at position 7 .aaabaacacb.a rule a-> at position 7 .aaabaaccb.a rule ba-> aaacb at position 8 .aaabaaccaaacb. rule cca-> b at position 6 .aaabaabaacb. rule ba-> aaacb at position 6 .aaabaaaaacbacb.