/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 1 0 2 0 , 2 2 1 -> 0 } 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 2 0 1 1 , 1 2 2 -> 0 } Loop of length 18 starting with a string of length 6 using right expansion and the encoding { 0->a, 1->b, ... }: .ba.aacc rule ba-> acabb at position 0 .acabb.aacc rule ba-> acabb at position 4 .acabacabb.acc rule ba-> acabb at position 3 .acaacabbcabb.acc rule a-> at position 9 .acaacabbcbb.acc rule ba-> acabb at position 10 .acaacabbcbacabb.cc rule ba-> acabb at position 9 .acaacabbcacabbcabb.cc rule a-> at position 9 .acaacabbccabbcabb.cc rule bcc-> a at position 7 .acaacabaabbcabb.cc rule a-> at position 12 .acaacabaabbcbb.cc rule bcc-> a at position 13 .acaacabaabbcba. rule ba-> acabb at position 12 .acaacabaabbcacabb. rule a-> at position 12 .acaacabaabbccabb. rule bcc-> a at position 10 .acaacabaabaabb. rule ba-> acabb at position 9 .acaacabaaacabbabb. rule a-> at position 11 .acaacabaaacbbabb. rule ba-> acabb at position 12 .acaacabaaacbacabbbb. rule ba-> acabb at position 11 .acaacabaaacacabbcabbbb. rule a-> at position 11 .acaacabaaaccabbcabbbb.