/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO After renaming modulo { a->0, b->1, c->2 }, it remains to prove termination of the 3-rule system { 0 -> 1 , 0 1 2 -> 2 2 0 0 , 2 2 -> 1 } Loop of length 12 starting with a string of length 9 using right expansion and the encoding { 0->a, 1->b, ... }: .abc.bccccc rule abc-> ccaa at position 0 .ccaa.bccccc rule abc-> ccaa at position 3 .ccaccaa.cccc rule cc-> b at position 3 .ccabaa.cccc rule a-> b at position 5 .ccabab.cccc rule abc-> ccaa at position 4 .ccabccaa.ccc rule a-> b at position 7 .ccabccab.ccc rule abc-> ccaa at position 6 .ccabccccaa.cc rule cc-> b at position 5 .ccabcbcaa.cc rule a-> b at position 8 .ccabcbcab.cc rule abc-> ccaa at position 7 .ccabcbcccaa.c rule a-> b at position 10 .ccabcbcccab.c rule abc-> ccaa at position 9 .ccabcbcccccaa.