/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 0 ,
  2 2 -> 1 1 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 ,
  2 2 -> 0 1 2 1 1 }


Loop of length 17 starting with a string of length 6
using right expansion and the encoding { 0->a, 1->b, ... }:

.cc.acac
	rule cc-> abcbb at position 0
.abcbb.acac
	rule ba-> ac at position 4
.abcbac.cac
	rule a->  at position 4
.abcbc.cac
	rule cc-> abcbb at position 4
.abcbabcbb.ac
	rule ba-> ac at position 3
.abcacbcbb.ac
	rule a->  at position 3
.abccbcbb.ac
	rule ba-> ac at position 7
.abccbcbac.c
	rule ba-> ac at position 6
.abccbcacc.c
	rule a->  at position 6
.abccbccc.c
	rule cc-> abcbb at position 5
.abccbabcbbc.c
	rule ba-> ac at position 4
.abccacbcbbc.c
	rule cc-> abcbb at position 10
.abccacbcbbabcbb.
	rule ba-> ac at position 9
.abccacbcbacbcbb.
	rule ba-> ac at position 8
.abccacbcaccbcbb.
	rule a->  at position 8
.abccacbcccbcbb.
	rule cc-> abcbb at position 7
.abccacbabcbbcbcbb.
	rule ba-> ac at position 6
.abccacacbcbbcbcbb.