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


Loop of length 21 starting with a string of length 8
using right expansion and the encoding { 0->a, 1->b, ... }:

.ac.cccbbc
	rule ac-> ccbb at position 0
.ccbb.cccbbc
	rule bc-> aa at position 3
.ccbaa.ccbbc
	rule ac-> ccbb at position 4
.ccbaccbb.cbbc
	rule ac-> ccbb at position 3
.ccbccbbcbb.cbbc
	rule bc-> aa at position 2
.ccaacbbcbb.cbbc
	rule ac-> ccbb at position 3
.ccaccbbbbcbb.cbbc
	rule bc-> aa at position 8
.ccaccbbbaabb.cbbc
	rule ab->  at position 9
.ccaccbbbab.cbbc
	rule bc-> aa at position 9
.ccaccbbbaaa.bbc
	rule ab->  at position 10
.ccaccbbbaa.bc
	rule ab->  at position 9
.ccaccbbba.c
	rule ac-> ccbb at position 8
.ccaccbbbccbb.
	rule bc-> aa at position 7
.ccaccbbaacbb.
	rule ac-> ccbb at position 8
.ccaccbbaccbbbb.
	rule ac-> ccbb at position 7
.ccaccbbccbbcbbbb.
	rule bc-> aa at position 6
.ccaccbaacbbcbbbb.
	rule ac-> ccbb at position 7
.ccaccbaccbbbbcbbbb.
	rule ac-> ccbb at position 6
.ccaccbccbbcbbbbcbbbb.
	rule bc-> aa at position 5
.ccaccaacbbcbbbbcbbbb.
	rule ac-> ccbb at position 6
.ccaccaccbbbbcbbbbcbbbb.
	rule ac-> ccbb at position 5
.ccaccccbbcbbbbcbbbbcbbbb.