/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO

After renaming modulo { a->0, c->1, b->2 }, 
it remains to prove termination of the 6-rule system
{ 0 1 0 -> 1 0 0 ,
  2 2 1 -> 2 0 2 ,
  1 0 0 -> 0 0 2 ,
  1 2 1 -> 2 1 0 ,
  0 2 0 ->= 0 1 1 ,
  2 0 2 ->= 2 2 2 }


Loop of length 22 starting with a string of length 11
using right expansion and the encoding { 0->a, 1->b, ... }:

.aba.aaaacbaa
	rule aba-> baa at position 0
.baa.aaaacbaa
	rule baa-> aac at position 0
.aac.aaaacbaa
	rule aca->= abb at position 1
.aabb.aaacbaa
	rule baa-> aac at position 3
.aabaac.acbaa
	rule aca->= abb at position 4
.aabaabb.cbaa
	rule baa-> aac at position 2
.aaaacbb.cbaa
	rule bcb-> cba at position 6
.aaaacbcba.aa
	rule bcb-> cba at position 5
.aaaaccbaa.aa
	rule ccb-> cac at position 4
.aaaacacaa.aa
	rule aca->= abb at position 5
.aaaacabba.aa
	rule aca->= abb at position 3
.aaaabbbba.aa
	rule baa-> aac at position 7
.aaaabbbaac.a
	rule aca->= abb at position 8
.aaaabbbaabb.
	rule baa-> aac at position 6
.aaaabbaacbb.
	rule baa-> aac at position 5
.aaaabaaccbb.
	rule aba-> baa at position 3
.aaabaaaccbb.
	rule aba-> baa at position 2
.aabaaaaccbb.
	rule aba-> baa at position 1
.abaaaaaccbb.
	rule ccb-> cac at position 7
.abaaaaacacb.
	rule aca->= abb at position 6
.abaaaaabbcb.
	rule bcb-> cba at position 8
.abaaaaabcba.
	rule bcb-> cba at position 7
.abaaaaacbaa.