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


Loop of length 12 starting with a string of length 7
using right expansion and the encoding { 0->a, 1->b, ... }:

.ba.aaaaa
	rule ba-> acb at position 0
.acb.aaaaa
	rule ba-> acb at position 2
.acacb.aaaa
	rule cac-> aab at position 1
.aaabb.aaaa
	rule ba-> acb at position 4
.aaabacb.aaa
	rule ba-> acb at position 6
.aaabacacb.aa
	rule cac-> aab at position 5
.aaabaaabb.aa
	rule ba-> acb at position 8
.aaabaaabacb.a
	rule ba-> acb at position 7
.aaabaaaacbcb.a
	rule ba-> acb at position 11
.aaabaaaacbcacb.
	rule cac-> aab at position 10
.aaabaaaacbaabb.
	rule ba-> acb at position 9
.aaabaaaacacbabb.
	rule cac-> aab at position 8
.aaabaaaaaabbabb.