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


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NO

After renaming modulo { a->0, c->1, b->2 }, 
it remains to prove termination of the 3-rule system
{ 0 0 0 -> 1 1 2 ,
  2 2 2 -> 1 1 1 ,
  1 1 1 -> 0 2 2 }


The system was reversed.

After renaming modulo { 0->0, 2->1, 1->2 }, 
it remains to prove termination of the 3-rule system
{ 0 0 0 -> 1 2 2 ,
  1 1 1 -> 2 2 2 ,
  2 2 2 -> 1 1 0 }


Loop of length 19 starting with a string of length 7
using right expansion and the encoding { 0->a, 1->b, ... }:

.bbb.aacc
	rule bbb-> ccc at position 0
.ccc.aacc
	rule ccc-> bba at position 0
.bba.aacc
	rule aaa-> bcc at position 2
.bbbcc.cc
	rule ccc-> bba at position 3
.bbbbba.c
	rule bbb-> ccc at position 2
.bbccca.c
	rule ccc-> bba at position 2
.bbbbaa.c
	rule bbb-> ccc at position 1
.bcccaa.c
	rule ccc-> bba at position 1
.bbbaaa.c
	rule aaa-> bcc at position 3
.bbbbcc.c
	rule ccc-> bba at position 4
.bbbbbba.
	rule bbb-> ccc at position 3
.bbbccca.
	rule ccc-> bba at position 3
.bbbbbaa.
	rule bbb-> ccc at position 2
.bbcccaa.
	rule ccc-> bba at position 2
.bbbbaaa.
	rule aaa-> bcc at position 4
.bbbbbcc.
	rule bbb-> ccc at position 2
.bbccccc.
	rule ccc-> bba at position 2
.bbbbacc.
	rule bbb-> ccc at position 1
.bcccacc.
	rule ccc-> bba at position 1
.bbbaacc.