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


Loop of length 23 starting with a string of length 4
using right expansion and the encoding { 0->a, 1->b, ... }:

.ba.aa
	rule ba-> cacbb at position 0
.cacbb.aa
	rule ba-> cacbb at position 4
.cacbcacbb.a
	rule a->  at position 5
.cacbccbb.a
	rule cc-> ab at position 4
.cacbabbb.a
	rule ba-> cacbb at position 3
.caccacbbbbb.a
	rule cc-> ab at position 2
.caabacbbbbb.a
	rule ba-> cacbb at position 10
.caabacbbbbcacbb.
	rule a->  at position 11
.caabacbbbbccbb.
	rule cc-> ab at position 10
.caabacbbbbabbb.
	rule ba-> cacbb at position 9
.caabacbbbcacbbbbb.
	rule a->  at position 10
.caabacbbbccbbbbb.
	rule cc-> ab at position 9
.caabacbbbabbbbbb.
	rule ba-> cacbb at position 8
.caabacbbcacbbbbbbbb.
	rule a->  at position 9
.caabacbbccbbbbbbbb.
	rule cc-> ab at position 8
.caabacbbabbbbbbbbb.
	rule ba-> cacbb at position 7
.caabacbcacbbbbbbbbbbb.
	rule a->  at position 8
.caabacbccbbbbbbbbbbb.
	rule cc-> ab at position 7
.caabacbabbbbbbbbbbbb.
	rule ba-> cacbb at position 6
.caabaccacbbbbbbbbbbbbbb.
	rule cc-> ab at position 5
.caabaabacbbbbbbbbbbbbbb.
	rule ba-> cacbb at position 6
.caabaacacbbcbbbbbbbbbbbbbb.
	rule a->  at position 7
.caabaaccbbcbbbbbbbbbbbbbb.
	rule cc-> ab at position 6
.caabaaabbbcbbbbbbbbbbbbbb.