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


--------------------------------------------------------------------------------


NO

After renaming modulo the bijection { 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 the bijection { 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.