/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 the bijection { a ↦ 0, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ ,
  0 1 ⟶ 1 2 0 0 ,
  2 2 ⟶ 1 }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  1 0 ⟶ ,
  1 0 ⟶ 0 0 2 1 ,
  2 2 ⟶ 1 }

Loop of length 10 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aaacaa
	rule ba-> aacb at position 0
.aacb.aaacaa
	rule ba-> aacb at position 3
.aacaacb.aacaa
	rule a->  at position 3
.aacacb.aacaa
	rule a->  at position 3
.aaccb.aacaa
	rule cc-> b at position 2
.aabb.aacaa
	rule ba-> aacb at position 3
.aabaacb.acaa
	rule ba->  at position 6
.aabaac.caa
	rule cc-> b at position 5
.aabaab.aa
	rule ba-> aacb at position 5
.aabaaaacb.a
	rule ba-> aacb at position 8
.aabaaaacaacb.