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

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 ⟶ 0 0 1 2 1 ,
  1 ⟶ 0 ,
  2 2 ⟶ }

Loop of length 9 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aa
	rule ba-> aabcb at position 0
.aabcb.aa
	rule ba-> aabcb at position 4
.aabcaabcb.a
	rule a->  at position 4
.aabcabcb.a
	rule a->  at position 4
.aabcbcb.a
	rule b-> a at position 4
.aabcacb.a
	rule a->  at position 4
.aabccb.a
	rule cc->  at position 3
.aabb.a
	rule ba-> aabcb at position 3
.aabaabcb.
	rule b-> a at position 5
.aabaaacb.