/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 3-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 2 1 0 2 0 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 3-rule system
{ 0 ⟶ ,
  1 0 ⟶ 0 0 2 0 1 2 1 ,
  2 2 ⟶ }

Loop of length 10 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aa
	rule ba-> aacabcb at position 0
.aacabcb.aa
	rule ba-> aacabcb at position 6
.aacabcaacabcb.a
	rule a->  at position 6
.aacabcacabcb.a
	rule a->  at position 6
.aacabccabcb.a
	rule cc->  at position 5
.aacababcb.a
	rule ba-> aacabcb at position 8
.aacababcaacabcb.
	rule a->  at position 8
.aacababcacabcb.
	rule a->  at position 8
.aacababccabcb.
	rule cc->  at position 7
.aacabababcb.
	rule ba-> aacabcb at position 6
.aacabaaacabcbbcb.