/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 5-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  0 1 ⟶ 1 0 2 0 ,
  1 ⟶ ,
  2 2 ⟶ }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 5-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  1 0 ⟶ 0 2 0 1 ,
  1 ⟶ ,
  2 2 ⟶ }

Loop of length 17 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.aaa
	rule a-> b at position 0
.b.aaa
	rule ba-> acab at position 0
.acab.aa
	rule a-> b at position 2
.acbb.aa
	rule ba-> acab at position 3
.acbacab.a
	rule ba-> acab at position 2
.acacabcab.a
	rule a->  at position 2
.accabcab.a
	rule cc->  at position 1
.aabcab.a
	rule a-> b at position 4
.aabcbb.a
	rule ba-> acab at position 5
.aabcbacab.
	rule ba-> acab at position 4
.aabcacabcab.
	rule a->  at position 4
.aabccabcab.
	rule cc->  at position 3
.aababcab.
	rule ba-> acab at position 2
.aaacabbcab.
	rule a->  at position 4
.aaacbbcab.
	rule b->  at position 4
.aaacbcab.
	rule b->  at position 4
.aaaccab.
	rule cc->  at position 3
.aaaab.