/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 2 ,
  2 1 1 1 ⟶ 1 0 1 0 }

Loop of length 15 starting with a string of length 10
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.bbbbbbbbb
	rule a-> bc at position 0
.bc.bbbbbbbbb
	rule cbbb-> baba at position 1
.bbaba.bbbbbb
	rule a-> bc at position 4
.bbabbc.bbbbbb
	rule cbbb-> baba at position 5
.bbabbbaba.bbb
	rule a-> bc at position 6
.bbabbbbcba.bbb
	rule a-> bc at position 9
.bbabbbbcbbc.bbb
	rule cbbb-> baba at position 10
.bbabbbbcbbbaba.
	rule cbbb-> baba at position 7
.bbabbbbbabaaba.
	rule a-> bc at position 8
.bbabbbbbbcbaaba.
	rule a->  at position 11
.bbabbbbbbcbaba.
	rule a->  at position 11
.bbabbbbbbcbba.
	rule a-> bc at position 12
.bbabbbbbbcbbbc.
	rule cbbb-> baba at position 9
.bbabbbbbbbabac.
	rule a->  at position 10
.bbabbbbbbbbac.
	rule a-> bc at position 11
.bbabbbbbbbbbcc.