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

Loop of length 16 starting with a string of length 11
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.aacaaccacc
	rule a-> b at position 0
.b.aacaaccacc
	rule baac-> ccaaaa at position 0
.ccaaaa.aaccacc
	rule a-> b at position 2
.ccbaaa.aaccacc
	rule a-> b at position 5
.ccbaab.aaccacc
	rule baac-> ccaaaa at position 5
.ccbaaccaaaa.cacc
	rule baac-> ccaaaa at position 2
.ccccaaaacaaaa.cacc
	rule a-> b at position 5
.ccccabaacaaaa.cacc
	rule baac-> ccaaaa at position 5
.ccccaccaaaaaaaa.cacc
	rule a-> b at position 9
.ccccaccaabaaaaa.cacc
	rule a-> b at position 12
.ccccaccaabaabaa.cacc
	rule baac-> ccaaaa at position 12
.ccccaccaabaaccaaaa.acc
	rule baac-> ccaaaa at position 9
.ccccaccaaccaaaacaaaa.acc
	rule a-> b at position 18
.ccccaccaaccaaaacaaba.acc
	rule baac-> ccaaaa at position 18
.ccccaccaaccaaaacaaccaaaa.c
	rule a-> b at position 21
.ccccaccaaccaaaacaaccabaa.c
	rule baac-> ccaaaa at position 21
.ccccaccaaccaaaacaaccaccaaaa.