/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 0 ⟶ 0 1 ,
  1 0 2 ⟶ 2 2 0 0 0 }

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

.aa.acaaccacc
	rule aa-> ab at position 0
.ab.acaaccacc
	rule bac-> ccaaa at position 1
.accaaa.aaccacc
	rule aa-> ab at position 3
.accaba.aaccacc
	rule aa-> ab at position 5
.accabab.accacc
	rule bac-> ccaaa at position 6
.accabaccaaa.cacc
	rule bac-> ccaaa at position 4
.accaccaaacaaa.cacc
	rule aa-> ab at position 6
.accaccabacaaa.cacc
	rule bac-> ccaaa at position 7
.accaccaccaaaaaa.cacc
	rule aa-> ab at position 10
.accaccaccaabaaa.cacc
	rule aa-> ab at position 12
.accaccaccaababa.cacc
	rule bac-> ccaaa at position 13
.accaccaccaabaccaaa.acc
	rule bac-> ccaaa at position 11
.accaccaccaaccaaacaaa.acc
	rule aa-> ab at position 18
.accaccaccaaccaaacaab.acc
	rule bac-> ccaaa at position 19
.accaccaccaaccaaacaaccaaa.c
	rule aa-> ab at position 21
.accaccaccaaccaaacaaccaba.c
	rule bac-> ccaaa at position 22
.accaccaccaaccaaacaaccaccaaa.