/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 ⟶ 1 ,
  1 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, ... }:

.aa.acaaccacc
	rule aa-> b at position 0
.b.acaaccacc
	rule bac-> ccaaaa at position 0
.ccaaaa.aaccacc
	rule aa-> b at position 2
.ccbaa.aaccacc
	rule aa-> b at position 4
.ccbab.accacc
	rule bac-> ccaaaa at position 4
.ccbaccaaaa.cacc
	rule bac-> ccaaaa at position 2
.ccccaaaacaaaa.cacc
	rule aa-> b at position 5
.ccccabacaaaa.cacc
	rule bac-> ccaaaa at position 5
.ccccaccaaaaaaaa.cacc
	rule aa-> b at position 9
.ccccaccaabaaaa.cacc
	rule aa-> b at position 11
.ccccaccaababa.cacc
	rule bac-> ccaaaa at position 11
.ccccaccaabaccaaaa.acc
	rule bac-> ccaaaa at position 9
.ccccaccaaccaaaacaaaa.acc
	rule aa-> b at position 18
.ccccaccaaccaaaacaab.acc
	rule bac-> ccaaaa at position 18
.ccccaccaaccaaaacaaccaaaa.c
	rule aa-> b at position 21
.ccccaccaaccaaaacaaccaba.c
	rule bac-> ccaaaa at position 21
.ccccaccaaccaaaacaaccaccaaaa.