/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 ,
  1 1 2 ⟶ 0 2 2 2 0 0 }

Loop of length 15 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.bcccbcc
	rule a-> b at position 0
.b.bcccbcc
	rule bbc-> acccaa at position 0
.acccaa.ccbcc
	rule a-> b at position 4
.acccba.ccbcc
	rule a-> b at position 5
.acccbb.ccbcc
	rule bbc-> acccaa at position 4
.acccacccaa.cbcc
	rule a-> b at position 8
.acccacccba.cbcc
	rule a-> b at position 9
.acccacccbb.cbcc
	rule bbc-> acccaa at position 8
.acccacccacccaa.bcc
	rule a-> b at position 13
.acccacccacccab.bcc
	rule bbc-> acccaa at position 13
.acccacccacccaacccaa.c
	rule a-> b at position 13
.acccacccacccabcccaa.c
	rule a-> b at position 17
.acccacccacccabcccba.c
	rule a-> b at position 18
.acccacccacccabcccbb.c
	rule bbc-> acccaa at position 17
.acccacccacccabcccacccaa.
	rule a-> b at position 17
.acccacccacccabcccbcccaa.