/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 ⟶ 1 ,
  0 2 ⟶ ,
  2 1 1 ⟶ 0 0 0 2 2 }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 3-rule system
{ 0 ⟶ 1 ,
  2 0 ⟶ ,
  1 1 2 ⟶ 2 2 0 0 0 }

Loop of length 12 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.bcbcc
	rule a-> b at position 0
.b.bcbcc
	rule bbc-> ccaaa at position 0
.ccaaa.bcc
	rule a-> b at position 2
.ccbaa.bcc
	rule a-> b at position 3
.ccbba.bcc
	rule a-> b at position 4
.ccbbb.bcc
	rule bbc-> ccaaa at position 4
.ccbbccaaa.c
	rule bbc-> ccaaa at position 2
.ccccaaacaaa.c
	rule a-> b at position 6
.ccccaabcaaa.c
	rule a-> b at position 8
.ccccaabcbaa.c
	rule a-> b at position 9
.ccccaabcbba.c
	rule a-> b at position 10
.ccccaabcbbb.c
	rule bbc-> ccaaa at position 9
.ccccaabcbccaaa.