/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 ,
  0 0 1 2 ⟶ 1 2 2 0 0 0 }

Loop of length 12 starting with a string of length 11
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aabc.abccbcc
	rule aabc-> bccaaa at position 0
.bccaaa.abccbcc
	rule aabc-> bccaaa at position 5
.bccaabccaaa.cbcc
	rule aabc-> bccaaa at position 3
.bccbccaaacaaa.cbcc
	rule a-> b at position 8
.bccbccaabcaaa.cbcc
	rule aabc-> bccaaa at position 6
.bccbccbccaaaaaa.cbcc
	rule a-> b at position 14
.bccbccbccaaaaab.cbcc
	rule aabc-> bccaaa at position 12
.bccbccbccaaabccaaa.bcc
	rule aabc-> bccaaa at position 10
.bccbccbccabccaaacaaa.bcc
	rule a-> b at position 15
.bccbccbccabccaabcaaa.bcc
	rule aabc-> bccaaa at position 18
.bccbccbccabccaabcabccaaa.c
	rule a-> b at position 23
.bccbccbccabccaabcabccaab.c
	rule aabc-> bccaaa at position 21
.bccbccbccabccaabcabccbccaaa.