/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 ⟶ 1 ,
  0 1 1 2 ⟶ 1 2 2 0 0 0 }

Loop of length 20 starting with a string of length 11
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abbc.bbccbcc
	rule abbc-> bccaaa at position 0
.bccaaa.bbccbcc
	rule a-> b at position 4
.bccaba.bbccbcc
	rule abbc-> bccaaa at position 5
.bccabbccaaa.cbcc
	rule abbc-> bccaaa at position 3
.bccbccaaacaaa.cbcc
	rule a-> b at position 7
.bccbccabacaaa.cbcc
	rule a-> b at position 8
.bccbccabbcaaa.cbcc
	rule abbc-> bccaaa at position 6
.bccbccbccaaaaaa.cbcc
	rule a-> b at position 11
.bccbccbccaabaaa.cbcc
	rule a-> b at position 13
.bccbccbccaababa.cbcc
	rule a-> b at position 14
.bccbccbccaababb.cbcc
	rule abbc-> bccaaa at position 12
.bccbccbccaabbccaaa.bcc
	rule abbc-> bccaaa at position 10
.bccbccbccabccaaacaaa.bcc
	rule a-> b at position 14
.bccbccbccabccabacaaa.bcc
	rule a-> b at position 15
.bccbccbccabccabbcaaa.bcc
	rule a-> b at position 17
.bccbccbccabccabbcbaa.bcc
	rule a-> b at position 19
.bccbccbccabccabbcbab.bcc
	rule abbc-> bccaaa at position 18
.bccbccbccabccabbcbbccaaa.c
	rule a-> b at position 22
.bccbccbccabccabbcbbccaba.c
	rule a-> b at position 23
.bccbccbccabccabbcbbccabb.c
	rule abbc-> bccaaa at position 21
.bccbccbccabccabbcbbccbccaaa.