/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 ⟶ 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, ... }:

.aabc.abccbcc
	rule aabc-> ccaaaa at position 0
.ccaaaa.abccbcc
	rule a-> b at position 4
.ccaaba.abccbcc
	rule aabc-> ccaaaa at position 5
.ccaabccaaaa.cbcc
	rule aabc-> ccaaaa at position 2
.ccccaaaacaaaa.cbcc
	rule a-> b at position 7
.ccccaaabcaaaa.cbcc
	rule aabc-> ccaaaa at position 5
.ccccaccaaaaaaaa.cbcc
	rule a-> b at position 11
.ccccaccaaaabaaa.cbcc
	rule a-> b at position 14
.ccccaccaaaabaab.cbcc
	rule aabc-> ccaaaa at position 12
.ccccaccaaaabccaaaa.bcc
	rule aabc-> ccaaaa at position 9
.ccccaccaaccaaaacaaaa.bcc
	rule a-> b at position 14
.ccccaccaaccaaabcaaaa.bcc
	rule a-> b at position 17
.ccccaccaaccaaabcabaa.bcc
	rule aabc-> ccaaaa at position 18
.ccccaccaaccaaabcabccaaaa.c
	rule a-> b at position 20
.ccccaccaaccaaabcabccbaaa.c
	rule a-> b at position 23
.ccccaccaaccaaabcabccbaab.c
	rule aabc-> ccaaaa at position 21
.ccccaccaaccaaabcabccbccaaaa.