/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 ,
  1 1 0 2 ⟶ 0 2 2 0 0 0 }

The system was reversed.

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

Loop of length 24 starting with a string of length 10
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cabb.bbbbbb
	rule cabb-> aaacca at position 0
.aaacca.bbbbbb
	rule cabb-> aaacca at position 4
.aaacaaacca.bbbb
	rule a-> b at position 5
.aaacabacca.bbbb
	rule a-> b at position 6
.aaacabbcca.bbbb
	rule cabb-> aaacca at position 8
.aaacabbcaaacca.bb
	rule a-> b at position 9
.aaacabbcabacca.bb
	rule a-> b at position 10
.aaacabbcabbcca.bb
	rule cabb-> aaacca at position 7
.aaacabbaaaccacca.bb
	rule a-> b at position 7
.aaacabbbaaccacca.bb
	rule a-> b at position 8
.aaacabbbbaccacca.bb
	rule a-> b at position 9
.aaacabbbbbccacca.bb
	rule cabb-> aaacca at position 14
.aaacabbbbbccacaaacca.
	rule a-> b at position 15
.aaacabbbbbccacabacca.
	rule a-> b at position 16
.aaacabbbbbccacabbcca.
	rule cabb-> aaacca at position 13
.aaacabbbbbccaaaaccacca.
	rule a-> b at position 13
.aaacabbbbbccabaaccacca.
	rule a-> b at position 14
.aaacabbbbbccabbaccacca.
	rule cabb-> aaacca at position 11
.aaacabbbbbcaaaccaaccacca.
	rule a-> b at position 12
.aaacabbbbbcabaccaaccacca.
	rule a-> b at position 13
.aaacabbbbbcabbccaaccacca.
	rule cabb-> aaacca at position 10
.aaacabbbbbaaaccaccaaccacca.
	rule a-> b at position 10
.aaacabbbbbbaaccaccaaccacca.
	rule a-> b at position 11
.aaacabbbbbbbaccaccaaccacca.
	rule a-> b at position 12
.aaacabbbbbbbbccaccaaccacca.