/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 ,
  1 0 1 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 1 0 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, ... }:

.cbab.ababab
	rule cbab-> aaacca at position 0
.aaacca.ababab
	rule a-> b at position 5
.aaaccb.ababab
	rule cbab-> aaacca at position 4
.aaacaaacca.abab
	rule a-> b at position 4
.aaacbaacca.abab
	rule a-> b at position 6
.aaacbabcca.abab
	rule a-> b at position 9
.aaacbabccb.abab
	rule cbab-> aaacca at position 8
.aaacbabcaaacca.ab
	rule a-> b at position 8
.aaacbabcbaacca.ab
	rule a-> b at position 10
.aaacbabcbabcca.ab
	rule cbab-> aaacca at position 7
.aaacbabaaaccacca.ab
	rule a-> b at position 8
.aaacbababaccacca.ab
	rule a-> b at position 12
.aaacbababaccbcca.ab
	rule a-> b at position 15
.aaacbababaccbccb.ab
	rule cbab-> aaacca at position 14
.aaacbababaccbcaaacca.
	rule a-> b at position 14
.aaacbababaccbcbaacca.
	rule a-> b at position 16
.aaacbababaccbcbabcca.
	rule cbab-> aaacca at position 13
.aaacbababaccbaaaccacca.
	rule a-> b at position 14
.aaacbababaccbabaccacca.
	rule cbab-> aaacca at position 11
.aaacbababacaaaccaaccacca.
	rule a-> b at position 11
.aaacbababacbaaccaaccacca.
	rule a-> b at position 13
.aaacbababacbabccaaccacca.
	rule cbab-> aaacca at position 10
.aaacbababaaaaccaccaaccacca.
	rule a-> b at position 10
.aaacbabababaaccaccaaccacca.
	rule a-> b at position 12
.aaacbababababccaccaaccacca.