/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 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 0 ⟶ 0 0 0 2 2 0 }

Loop of length 16 starting with a string of length 10
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.caba.bababa
	rule caba-> aaacca at position 0
.aaacca.bababa
	rule caba-> aaacca at position 4
.aaacaaacca.baba
	rule a-> b at position 5
.aaacabacca.baba
	rule caba-> aaacca at position 8
.aaacabacaaacca.ba
	rule a-> b at position 9
.aaacabacabacca.ba
	rule caba-> aaacca at position 7
.aaacabaaaaccacca.ba
	rule a-> b at position 7
.aaacababaaccacca.ba
	rule a-> b at position 9
.aaacabababccacca.ba
	rule caba-> aaacca at position 14
.aaacabababccacaaacca.
	rule a-> b at position 15
.aaacabababccacabacca.
	rule caba-> aaacca at position 13
.aaacabababccaaaaccacca.
	rule a-> b at position 13
.aaacabababccabaaccacca.
	rule caba-> aaacca at position 11
.aaacabababcaaaccaaccacca.
	rule a-> b at position 12
.aaacabababcabaccaaccacca.
	rule caba-> aaacca at position 10
.aaacabababaaaccaccaaccacca.
	rule a-> b at position 11
.aaacababababaccaccaaccacca.