/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 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 0 1 ⟶ 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, ... }:

.caab.ababab
	rule caab-> aaacca at position 0
.aaacca.ababab
	rule caab-> aaacca at position 4
.aaacaaacca.abab
	rule a-> b at position 6
.aaacaabcca.abab
	rule caab-> aaacca at position 8
.aaacaabcaaacca.ab
	rule a-> b at position 10
.aaacaabcaabcca.ab
	rule caab-> aaacca at position 7
.aaacaabaaaccacca.ab
	rule a-> b at position 8
.aaacaababaccacca.ab
	rule caab-> aaacca at position 14
.aaacaababaccacaaacca.
	rule a-> b at position 16
.aaacaababaccacaabcca.
	rule caab-> aaacca at position 13
.aaacaababaccaaaaccacca.
	rule a-> b at position 14
.aaacaababaccaabaccacca.
	rule caab-> aaacca at position 11
.aaacaababacaaaccaaccacca.
	rule a-> b at position 13
.aaacaababacaabccaaccacca.
	rule caab-> aaacca at position 10
.aaacaababaaaaccaccaaccacca.
	rule a-> b at position 10
.aaacaabababaaccaccaaccacca.
	rule a-> b at position 12
.aaacaababababccaccaaccacca.