/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2, d ↦ 3, f ↦ 4, g ↦ 5 }, it remains to prove termination of the 9-rule system { 0 1 2 0 ⟶ 1 0 2 1 0 1 , 0 3 ⟶ 2 , 0 4 4 ⟶ 5 , 1 5 ⟶ 5 1 , 2 ⟶ 4 4 , 2 0 2 ⟶ 1 2 0 1 2 , 2 3 ⟶ 0 0 , 5 ⟶ 2 0 , 5 ⟶ 3 3 3 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 12 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 18 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 7 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 9 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 30 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 4 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 6-rule system { 0 1 2 0 ⟶ 1 0 2 1 0 1 , 0 3 3 ⟶ 4 , 1 4 ⟶ 4 1 , 2 ⟶ 3 3 , 2 0 2 ⟶ 1 2 0 1 2 , 4 ⟶ 2 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (2,↓) ↦ 2, (0,↓) ↦ 3, (1,↑) ↦ 4, (2,↑) ↦ 5, (3,↓) ↦ 6, (4,↑) ↦ 7, (4,↓) ↦ 8 }, it remains to prove termination of the 21-rule system { 0 1 2 3 ⟶ 4 3 2 1 3 1 , 0 1 2 3 ⟶ 0 2 1 3 1 , 0 1 2 3 ⟶ 5 1 3 1 , 0 1 2 3 ⟶ 4 3 1 , 0 1 2 3 ⟶ 0 1 , 0 1 2 3 ⟶ 4 , 0 6 6 ⟶ 7 , 4 8 ⟶ 7 1 , 4 8 ⟶ 4 , 5 3 2 ⟶ 4 2 3 1 2 , 5 3 2 ⟶ 5 3 1 2 , 5 3 2 ⟶ 0 1 2 , 5 3 2 ⟶ 4 2 , 7 ⟶ 5 3 , 7 ⟶ 0 , 3 1 2 3 →= 1 3 2 1 3 1 , 3 6 6 →= 8 , 1 8 →= 8 1 , 2 →= 6 6 , 2 3 2 →= 1 2 3 1 2 , 8 →= 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 8 ↦ 5, 7 ↦ 6, 5 ↦ 7, 6 ↦ 8 }, it remains to prove termination of the 11-rule system { 0 1 2 3 ⟶ 0 2 1 3 1 , 4 5 ⟶ 6 1 , 7 3 2 ⟶ 4 2 3 1 2 , 7 3 2 ⟶ 7 3 1 2 , 6 ⟶ 7 3 , 3 1 2 3 →= 1 3 2 1 3 1 , 3 8 8 →= 5 , 1 5 →= 5 1 , 2 →= 8 8 , 2 3 2 →= 1 2 3 1 2 , 5 →= 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 3: 0 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 1 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 1 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 1 0 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 7 ↦ 4, 4 ↦ 5, 6 ↦ 6, 8 ↦ 7, 5 ↦ 8 }, it remains to prove termination of the 10-rule system { 0 1 2 3 ⟶ 0 2 1 3 1 , 4 3 2 ⟶ 5 2 3 1 2 , 4 3 2 ⟶ 4 3 1 2 , 6 ⟶ 4 3 , 3 1 2 3 →= 1 3 2 1 3 1 , 3 7 7 →= 8 , 1 8 →= 8 1 , 2 →= 7 7 , 2 3 2 →= 1 2 3 1 2 , 8 →= 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5, 7 ↦ 6, 8 ↦ 7 }, it remains to prove termination of the 9-rule system { 0 1 2 3 ⟶ 0 2 1 3 1 , 4 3 2 ⟶ 5 2 3 1 2 , 4 3 2 ⟶ 4 3 1 2 , 3 1 2 3 →= 1 3 2 1 3 1 , 3 6 6 →= 7 , 1 7 →= 7 1 , 2 →= 6 6 , 2 3 2 →= 1 2 3 1 2 , 7 →= 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 6 ↦ 5, 7 ↦ 6 }, it remains to prove termination of the 8-rule system { 0 1 2 3 ⟶ 0 2 1 3 1 , 4 3 2 ⟶ 4 3 1 2 , 3 1 2 3 →= 1 3 2 1 3 1 , 3 5 5 →= 6 , 1 6 →= 6 1 , 2 →= 5 5 , 2 3 2 →= 1 2 3 1 2 , 6 →= 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 5: 0 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 1 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 5 ↦ 4, 6 ↦ 5 }, it remains to prove termination of the 7-rule system { 0 1 2 3 ⟶ 0 2 1 3 1 , 3 1 2 3 →= 1 3 2 1 3 1 , 3 4 4 →= 5 , 1 5 →= 5 1 , 2 →= 4 4 , 2 3 2 →= 1 2 3 1 2 , 5 →= 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 5: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 1 0 ⎟ ⎜ 0 0 0 1 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 3 ↦ 0, 1 ↦ 1, 2 ↦ 2, 4 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 6-rule system { 0 1 2 0 →= 1 0 2 1 0 1 , 0 3 3 →= 4 , 1 4 →= 4 1 , 2 →= 3 3 , 2 0 2 →= 1 2 0 1 2 , 4 →= 2 0 } The system is trivially terminating.