/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 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 0 1 ⟶ 2 1 1 0 0 , 1 1 ⟶ , 2 2 ⟶ } 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 }, it remains to prove termination of the 9-rule system { 0 1 ⟶ 2 1 1 3 3 , 0 1 ⟶ 4 1 3 3 , 0 1 ⟶ 4 3 3 , 0 1 ⟶ 0 3 , 0 1 ⟶ 0 , 3 →= , 3 1 →= 5 1 1 3 3 , 1 1 →= , 5 5 →= } 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 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 3 ↦ 2, 5 ↦ 3 }, it remains to prove termination of the 6-rule system { 0 1 ⟶ 0 2 , 0 1 ⟶ 0 , 2 →= , 2 1 →= 3 1 1 2 2 , 1 1 →= , 3 3 →= } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (4,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (0,2) ↦ 3, (2,1) ↦ 4, (1,2) ↦ 5, (2,2) ↦ 6, (1,3) ↦ 7, (2,3) ↦ 8, (1,5) ↦ 9, (2,5) ↦ 10, (0,3) ↦ 11, (0,5) ↦ 12, (3,2) ↦ 13, (3,1) ↦ 14, (3,3) ↦ 15, (3,5) ↦ 16, (4,2) ↦ 17, (4,1) ↦ 18, (4,3) ↦ 19, (4,5) ↦ 20 }, it remains to prove termination of the 88-rule system { 0 1 2 ⟶ 0 3 4 , 0 1 5 ⟶ 0 3 6 , 0 1 7 ⟶ 0 3 8 , 0 1 9 ⟶ 0 3 10 , 0 1 2 ⟶ 0 1 , 0 1 5 ⟶ 0 3 , 0 1 7 ⟶ 0 11 , 0 1 9 ⟶ 0 12 , 3 4 →= 1 , 3 6 →= 3 , 3 8 →= 11 , 3 10 →= 12 , 5 4 →= 2 , 5 6 →= 5 , 5 8 →= 7 , 5 10 →= 9 , 6 4 →= 4 , 6 6 →= 6 , 6 8 →= 8 , 6 10 →= 10 , 13 4 →= 14 , 13 6 →= 13 , 13 8 →= 15 , 13 10 →= 16 , 17 4 →= 18 , 17 6 →= 17 , 17 8 →= 19 , 17 10 →= 20 , 3 4 2 →= 11 14 2 5 6 4 , 3 4 5 →= 11 14 2 5 6 6 , 3 4 7 →= 11 14 2 5 6 8 , 3 4 9 →= 11 14 2 5 6 10 , 5 4 2 →= 7 14 2 5 6 4 , 5 4 5 →= 7 14 2 5 6 6 , 5 4 7 →= 7 14 2 5 6 8 , 5 4 9 →= 7 14 2 5 6 10 , 6 4 2 →= 8 14 2 5 6 4 , 6 4 5 →= 8 14 2 5 6 6 , 6 4 7 →= 8 14 2 5 6 8 , 6 4 9 →= 8 14 2 5 6 10 , 13 4 2 →= 15 14 2 5 6 4 , 13 4 5 →= 15 14 2 5 6 6 , 13 4 7 →= 15 14 2 5 6 8 , 13 4 9 →= 15 14 2 5 6 10 , 17 4 2 →= 19 14 2 5 6 4 , 17 4 5 →= 19 14 2 5 6 6 , 17 4 7 →= 19 14 2 5 6 8 , 17 4 9 →= 19 14 2 5 6 10 , 1 2 2 →= 1 , 1 2 5 →= 3 , 1 2 7 →= 11 , 1 2 9 →= 12 , 2 2 2 →= 2 , 2 2 5 →= 5 , 2 2 7 →= 7 , 2 2 9 →= 9 , 4 2 2 →= 4 , 4 2 5 →= 6 , 4 2 7 →= 8 , 4 2 9 →= 10 , 14 2 2 →= 14 , 14 2 5 →= 13 , 14 2 7 →= 15 , 14 2 9 →= 16 , 18 2 2 →= 18 , 18 2 5 →= 17 , 18 2 7 →= 19 , 18 2 9 →= 20 , 11 15 14 →= 1 , 11 15 13 →= 3 , 11 15 15 →= 11 , 11 15 16 →= 12 , 7 15 14 →= 2 , 7 15 13 →= 5 , 7 15 15 →= 7 , 7 15 16 →= 9 , 8 15 14 →= 4 , 8 15 13 →= 6 , 8 15 15 →= 8 , 8 15 16 →= 10 , 15 15 14 →= 14 , 15 15 13 →= 13 , 15 15 15 →= 15 , 15 15 16 →= 16 , 19 15 14 →= 18 , 19 15 13 →= 17 , 19 15 15 →= 19 , 19 15 16 →= 20 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 9 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 10 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 11 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 12 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 13 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 14 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 15 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 16 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 17 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 18 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 19 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 20 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 3 ↦ 0, 6 ↦ 1, 5 ↦ 2, 4 ↦ 3, 2 ↦ 4, 8 ↦ 5, 7 ↦ 6, 10 ↦ 7, 9 ↦ 8, 13 ↦ 9, 15 ↦ 10, 17 ↦ 11, 14 ↦ 12 }, it remains to prove termination of the 28-rule system { 0 1 →= 0 , 2 3 →= 4 , 2 1 →= 2 , 2 5 →= 6 , 2 7 →= 8 , 1 3 →= 3 , 1 1 →= 1 , 1 5 →= 5 , 1 7 →= 7 , 9 1 →= 9 , 9 5 →= 10 , 11 1 →= 11 , 2 3 4 →= 6 12 4 2 1 3 , 2 3 2 →= 6 12 4 2 1 1 , 2 3 6 →= 6 12 4 2 1 5 , 2 3 8 →= 6 12 4 2 1 7 , 1 3 4 →= 5 12 4 2 1 3 , 1 3 2 →= 5 12 4 2 1 1 , 1 3 6 →= 5 12 4 2 1 5 , 1 3 8 →= 5 12 4 2 1 7 , 9 3 4 →= 10 12 4 2 1 3 , 9 3 2 →= 10 12 4 2 1 1 , 9 3 6 →= 10 12 4 2 1 5 , 9 3 8 →= 10 12 4 2 1 7 , 12 4 2 →= 9 , 12 4 6 →= 10 , 6 10 12 →= 4 , 5 10 12 →= 3 } The system is trivially terminating.