/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 3-rule system { 0 ⟶ , 0 0 ⟶ 0 1 2 , 2 1 ⟶ 0 2 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (2,↓) ↦ 3, (2,↑) ↦ 4 }, it remains to prove termination of the 8-rule system { 0 1 ⟶ 0 2 3 , 0 1 ⟶ 4 , 4 2 ⟶ 0 3 1 , 4 2 ⟶ 4 1 , 4 2 ⟶ 0 , 1 →= , 1 1 →= 1 2 3 , 3 2 →= 1 3 1 } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (5,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (0,2) ↦ 3, (2,3) ↦ 4, (3,1) ↦ 5, (1,2) ↦ 6, (3,2) ↦ 7, (1,3) ↦ 8, (3,3) ↦ 9, (1,6) ↦ 10, (3,6) ↦ 11, (5,4) ↦ 12, (4,1) ↦ 13, (4,2) ↦ 14, (4,3) ↦ 15, (4,6) ↦ 16, (2,1) ↦ 17, (0,3) ↦ 18, (2,2) ↦ 19, (2,6) ↦ 20, (0,6) ↦ 21, (5,1) ↦ 22, (5,2) ↦ 23, (5,3) ↦ 24, (5,6) ↦ 25 }, it remains to prove termination of the 92-rule system { 0 1 2 ⟶ 0 3 4 5 , 0 1 6 ⟶ 0 3 4 7 , 0 1 8 ⟶ 0 3 4 9 , 0 1 10 ⟶ 0 3 4 11 , 0 1 2 ⟶ 12 13 , 0 1 6 ⟶ 12 14 , 0 1 8 ⟶ 12 15 , 0 1 10 ⟶ 12 16 , 12 14 17 ⟶ 0 18 5 2 , 12 14 19 ⟶ 0 18 5 6 , 12 14 4 ⟶ 0 18 5 8 , 12 14 20 ⟶ 0 18 5 10 , 12 14 17 ⟶ 12 13 2 , 12 14 19 ⟶ 12 13 6 , 12 14 4 ⟶ 12 13 8 , 12 14 20 ⟶ 12 13 10 , 12 14 17 ⟶ 0 1 , 12 14 19 ⟶ 0 3 , 12 14 4 ⟶ 0 18 , 12 14 20 ⟶ 0 21 , 1 2 →= 1 , 1 6 →= 3 , 1 8 →= 18 , 1 10 →= 21 , 2 2 →= 2 , 2 6 →= 6 , 2 8 →= 8 , 2 10 →= 10 , 17 2 →= 17 , 17 6 →= 19 , 17 8 →= 4 , 17 10 →= 20 , 5 2 →= 5 , 5 6 →= 7 , 5 8 →= 9 , 5 10 →= 11 , 13 2 →= 13 , 13 6 →= 14 , 13 8 →= 15 , 13 10 →= 16 , 22 2 →= 22 , 22 6 →= 23 , 22 8 →= 24 , 22 10 →= 25 , 1 2 2 →= 1 6 4 5 , 1 2 6 →= 1 6 4 7 , 1 2 8 →= 1 6 4 9 , 1 2 10 →= 1 6 4 11 , 2 2 2 →= 2 6 4 5 , 2 2 6 →= 2 6 4 7 , 2 2 8 →= 2 6 4 9 , 2 2 10 →= 2 6 4 11 , 17 2 2 →= 17 6 4 5 , 17 2 6 →= 17 6 4 7 , 17 2 8 →= 17 6 4 9 , 17 2 10 →= 17 6 4 11 , 5 2 2 →= 5 6 4 5 , 5 2 6 →= 5 6 4 7 , 5 2 8 →= 5 6 4 9 , 5 2 10 →= 5 6 4 11 , 13 2 2 →= 13 6 4 5 , 13 2 6 →= 13 6 4 7 , 13 2 8 →= 13 6 4 9 , 13 2 10 →= 13 6 4 11 , 22 2 2 →= 22 6 4 5 , 22 2 6 →= 22 6 4 7 , 22 2 8 →= 22 6 4 9 , 22 2 10 →= 22 6 4 11 , 18 7 17 →= 1 8 5 2 , 18 7 19 →= 1 8 5 6 , 18 7 4 →= 1 8 5 8 , 18 7 20 →= 1 8 5 10 , 8 7 17 →= 2 8 5 2 , 8 7 19 →= 2 8 5 6 , 8 7 4 →= 2 8 5 8 , 8 7 20 →= 2 8 5 10 , 4 7 17 →= 17 8 5 2 , 4 7 19 →= 17 8 5 6 , 4 7 4 →= 17 8 5 8 , 4 7 20 →= 17 8 5 10 , 9 7 17 →= 5 8 5 2 , 9 7 19 →= 5 8 5 6 , 9 7 4 →= 5 8 5 8 , 9 7 20 →= 5 8 5 10 , 15 7 17 →= 13 8 5 2 , 15 7 19 →= 13 8 5 6 , 15 7 4 →= 13 8 5 8 , 15 7 20 →= 13 8 5 10 , 24 7 17 →= 22 8 5 2 , 24 7 19 →= 22 8 5 6 , 24 7 4 →= 22 8 5 8 , 24 7 20 →= 22 8 5 10 } 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 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 11 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 12 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 13 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 14 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 15 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 16 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 17 ↦ ⎛ ⎞ ⎜ 1 4 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 18 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 19 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 20 ↦ ⎛ ⎞ ⎜ 1 4 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 21 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 22 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 23 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 24 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 25 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 5 ↦ 0, 6 ↦ 1, 7 ↦ 2, 8 ↦ 3, 9 ↦ 4, 1 ↦ 5, 2 ↦ 6, 4 ↦ 7, 17 ↦ 8, 13 ↦ 9, 22 ↦ 10 }, it remains to prove termination of the 18-rule system { 0 1 →= 2 , 0 3 →= 4 , 5 6 1 →= 5 1 7 2 , 5 6 3 →= 5 1 7 4 , 6 6 1 →= 6 1 7 2 , 6 6 3 →= 6 1 7 4 , 8 6 1 →= 8 1 7 2 , 8 6 3 →= 8 1 7 4 , 0 6 1 →= 0 1 7 2 , 0 6 3 →= 0 1 7 4 , 9 6 1 →= 9 1 7 2 , 9 6 3 →= 9 1 7 4 , 10 6 1 →= 10 1 7 2 , 10 6 3 →= 10 1 7 4 , 3 2 8 →= 6 3 0 6 , 3 2 7 →= 6 3 0 3 , 7 2 8 →= 8 3 0 6 , 7 2 7 →= 8 3 0 3 } The system is trivially terminating.