/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 { c ↦ 0, a ↦ 1, b ↦ 2 }, it remains to prove termination of the 1-rule system { 0 1 2 1 2 ⟶ 1 2 1 2 2 1 2 2 0 1 2 0 1 } Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (1,0) ↦ 0, (0,1) ↦ 1, (1,2) ↦ 2, (2,1) ↦ 3, (2,0) ↦ 4, (1,1) ↦ 5, (2,2) ↦ 6 }, it remains to prove termination of the 6-rule system { 0 1 2 3 2 4 ⟶ 5 2 3 2 6 3 2 6 4 1 2 4 1 0 , 0 1 2 3 2 3 ⟶ 5 2 3 2 6 3 2 6 4 1 2 4 1 5 , 0 1 2 3 2 6 ⟶ 5 2 3 2 6 3 2 6 4 1 2 4 1 2 , 4 1 2 3 2 4 ⟶ 3 2 3 2 6 3 2 6 4 1 2 4 1 0 , 4 1 2 3 2 3 ⟶ 3 2 3 2 6 3 2 6 4 1 2 4 1 5 , 4 1 2 3 2 6 ⟶ 3 2 3 2 6 3 2 6 4 1 2 4 1 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 7: 0 ↦ ⎛ ⎞ ⎜ 1 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 0 ⎟ ⎜ 0 0 0 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 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 1 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 1 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 1 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 1 0 0 0 0 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 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 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 5 ↦ 4, 6 ↦ 5, 4 ↦ 6 }, it remains to prove termination of the 4-rule system { 0 1 2 3 2 3 ⟶ 4 2 3 2 5 3 2 5 6 1 2 6 1 4 , 0 1 2 3 2 5 ⟶ 4 2 3 2 5 3 2 5 6 1 2 6 1 2 , 6 1 2 3 2 3 ⟶ 3 2 3 2 5 3 2 5 6 1 2 6 1 4 , 6 1 2 3 2 5 ⟶ 3 2 3 2 5 3 2 5 6 1 2 6 1 2 } 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 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 6 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 5 ↦ 4, 4 ↦ 5 }, it remains to prove termination of the 2-rule system { 0 1 2 3 2 3 ⟶ 3 2 3 2 4 3 2 4 0 1 2 0 1 5 , 0 1 2 3 2 4 ⟶ 3 2 3 2 4 3 2 4 0 1 2 0 1 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 7: 0 ↦ ⎛ ⎞ ⎜ 1 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 0 ⎟ ⎜ 0 0 0 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 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 1 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 1 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 1 0 0 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 1-rule system { 0 1 2 3 2 3 ⟶ 3 2 3 2 4 3 2 4 0 1 2 0 1 5 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 7: 0 ↦ ⎛ ⎞ ⎜ 1 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 0 ⎟ ⎜ 0 0 1 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 1 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 1 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 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 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { }, it remains to prove termination of the 0-rule system { } The system is trivially terminating.