/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { b ↦ 0, a ↦ 1 }, it remains to prove termination of the 3-rule system { 0 0 0 1 0 ⟶ 0 1 0 0 1 0 , 0 1 0 0 ⟶ 0 0 1 0 1 0 , 0 1 0 1 1 0 0 ⟶ 0 1 1 0 1 1 0 0 0 1 0 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 8: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 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 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 1 ⎟ ⎜ 0 1 0 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 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 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 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 2-rule system { 0 0 0 1 0 ⟶ 0 1 0 0 1 0 , 0 1 0 0 ⟶ 0 0 1 0 1 0 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 5: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 1 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 1 ⎟ ⎜ 0 1 0 0 2 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 1 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 1 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 1-rule system { 0 0 0 1 0 ⟶ 0 1 0 0 1 0 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 6: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 ⎟ ⎜ 0 0 0 1 0 0 ⎟ ⎜ 0 0 1 0 1 0 ⎟ ⎜ 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 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 1 ⎟ ⎜ 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.