NO ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f32_0_increase_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1>0 && arg2>-1 && arg2==arg1P_1 ], cost: 1 1: f32_0_increase_LE -> f32_0_increase_LE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1>9 && arg1==arg1P_2 ], cost: 1 2: f32_0_increase_LE -> f32_0_increase_LE : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1>5 && arg1<10 && -1+arg1==arg1P_3 ], cost: 1 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1_0_main_Load -> f32_0_increase_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1 1: f32_0_increase_LE -> f32_0_increase_LE : arg2'=arg2P_2, [ arg1>9 ], cost: 1 2: f32_0_increase_LE -> f32_0_increase_LE : arg1'=-1+arg1, arg2'=arg2P_3, [ arg1>5 && arg1<10 ], cost: 1 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f32_0_increase_LE -> f32_0_increase_LE : arg2'=arg2P_2, [ arg1>9 ], cost: 1 2: f32_0_increase_LE -> f32_0_increase_LE : arg1'=-1+arg1, arg2'=arg2P_3, [ arg1>5 && arg1<10 ], cost: 1 Accelerated rule 1 with non-termination, yielding the new rule 4. Accelerated rule 2 with backward acceleration, yielding the new rule 5. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 1 2. Accelerated all simple loops using metering functions (where possible): Start location: __init 0: f1_0_main_Load -> f32_0_increase_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1 4: f32_0_increase_LE -> [3] : [ arg1>9 ], cost: NONTERM 5: f32_0_increase_LE -> f32_0_increase_LE : arg1'=5, arg2'=arg2P_3, [ arg1<10 && -5+arg1>=1 ], cost: -5+arg1 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1_0_main_Load -> f32_0_increase_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1 6: f1_0_main_Load -> [3] : [ arg1>0 && arg2>9 ], cost: NONTERM 7: f1_0_main_Load -> f32_0_increase_LE : arg1'=5, arg2'=arg2P_3, [ arg1>0 && arg2<10 && -5+arg2>=1 ], cost: -4+arg2 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 6: f1_0_main_Load -> [3] : [ arg1>0 && arg2>9 ], cost: NONTERM 7: f1_0_main_Load -> f32_0_increase_LE : arg1'=5, arg2'=arg2P_3, [ arg1>0 && arg2<10 && -5+arg2>=1 ], cost: -4+arg2 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: __init 8: __init -> [3] : [ arg1P_4>0 && arg2P_4>9 ], cost: NONTERM 9: __init -> f32_0_increase_LE : arg1'=5, arg2'=arg2P_3, [ arg1P_4>0 && arg2P_4<10 && -5+arg2P_4>=1 ], cost: -3+arg2P_4 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 8: __init -> [3] : [ arg1P_4>0 && arg2P_4>9 ], cost: NONTERM 9: __init -> f32_0_increase_LE : arg1'=5, arg2'=arg2P_3, [ arg1P_4>0 && arg2P_4<10 && -5+arg2P_4>=1 ], cost: -3+arg2P_4 Computing asymptotic complexity for rule 8 Guard is satisfiable, yielding nontermination Resulting cost NONTERM has complexity: Nonterm Found new complexity Nonterm. Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Nonterm Cpx degree: Nonterm Solved cost: NONTERM Rule cost: NONTERM Rule guard: [ arg1P_4>0 && arg2P_4>9 ] NO