NO ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f86_0_sum_NE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1>0 && arg2>4 && arg2==arg1P_1 ], cost: 1 1: f1_0_main_Load -> f86_0_sum_NE : arg1'=arg1P_2, arg2'=arg2P_2, [ 1>-arg2 && arg2>-1 && arg2<5 && arg1>0 && -arg2==arg1P_2 ], cost: 1 2: f86_0_sum_NE -> f86_0_sum_NE : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1<0 && -1+arg1 f86_0_sum_NE : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1>0 && -1+arg1 f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1_0_main_Load -> f86_0_sum_NE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>4 ], cost: 1 1: f1_0_main_Load -> f86_0_sum_NE : arg1'=-arg2, arg2'=arg2P_2, [ 1>-arg2 && arg2<5 && arg1>0 ], cost: 1 2: f86_0_sum_NE -> f86_0_sum_NE : arg1'=-1+arg1, arg2'=arg2P_3, [ arg1<0 ], cost: 1 3: f86_0_sum_NE -> f86_0_sum_NE : arg1'=-1+arg1, arg2'=arg2P_4, [ arg1>0 ], cost: 1 4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 2: f86_0_sum_NE -> f86_0_sum_NE : arg1'=-1+arg1, arg2'=arg2P_3, [ arg1<0 ], cost: 1 3: f86_0_sum_NE -> f86_0_sum_NE : arg1'=-1+arg1, arg2'=arg2P_4, [ arg1>0 ], cost: 1 Accelerated rule 2 with non-termination, yielding the new rule 5. Accelerated rule 3 with backward acceleration, yielding the new rule 6. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 2 3. Accelerated all simple loops using metering functions (where possible): Start location: __init 0: f1_0_main_Load -> f86_0_sum_NE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>4 ], cost: 1 1: f1_0_main_Load -> f86_0_sum_NE : arg1'=-arg2, arg2'=arg2P_2, [ 1>-arg2 && arg2<5 && arg1>0 ], cost: 1 5: f86_0_sum_NE -> [3] : [ arg1<0 ], cost: NONTERM 6: f86_0_sum_NE -> f86_0_sum_NE : arg1'=0, arg2'=arg2P_4, [ arg1>=1 ], cost: arg1 4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1_0_main_Load -> f86_0_sum_NE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>4 ], cost: 1 1: f1_0_main_Load -> f86_0_sum_NE : arg1'=-arg2, arg2'=arg2P_2, [ 1>-arg2 && arg2<5 && arg1>0 ], cost: 1 7: f1_0_main_Load -> [3] : [ arg2<5 && arg1>0 && -arg2<0 ], cost: NONTERM 8: f1_0_main_Load -> f86_0_sum_NE : arg1'=0, arg2'=arg2P_4, [ arg1>0 && arg2>4 ], cost: 1+arg2 4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 7: f1_0_main_Load -> [3] : [ arg2<5 && arg1>0 && -arg2<0 ], cost: NONTERM 8: f1_0_main_Load -> f86_0_sum_NE : arg1'=0, arg2'=arg2P_4, [ arg1>0 && arg2>4 ], cost: 1+arg2 4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: __init 9: __init -> [3] : [ arg2P_5<5 && arg1P_5>0 && -arg2P_5<0 ], cost: NONTERM 10: __init -> f86_0_sum_NE : arg1'=0, arg2'=arg2P_4, [ arg1P_5>0 && arg2P_5>4 ], cost: 2+arg2P_5 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 9: __init -> [3] : [ arg2P_5<5 && arg1P_5>0 && -arg2P_5<0 ], cost: NONTERM 10: __init -> f86_0_sum_NE : arg1'=0, arg2'=arg2P_4, [ arg1P_5>0 && arg2P_5>4 ], cost: 2+arg2P_5 Computing asymptotic complexity for rule 9 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: [ arg2P_5<5 && arg1P_5>0 && -arg2P_5<0 ] NO