WORST_CASE(INF,?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f42_0_decrease_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1>0 && arg2>-1 && arg2==arg1P_1 ], cost: 1 1: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1>5 && -1+arg1==arg1P_2 ], cost: 1 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1_0_main_Load -> f42_0_decrease_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1 1: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=-1+arg1, arg2'=arg2P_2, [ arg1>5 ], cost: 1 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=-1+arg1, arg2'=arg2P_2, [ arg1>5 ], cost: 1 Accelerated rule 1 with metering function -5+arg1, yielding the new rule 3. Removing the simple loops: 1. Accelerated all simple loops using metering functions (where possible): Start location: __init 0: f1_0_main_Load -> f42_0_decrease_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1 3: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1>5 ], cost: -5+arg1 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1_0_main_Load -> f42_0_decrease_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1 4: f1_0_main_Load -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1>0 && arg2>5 ], cost: -4+arg2 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 4: f1_0_main_Load -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1>0 && arg2>5 ], cost: -4+arg2 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Eliminated locations (on linear paths): Start location: __init 5: __init -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1P_3>0 && arg2P_3>5 ], cost: -3+arg2P_3 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 5: __init -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1P_3>0 && arg2P_3>5 ], cost: -3+arg2P_3 Computing asymptotic complexity for rule 5 Solved the limit problem by the following transformations: Created initial limit problem: -5+arg2P_3 (+/+!), -3+arg2P_3 (+), arg1P_3 (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {arg2P_3==n,arg1P_3==n} resulting limit problem: [solved] Solution: arg2P_3 / n arg1P_3 / n Resulting cost -3+n has complexity: Unbounded Found new complexity Unbounded. Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Unbounded Cpx degree: Unbounded Solved cost: -3+n Rule cost: -3+arg2P_3 Rule guard: [ arg1P_3>0 && arg2P_3>5 ] WORST_CASE(INF,?)