NO ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f182_0_middle_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>1 && arg1P_1>-1 && arg1>0 ], cost: 1 1: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=arg1P_2, arg2'=arg2P_2, [ arg2>arg1 && -1+arg1==arg1P_2 && 1+arg2==arg2P_2 ], cost: 1 2: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2 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 -> f182_0_middle_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>1 && arg1P_1>-1 && arg1>0 ], cost: 1 1: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=-1+arg1, arg2'=1+arg2, [ arg2>arg1 ], cost: 1 2: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=-1+arg1, arg2'=1+arg2, [ arg2 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: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=-1+arg1, arg2'=1+arg2, [ arg2>arg1 ], cost: 1 2: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=-1+arg1, arg2'=1+arg2, [ arg2 f182_0_middle_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>1 && arg1P_1>-1 && arg1>0 ], cost: 1 4: f182_0_middle_EQ -> [3] : [ arg2>arg1 ], cost: NONTERM 5: f182_0_middle_EQ -> f182_0_middle_EQ : arg1'=-k+arg1, arg2'=arg2+k, [ k>=0 && -1+arg2+k<1-k+arg1 ], cost: k 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 -> f182_0_middle_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>1 && arg1P_1>-1 && arg1>0 ], cost: 1 6: f1_0_main_Load -> [3] : [ arg2>1 && arg1>0 ], cost: NONTERM 7: f1_0_main_Load -> f182_0_middle_EQ : arg1'=arg1P_1-k, arg2'=arg2P_1+k, [ arg2P_1>-1 && arg2>1 && arg1P_1>-1 && arg1>0 && k>=0 && -1+arg2P_1+k<1+arg1P_1-k ], cost: 1+k 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] : [ arg2>1 && arg1>0 ], cost: NONTERM 7: f1_0_main_Load -> f182_0_middle_EQ : arg1'=arg1P_1-k, arg2'=arg2P_1+k, [ arg2P_1>-1 && arg2>1 && arg1P_1>-1 && arg1>0 && k>=0 && -1+arg2P_1+k<1+arg1P_1-k ], cost: 1+k 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] : [ arg2P_4>1 && arg1P_4>0 ], cost: NONTERM 9: __init -> f182_0_middle_EQ : arg1'=arg1P_1-k, arg2'=arg2P_1+k, [ arg2P_1>-1 && arg2P_4>1 && arg1P_1>-1 && arg1P_4>0 && k>=0 && -1+arg2P_1+k<1+arg1P_1-k ], cost: 2+k ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 8: __init -> [3] : [ arg2P_4>1 && arg1P_4>0 ], cost: NONTERM 9: __init -> f182_0_middle_EQ : arg1'=arg1P_1-k, arg2'=arg2P_1+k, [ arg2P_1>-1 && arg2P_4>1 && arg1P_1>-1 && arg1P_4>0 && k>=0 && -1+arg2P_1+k<1+arg1P_1-k ], cost: 2+k 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: [ arg2P_4>1 && arg1P_4>0 ] NO