YES(O(1),O(n^3)) 190.24/60.03 YES(O(1),O(n^3)) 190.24/60.03 190.24/60.03 We are left with following problem, upon which TcT provides the 190.24/60.03 certificate YES(O(1),O(n^3)). 190.24/60.03 190.24/60.03 Strict Trs: 190.24/60.03 { active(f(f(a()))) -> mark(f(g(f(a())))) 190.24/60.03 , active(g(X)) -> g(active(X)) 190.24/60.03 , f(ok(X)) -> ok(f(X)) 190.24/60.03 , g(mark(X)) -> mark(g(X)) 190.24/60.03 , g(ok(X)) -> ok(g(X)) 190.24/60.03 , proper(f(X)) -> f(proper(X)) 190.24/60.03 , proper(a()) -> ok(a()) 190.24/60.03 , proper(g(X)) -> g(proper(X)) 190.24/60.03 , top(mark(X)) -> top(proper(X)) 190.24/60.03 , top(ok(X)) -> top(active(X)) } 190.24/60.03 Obligation: 190.24/60.03 derivational complexity 190.24/60.03 Answer: 190.24/60.03 YES(O(1),O(n^3)) 190.24/60.03 190.24/60.03 We use the processor 'matrix interpretation of dimension 2' to 190.24/60.03 orient following rules strictly. 190.24/60.03 190.24/60.03 Trs: 190.24/60.03 { active(f(f(a()))) -> mark(f(g(f(a())))) 190.24/60.03 , proper(a()) -> ok(a()) 190.24/60.03 , top(mark(X)) -> top(proper(X)) } 190.24/60.03 190.24/60.03 The induced complexity on above rules (modulo remaining rules) is 190.24/60.03 YES(?,O(n^1)) . These rules are removed from the problem. Note that 190.24/60.03 none of the weakly oriented rules is size-increasing. The overall 190.24/60.03 complexity is obtained by composition . 190.24/60.03 190.24/60.03 Sub-proof: 190.24/60.03 ---------- 190.24/60.03 TcT has computed the following triangular matrix interpretation. 190.24/60.03 190.24/60.03 [active](x1) = [1 0] x1 + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 190.24/60.03 [f](x1) = [1 2] x1 + [0] 190.24/60.03 [0 0] [2] 190.24/60.03 190.24/60.03 [a] = [1] 190.24/60.03 [0] 190.24/60.03 190.24/60.03 [mark](x1) = [1 0] x1 + [2] 190.24/60.03 [0 0] [0] 190.24/60.03 190.24/60.03 [g](x1) = [1 0] x1 + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 190.24/60.03 [proper](x1) = [1 0] x1 + [1] 190.24/60.03 [0 1] [0] 190.24/60.03 190.24/60.03 [ok](x1) = [1 0] x1 + [0] 190.24/60.03 [0 1] [0] 190.24/60.03 190.24/60.03 [top](x1) = [1 0] x1 + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 190.24/60.03 The order satisfies the following ordering constraints: 190.24/60.03 190.24/60.03 [active(f(f(a())))] = [5] 190.24/60.03 [0] 190.24/60.03 > [3] 190.24/60.03 [0] 190.24/60.03 = [mark(f(g(f(a()))))] 190.24/60.03 190.24/60.03 [active(g(X))] = [1 0] X + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 >= [1 0] X + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 = [g(active(X))] 190.24/60.03 190.24/60.03 [f(ok(X))] = [1 2] X + [0] 190.24/60.03 [0 0] [2] 190.24/60.03 >= [1 2] X + [0] 190.24/60.03 [0 0] [2] 190.24/60.03 = [ok(f(X))] 190.24/60.03 190.24/60.03 [g(mark(X))] = [1 0] X + [2] 190.24/60.03 [0 0] [0] 190.24/60.03 >= [1 0] X + [2] 190.24/60.03 [0 0] [0] 190.24/60.03 = [mark(g(X))] 190.24/60.03 190.24/60.03 [g(ok(X))] = [1 0] X + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 >= [1 0] X + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 = [ok(g(X))] 190.24/60.03 190.24/60.03 [proper(f(X))] = [1 2] X + [1] 190.24/60.03 [0 0] [2] 190.24/60.03 >= [1 2] X + [1] 190.24/60.03 [0 0] [2] 190.24/60.03 = [f(proper(X))] 190.24/60.03 190.24/60.03 [proper(a())] = [2] 190.24/60.03 [0] 190.24/60.03 > [1] 190.24/60.03 [0] 190.24/60.03 = [ok(a())] 190.24/60.03 190.24/60.03 [proper(g(X))] = [1 0] X + [1] 190.24/60.03 [0 0] [0] 190.24/60.03 >= [1 0] X + [1] 190.24/60.03 [0 0] [0] 190.24/60.03 = [g(proper(X))] 190.24/60.03 190.24/60.03 [top(mark(X))] = [1 0] X + [2] 190.24/60.03 [0 0] [0] 190.24/60.03 > [1 0] X + [1] 190.24/60.03 [0 0] [0] 190.24/60.03 = [top(proper(X))] 190.24/60.03 190.24/60.03 [top(ok(X))] = [1 0] X + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 >= [1 0] X + [0] 190.24/60.03 [0 0] [0] 190.24/60.03 = [top(active(X))] 190.24/60.03 190.24/60.03 190.24/60.03 We return to the main proof. 190.24/60.03 190.24/60.03 We are left with following problem, upon which TcT provides the 190.24/60.03 certificate YES(?,O(n^2)). 190.24/60.03 190.24/60.03 Strict Trs: 190.24/60.03 { active(g(X)) -> g(active(X)) 190.24/60.03 , f(ok(X)) -> ok(f(X)) 190.24/60.03 , g(mark(X)) -> mark(g(X)) 190.24/60.03 , g(ok(X)) -> ok(g(X)) 190.24/60.03 , proper(f(X)) -> f(proper(X)) 190.24/60.03 , proper(g(X)) -> g(proper(X)) 190.24/60.03 , top(ok(X)) -> top(active(X)) } 190.24/60.03 Obligation: 190.24/60.03 derivational complexity 190.24/60.03 Answer: 190.24/60.03 YES(?,O(n^2)) 190.24/60.03 190.24/60.03 TcT has computed the following triangular matrix interpretation. 190.24/60.03 190.24/60.03 [active](x1) = [1 1] x1 + [0] 190.24/60.03 [0 1] [0] 190.24/60.03 190.24/60.03 [f](x1) = [1 1] x1 + [4] 190.24/60.03 [0 1] [3] 190.24/60.03 190.24/60.03 [mark](x1) = [1 0] x1 + [0] 190.24/60.03 [0 1] [1] 190.24/60.03 190.24/60.03 [g](x1) = [1 1] x1 + [0] 190.24/60.03 [0 1] [4] 190.24/60.03 190.24/60.03 [proper](x1) = [1 2] x1 + [0] 190.24/60.03 [0 1] [0] 190.24/60.03 190.24/60.03 [ok](x1) = [1 1] x1 + [1] 190.24/60.03 [0 1] [7] 190.24/60.03 190.24/60.03 [top](x1) = [1 0] x1 + [0] 190.24/60.03 [0 0] [5] 190.24/60.03 190.24/60.03 The order satisfies the following ordering constraints: 190.24/60.03 190.24/60.03 [active(g(X))] = [1 2] X + [4] 190.24/60.03 [0 1] [4] 190.24/60.03 > [1 2] X + [0] 190.24/60.03 [0 1] [4] 190.24/60.03 = [g(active(X))] 190.24/60.03 190.24/60.03 [f(ok(X))] = [1 2] X + [12] 190.24/60.03 [0 1] [10] 190.24/60.03 > [1 2] X + [8] 190.24/60.03 [0 1] [10] 190.24/60.03 = [ok(f(X))] 190.24/60.03 190.24/60.03 [g(mark(X))] = [1 1] X + [1] 190.24/60.03 [0 1] [5] 190.24/60.03 > [1 1] X + [0] 190.24/60.03 [0 1] [5] 190.24/60.03 = [mark(g(X))] 190.24/60.03 190.24/60.03 [g(ok(X))] = [1 2] X + [8] 190.24/60.03 [0 1] [11] 190.24/60.03 > [1 2] X + [5] 190.24/60.03 [0 1] [11] 190.24/60.03 = [ok(g(X))] 190.24/60.03 190.24/60.03 [proper(f(X))] = [1 3] X + [10] 190.24/60.03 [0 1] [3] 190.24/60.03 > [1 3] X + [4] 190.24/60.03 [0 1] [3] 190.24/60.03 = [f(proper(X))] 190.24/60.03 190.24/60.03 [proper(g(X))] = [1 3] X + [8] 190.24/60.03 [0 1] [4] 190.24/60.03 > [1 3] X + [0] 190.24/60.03 [0 1] [4] 190.24/60.03 = [g(proper(X))] 190.24/60.03 190.24/60.03 [top(ok(X))] = [1 1] X + [1] 190.24/60.03 [0 0] [5] 190.24/60.03 > [1 1] X + [0] 190.24/60.03 [0 0] [5] 190.24/60.03 = [top(active(X))] 190.24/60.03 190.24/60.03 190.24/60.03 Hurray, we answered YES(O(1),O(n^3)) 190.24/60.04 EOF