YES(O(1),O(n^2)) 281.01/101.23 YES(O(1),O(n^2)) 281.01/101.23 281.01/101.23 We are left with following problem, upon which TcT provides the 281.01/101.23 certificate YES(O(1),O(n^2)). 281.01/101.23 281.01/101.23 Strict Trs: 281.01/101.23 { active(f(f(a()))) -> mark(c(f(g(f(a()))))) 281.01/101.23 , f(active(X)) -> f(X) 281.01/101.23 , f(mark(X)) -> f(X) 281.01/101.23 , mark(f(X)) -> active(f(mark(X))) 281.01/101.23 , mark(a()) -> active(a()) 281.01/101.23 , mark(c(X)) -> active(c(X)) 281.01/101.23 , mark(g(X)) -> active(g(mark(X))) 281.01/101.23 , c(active(X)) -> c(X) 281.01/101.23 , c(mark(X)) -> c(X) 281.01/101.23 , g(active(X)) -> g(X) 281.01/101.23 , g(mark(X)) -> g(X) } 281.01/101.23 Obligation: 281.01/101.23 derivational complexity 281.01/101.23 Answer: 281.01/101.23 YES(O(1),O(n^2)) 281.01/101.23 281.01/101.23 The weightgap principle applies (using the following nonconstant 281.01/101.23 growth matrix-interpretation) 281.01/101.23 281.01/101.23 TcT has computed the following triangular matrix interpretation. 281.01/101.23 Note that the diagonal of the component-wise maxima of 281.01/101.23 interpretation-entries contains no more than 1 non-zero entries. 281.01/101.23 281.01/101.23 [active](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 [f](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 [a] = [0] 281.01/101.23 281.01/101.23 [mark](x1) = [1] x1 + [1] 281.01/101.23 281.01/101.23 [c](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 [g](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 The order satisfies the following ordering constraints: 281.01/101.23 281.01/101.23 [active(f(f(a())))] = [0] 281.01/101.23 ? [1] 281.01/101.23 = [mark(c(f(g(f(a())))))] 281.01/101.23 281.01/101.23 [f(active(X))] = [1] X + [0] 281.01/101.23 >= [1] X + [0] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [f(mark(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [mark(f(X))] = [1] X + [1] 281.01/101.23 >= [1] X + [1] 281.01/101.23 = [active(f(mark(X)))] 281.01/101.23 281.01/101.23 [mark(a())] = [1] 281.01/101.23 > [0] 281.01/101.23 = [active(a())] 281.01/101.23 281.01/101.23 [mark(c(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [active(c(X))] 281.01/101.23 281.01/101.23 [mark(g(X))] = [1] X + [1] 281.01/101.23 >= [1] X + [1] 281.01/101.23 = [active(g(mark(X)))] 281.01/101.23 281.01/101.23 [c(active(X))] = [1] X + [0] 281.01/101.23 >= [1] X + [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [c(mark(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [g(active(X))] = [1] X + [0] 281.01/101.23 >= [1] X + [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 [g(mark(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 281.01/101.23 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 281.01/101.23 281.01/101.23 We are left with following problem, upon which TcT provides the 281.01/101.23 certificate YES(O(1),O(n^2)). 281.01/101.23 281.01/101.23 Strict Trs: 281.01/101.23 { active(f(f(a()))) -> mark(c(f(g(f(a()))))) 281.01/101.23 , f(active(X)) -> f(X) 281.01/101.23 , mark(f(X)) -> active(f(mark(X))) 281.01/101.23 , mark(g(X)) -> active(g(mark(X))) 281.01/101.23 , c(active(X)) -> c(X) 281.01/101.23 , g(active(X)) -> g(X) } 281.01/101.23 Weak Trs: 281.01/101.23 { f(mark(X)) -> f(X) 281.01/101.23 , mark(a()) -> active(a()) 281.01/101.23 , mark(c(X)) -> active(c(X)) 281.01/101.23 , c(mark(X)) -> c(X) 281.01/101.23 , g(mark(X)) -> g(X) } 281.01/101.23 Obligation: 281.01/101.23 derivational complexity 281.01/101.23 Answer: 281.01/101.23 YES(O(1),O(n^2)) 281.01/101.23 281.01/101.23 The weightgap principle applies (using the following nonconstant 281.01/101.23 growth matrix-interpretation) 281.01/101.23 281.01/101.23 TcT has computed the following triangular matrix interpretation. 281.01/101.23 Note that the diagonal of the component-wise maxima of 281.01/101.23 interpretation-entries contains no more than 1 non-zero entries. 281.01/101.23 281.01/101.23 [active](x1) = [1] x1 + [1] 281.01/101.23 281.01/101.23 [f](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 [a] = [0] 281.01/101.23 281.01/101.23 [mark](x1) = [1] x1 + [1] 281.01/101.23 281.01/101.23 [c](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 [g](x1) = [1] x1 + [0] 281.01/101.23 281.01/101.23 The order satisfies the following ordering constraints: 281.01/101.23 281.01/101.23 [active(f(f(a())))] = [1] 281.01/101.23 >= [1] 281.01/101.23 = [mark(c(f(g(f(a())))))] 281.01/101.23 281.01/101.23 [f(active(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [f(mark(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [mark(f(X))] = [1] X + [1] 281.01/101.23 ? [1] X + [2] 281.01/101.23 = [active(f(mark(X)))] 281.01/101.23 281.01/101.23 [mark(a())] = [1] 281.01/101.23 >= [1] 281.01/101.23 = [active(a())] 281.01/101.23 281.01/101.23 [mark(c(X))] = [1] X + [1] 281.01/101.23 >= [1] X + [1] 281.01/101.23 = [active(c(X))] 281.01/101.23 281.01/101.23 [mark(g(X))] = [1] X + [1] 281.01/101.23 ? [1] X + [2] 281.01/101.23 = [active(g(mark(X)))] 281.01/101.23 281.01/101.23 [c(active(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [c(mark(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [g(active(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 [g(mark(X))] = [1] X + [1] 281.01/101.23 > [1] X + [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 281.01/101.23 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 281.01/101.23 281.01/101.23 We are left with following problem, upon which TcT provides the 281.01/101.23 certificate YES(O(1),O(n^2)). 281.01/101.23 281.01/101.23 Strict Trs: 281.01/101.23 { active(f(f(a()))) -> mark(c(f(g(f(a()))))) 281.01/101.23 , mark(f(X)) -> active(f(mark(X))) 281.01/101.23 , mark(g(X)) -> active(g(mark(X))) } 281.01/101.23 Weak Trs: 281.01/101.23 { f(active(X)) -> f(X) 281.01/101.23 , f(mark(X)) -> f(X) 281.01/101.23 , mark(a()) -> active(a()) 281.01/101.23 , mark(c(X)) -> active(c(X)) 281.01/101.23 , c(active(X)) -> c(X) 281.01/101.23 , c(mark(X)) -> c(X) 281.01/101.23 , g(active(X)) -> g(X) 281.01/101.23 , g(mark(X)) -> g(X) } 281.01/101.23 Obligation: 281.01/101.23 derivational complexity 281.01/101.23 Answer: 281.01/101.23 YES(O(1),O(n^2)) 281.01/101.23 281.01/101.23 We use the processor 'matrix interpretation of dimension 2' to 281.01/101.23 orient following rules strictly. 281.01/101.23 281.01/101.23 Trs: { active(f(f(a()))) -> mark(c(f(g(f(a()))))) } 281.01/101.23 281.01/101.23 The induced complexity on above rules (modulo remaining rules) is 281.01/101.23 YES(?,O(n^1)) . These rules are moved into the corresponding weak 281.01/101.23 component(s). 281.01/101.23 281.01/101.23 Sub-proof: 281.01/101.23 ---------- 281.01/101.23 TcT has computed the following triangular matrix interpretation. 281.01/101.23 Note that the diagonal of the component-wise maxima of 281.01/101.23 interpretation-entries contains no more than 1 non-zero entries. 281.01/101.23 281.01/101.23 [active](x1) = [1 1] x1 + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 281.01/101.23 [f](x1) = [1 1] x1 + [0] 281.01/101.23 [0 0] [2] 281.01/101.23 281.01/101.23 [a] = [0] 281.01/101.23 [0] 281.01/101.23 281.01/101.23 [mark](x1) = [1 1] x1 + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 281.01/101.23 [c](x1) = [1 0] x1 + [0] 281.01/101.23 [0 0] [1] 281.01/101.23 281.01/101.23 [g](x1) = [1 1] x1 + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 281.01/101.23 The order satisfies the following ordering constraints: 281.01/101.23 281.01/101.23 [active(f(f(a())))] = [4] 281.01/101.23 [0] 281.01/101.23 > [3] 281.01/101.23 [0] 281.01/101.23 = [mark(c(f(g(f(a())))))] 281.01/101.23 281.01/101.23 [f(active(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [2] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 0] [2] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [f(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [2] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 0] [2] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [mark(f(X))] = [1 1] X + [2] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 1] X + [2] 281.01/101.23 [0 0] [0] 281.01/101.23 = [active(f(mark(X)))] 281.01/101.23 281.01/101.23 [mark(a())] = [0] 281.01/101.23 [0] 281.01/101.23 >= [0] 281.01/101.23 [0] 281.01/101.23 = [active(a())] 281.01/101.23 281.01/101.23 [mark(c(X))] = [1 0] X + [1] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [1] 281.01/101.23 [0 0] [0] 281.01/101.23 = [active(c(X))] 281.01/101.23 281.01/101.23 [mark(g(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [active(g(mark(X)))] 281.01/101.23 281.01/101.23 [c(active(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [1] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [1] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [c(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [1] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [1] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [g(active(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 [g(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 281.01/101.23 We return to the main proof. 281.01/101.23 281.01/101.23 We are left with following problem, upon which TcT provides the 281.01/101.23 certificate YES(O(1),O(n^2)). 281.01/101.23 281.01/101.23 Strict Trs: 281.01/101.23 { mark(f(X)) -> active(f(mark(X))) 281.01/101.23 , mark(g(X)) -> active(g(mark(X))) } 281.01/101.23 Weak Trs: 281.01/101.23 { active(f(f(a()))) -> mark(c(f(g(f(a()))))) 281.01/101.23 , f(active(X)) -> f(X) 281.01/101.23 , f(mark(X)) -> f(X) 281.01/101.23 , mark(a()) -> active(a()) 281.01/101.23 , mark(c(X)) -> active(c(X)) 281.01/101.23 , c(active(X)) -> c(X) 281.01/101.23 , c(mark(X)) -> c(X) 281.01/101.23 , g(active(X)) -> g(X) 281.01/101.23 , g(mark(X)) -> g(X) } 281.01/101.23 Obligation: 281.01/101.23 derivational complexity 281.01/101.23 Answer: 281.01/101.23 YES(O(1),O(n^2)) 281.01/101.23 281.01/101.23 We use the processor 'matrix interpretation of dimension 2' to 281.01/101.23 orient following rules strictly. 281.01/101.23 281.01/101.23 Trs: { mark(g(X)) -> active(g(mark(X))) } 281.01/101.23 281.01/101.23 The induced complexity on above rules (modulo remaining rules) is 281.01/101.23 YES(?,O(n^2)) . These rules are moved into the corresponding weak 281.01/101.23 component(s). 281.01/101.23 281.01/101.23 Sub-proof: 281.01/101.23 ---------- 281.01/101.23 TcT has computed the following triangular matrix interpretation. 281.01/101.23 281.01/101.23 [active](x1) = [1 0] x1 + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 281.01/101.23 [f](x1) = [1 0] x1 + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 281.01/101.23 [a] = [0] 281.01/101.23 [0] 281.01/101.23 281.01/101.23 [mark](x1) = [1 1] x1 + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 281.01/101.23 [c](x1) = [1 0] x1 + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 281.01/101.23 [g](x1) = [1 0] x1 + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 281.01/101.23 The order satisfies the following ordering constraints: 281.01/101.23 281.01/101.23 [active(f(f(a())))] = [0] 281.01/101.23 [0] 281.01/101.23 >= [0] 281.01/101.23 [0] 281.01/101.23 = [mark(c(f(g(f(a())))))] 281.01/101.23 281.01/101.23 [f(active(X))] = [1 0] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [f(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [mark(f(X))] = [1 1] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 = [active(f(mark(X)))] 281.01/101.23 281.01/101.23 [mark(a())] = [0] 281.01/101.23 [0] 281.01/101.23 >= [0] 281.01/101.23 [0] 281.01/101.23 = [active(a())] 281.01/101.23 281.01/101.23 [mark(c(X))] = [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [active(c(X))] 281.01/101.23 281.01/101.23 [mark(g(X))] = [1 1] X + [1] 281.01/101.23 [0 1] [1] 281.01/101.23 > [1 1] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 = [active(g(mark(X)))] 281.01/101.23 281.01/101.23 [c(active(X))] = [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [c(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [g(active(X))] = [1 0] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 [g(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 281.01/101.23 We return to the main proof. 281.01/101.23 281.01/101.23 We are left with following problem, upon which TcT provides the 281.01/101.23 certificate YES(O(1),O(n^2)). 281.01/101.23 281.01/101.23 Strict Trs: { mark(f(X)) -> active(f(mark(X))) } 281.01/101.23 Weak Trs: 281.01/101.23 { active(f(f(a()))) -> mark(c(f(g(f(a()))))) 281.01/101.23 , f(active(X)) -> f(X) 281.01/101.23 , f(mark(X)) -> f(X) 281.01/101.23 , mark(a()) -> active(a()) 281.01/101.23 , mark(c(X)) -> active(c(X)) 281.01/101.23 , mark(g(X)) -> active(g(mark(X))) 281.01/101.23 , c(active(X)) -> c(X) 281.01/101.23 , c(mark(X)) -> c(X) 281.01/101.23 , g(active(X)) -> g(X) 281.01/101.23 , g(mark(X)) -> g(X) } 281.01/101.23 Obligation: 281.01/101.23 derivational complexity 281.01/101.23 Answer: 281.01/101.23 YES(O(1),O(n^2)) 281.01/101.23 281.01/101.23 We use the processor 'matrix interpretation of dimension 2' to 281.01/101.23 orient following rules strictly. 281.01/101.23 281.01/101.23 Trs: { mark(f(X)) -> active(f(mark(X))) } 281.01/101.23 281.01/101.23 The induced complexity on above rules (modulo remaining rules) is 281.01/101.23 YES(?,O(n^2)) . These rules are moved into the corresponding weak 281.01/101.23 component(s). 281.01/101.23 281.01/101.23 Sub-proof: 281.01/101.23 ---------- 281.01/101.23 TcT has computed the following triangular matrix interpretation. 281.01/101.23 281.01/101.23 [active](x1) = [1 0] x1 + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 281.01/101.23 [f](x1) = [1 0] x1 + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 281.01/101.23 [a] = [0] 281.01/101.23 [2] 281.01/101.23 281.01/101.23 [mark](x1) = [1 1] x1 + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 281.01/101.23 [c](x1) = [1 0] x1 + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 281.01/101.23 [g](x1) = [1 0] x1 + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 281.01/101.23 The order satisfies the following ordering constraints: 281.01/101.23 281.01/101.23 [active(f(f(a())))] = [0] 281.01/101.23 [4] 281.01/101.23 >= [0] 281.01/101.23 [0] 281.01/101.23 = [mark(c(f(g(f(a())))))] 281.01/101.23 281.01/101.23 [f(active(X))] = [1 0] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [f(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 = [f(X)] 281.01/101.23 281.01/101.23 [mark(f(X))] = [1 1] X + [1] 281.01/101.23 [0 1] [1] 281.01/101.23 > [1 1] X + [0] 281.01/101.23 [0 1] [1] 281.01/101.23 = [active(f(mark(X)))] 281.01/101.23 281.01/101.23 [mark(a())] = [2] 281.01/101.23 [2] 281.01/101.23 > [0] 281.01/101.23 [2] 281.01/101.23 = [active(a())] 281.01/101.23 281.01/101.23 [mark(c(X))] = [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [active(c(X))] 281.01/101.23 281.01/101.23 [mark(g(X))] = [1 1] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 >= [1 1] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 = [active(g(mark(X)))] 281.01/101.23 281.01/101.23 [c(active(X))] = [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [c(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 0] [0] 281.01/101.23 = [c(X)] 281.01/101.23 281.01/101.23 [g(active(X))] = [1 0] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 [g(mark(X))] = [1 1] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 >= [1 0] X + [0] 281.01/101.23 [0 1] [0] 281.01/101.23 = [g(X)] 281.01/101.23 281.01/101.23 281.01/101.23 We return to the main proof. 281.01/101.23 281.01/101.23 We are left with following problem, upon which TcT provides the 281.01/101.23 certificate YES(O(1),O(1)). 281.01/101.23 281.01/101.23 Weak Trs: 281.01/101.23 { active(f(f(a()))) -> mark(c(f(g(f(a()))))) 281.01/101.23 , f(active(X)) -> f(X) 281.01/101.23 , f(mark(X)) -> f(X) 281.01/101.23 , mark(f(X)) -> active(f(mark(X))) 281.01/101.23 , mark(a()) -> active(a()) 281.01/101.23 , mark(c(X)) -> active(c(X)) 281.01/101.23 , mark(g(X)) -> active(g(mark(X))) 281.01/101.23 , c(active(X)) -> c(X) 281.01/101.23 , c(mark(X)) -> c(X) 281.01/101.23 , g(active(X)) -> g(X) 281.01/101.23 , g(mark(X)) -> g(X) } 281.01/101.23 Obligation: 281.01/101.23 derivational complexity 281.01/101.23 Answer: 281.01/101.23 YES(O(1),O(1)) 281.01/101.23 281.01/101.23 Empty rules are trivially bounded 281.01/101.23 281.01/101.23 Hurray, we answered YES(O(1),O(n^2)) 281.01/101.27 EOF