YES(O(1),O(n^2)) 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(n^2)). 162.41/60.06 162.41/60.06 Strict Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , a() -> n__a() 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() 162.41/60.06 , activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 We use the processor 'matrix interpretation of dimension 1' to 162.41/60.06 orient following rules strictly. 162.41/60.06 162.41/60.06 Trs: 162.41/60.06 { activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() } 162.41/60.06 162.41/60.06 The induced complexity on above rules (modulo remaining rules) is 162.41/60.06 YES(?,O(n^1)) . These rules are moved into the corresponding weak 162.41/60.06 component(s). 162.41/60.06 162.41/60.06 Sub-proof: 162.41/60.06 ---------- 162.41/60.06 TcT has computed the following triangular matrix interpretation. 162.41/60.06 162.41/60.06 [f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__a] = [0] 162.41/60.06 162.41/60.06 [n__g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [a] = [0] 162.41/60.06 162.41/60.06 [g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [activate](x1) = [1] x1 + [1] 162.41/60.06 162.41/60.06 The order satisfies the following ordering constraints: 162.41/60.06 162.41/60.06 [f(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [n__f(X)] 162.41/60.06 162.41/60.06 [f(n__f(n__a()))] = [0] 162.41/60.06 >= [0] 162.41/60.06 = [f(n__g(n__f(n__a())))] 162.41/60.06 162.41/60.06 [a()] = [0] 162.41/60.06 >= [0] 162.41/60.06 = [n__a()] 162.41/60.06 162.41/60.06 [g(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [n__g(X)] 162.41/60.06 162.41/60.06 [activate(X)] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [X] 162.41/60.06 162.41/60.06 [activate(n__f(X))] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [f(X)] 162.41/60.06 162.41/60.06 [activate(n__a())] = [1] 162.41/60.06 > [0] 162.41/60.06 = [a()] 162.41/60.06 162.41/60.06 [activate(n__g(X))] = [1] X + [1] 162.41/60.06 >= [1] X + [1] 162.41/60.06 = [g(activate(X))] 162.41/60.06 162.41/60.06 162.41/60.06 We return to the main proof. 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(n^2)). 162.41/60.06 162.41/60.06 Strict Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , a() -> n__a() 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Weak Trs: 162.41/60.06 { activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 We use the processor 'matrix interpretation of dimension 1' to 162.41/60.06 orient following rules strictly. 162.41/60.06 162.41/60.06 Trs: { a() -> n__a() } 162.41/60.06 162.41/60.06 The induced complexity on above rules (modulo remaining rules) is 162.41/60.06 YES(?,O(n^1)) . These rules are moved into the corresponding weak 162.41/60.06 component(s). 162.41/60.06 162.41/60.06 Sub-proof: 162.41/60.06 ---------- 162.41/60.06 TcT has computed the following triangular matrix interpretation. 162.41/60.06 162.41/60.06 [f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__a] = [0] 162.41/60.06 162.41/60.06 [n__g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [a] = [1] 162.41/60.06 162.41/60.06 [g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [activate](x1) = [1] x1 + [1] 162.41/60.06 162.41/60.06 The order satisfies the following ordering constraints: 162.41/60.06 162.41/60.06 [f(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [n__f(X)] 162.41/60.06 162.41/60.06 [f(n__f(n__a()))] = [0] 162.41/60.06 >= [0] 162.41/60.06 = [f(n__g(n__f(n__a())))] 162.41/60.06 162.41/60.06 [a()] = [1] 162.41/60.06 > [0] 162.41/60.06 = [n__a()] 162.41/60.06 162.41/60.06 [g(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [n__g(X)] 162.41/60.06 162.41/60.06 [activate(X)] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [X] 162.41/60.06 162.41/60.06 [activate(n__f(X))] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [f(X)] 162.41/60.06 162.41/60.06 [activate(n__a())] = [1] 162.41/60.06 >= [1] 162.41/60.06 = [a()] 162.41/60.06 162.41/60.06 [activate(n__g(X))] = [1] X + [1] 162.41/60.06 >= [1] X + [1] 162.41/60.06 = [g(activate(X))] 162.41/60.06 162.41/60.06 162.41/60.06 We return to the main proof. 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(n^2)). 162.41/60.06 162.41/60.06 Strict Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Weak Trs: 162.41/60.06 { a() -> n__a() 162.41/60.06 , activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 We use the processor 'matrix interpretation of dimension 1' to 162.41/60.06 orient following rules strictly. 162.41/60.06 162.41/60.06 Trs: { f(X) -> n__f(X) } 162.41/60.06 162.41/60.06 The induced complexity on above rules (modulo remaining rules) is 162.41/60.06 YES(?,O(n^1)) . These rules are moved into the corresponding weak 162.41/60.06 component(s). 162.41/60.06 162.41/60.06 Sub-proof: 162.41/60.06 ---------- 162.41/60.06 TcT has computed the following triangular matrix interpretation. 162.41/60.06 162.41/60.06 [f](x1) = [1] x1 + [1] 162.41/60.06 162.41/60.06 [n__f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__a] = [0] 162.41/60.06 162.41/60.06 [n__g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [a] = [1] 162.41/60.06 162.41/60.06 [g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [activate](x1) = [1] x1 + [1] 162.41/60.06 162.41/60.06 The order satisfies the following ordering constraints: 162.41/60.06 162.41/60.06 [f(X)] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [n__f(X)] 162.41/60.06 162.41/60.06 [f(n__f(n__a()))] = [1] 162.41/60.06 >= [1] 162.41/60.06 = [f(n__g(n__f(n__a())))] 162.41/60.06 162.41/60.06 [a()] = [1] 162.41/60.06 > [0] 162.41/60.06 = [n__a()] 162.41/60.06 162.41/60.06 [g(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [n__g(X)] 162.41/60.06 162.41/60.06 [activate(X)] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [X] 162.41/60.06 162.41/60.06 [activate(n__f(X))] = [1] X + [1] 162.41/60.06 >= [1] X + [1] 162.41/60.06 = [f(X)] 162.41/60.06 162.41/60.06 [activate(n__a())] = [1] 162.41/60.06 >= [1] 162.41/60.06 = [a()] 162.41/60.06 162.41/60.06 [activate(n__g(X))] = [1] X + [1] 162.41/60.06 >= [1] X + [1] 162.41/60.06 = [g(activate(X))] 162.41/60.06 162.41/60.06 162.41/60.06 We return to the main proof. 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(n^2)). 162.41/60.06 162.41/60.06 Strict Trs: 162.41/60.06 { f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Weak Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , a() -> n__a() 162.41/60.06 , activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 The weightgap principle applies (using the following nonconstant 162.41/60.06 growth matrix-interpretation) 162.41/60.06 162.41/60.06 TcT has computed the following triangular matrix interpretation. 162.41/60.06 Note that the diagonal of the component-wise maxima of 162.41/60.06 interpretation-entries contains no more than 1 non-zero entries. 162.41/60.06 162.41/60.06 [f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__f](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [n__a] = [0] 162.41/60.06 162.41/60.06 [n__g](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 [a] = [0] 162.41/60.06 162.41/60.06 [g](x1) = [1] x1 + [1] 162.41/60.06 162.41/60.06 [activate](x1) = [1] x1 + [0] 162.41/60.06 162.41/60.06 The order satisfies the following ordering constraints: 162.41/60.06 162.41/60.06 [f(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [n__f(X)] 162.41/60.06 162.41/60.06 [f(n__f(n__a()))] = [0] 162.41/60.06 >= [0] 162.41/60.06 = [f(n__g(n__f(n__a())))] 162.41/60.06 162.41/60.06 [a()] = [0] 162.41/60.06 >= [0] 162.41/60.06 = [n__a()] 162.41/60.06 162.41/60.06 [g(X)] = [1] X + [1] 162.41/60.06 > [1] X + [0] 162.41/60.06 = [n__g(X)] 162.41/60.06 162.41/60.06 [activate(X)] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [X] 162.41/60.06 162.41/60.06 [activate(n__f(X))] = [1] X + [0] 162.41/60.06 >= [1] X + [0] 162.41/60.06 = [f(X)] 162.41/60.06 162.41/60.06 [activate(n__a())] = [0] 162.41/60.06 >= [0] 162.41/60.06 = [a()] 162.41/60.06 162.41/60.06 [activate(n__g(X))] = [1] X + [0] 162.41/60.06 ? [1] X + [1] 162.41/60.06 = [g(activate(X))] 162.41/60.06 162.41/60.06 162.41/60.06 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(n^2)). 162.41/60.06 162.41/60.06 Strict Trs: 162.41/60.06 { f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Weak Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , a() -> n__a() 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 We use the processor 'matrix interpretation of dimension 2' to 162.41/60.06 orient following rules strictly. 162.41/60.06 162.41/60.06 Trs: { f(n__f(n__a())) -> f(n__g(n__f(n__a()))) } 162.41/60.06 162.41/60.06 The induced complexity on above rules (modulo remaining rules) is 162.41/60.06 YES(?,O(n^1)) . These rules are moved into the corresponding weak 162.41/60.06 component(s). 162.41/60.06 162.41/60.06 Sub-proof: 162.41/60.06 ---------- 162.41/60.06 TcT has computed the following triangular matrix interpretation. 162.41/60.06 162.41/60.06 [f](x1) = [1 1] x1 + [0] 162.41/60.06 [0 0] [1] 162.41/60.06 162.41/60.06 [n__f](x1) = [1 1] x1 + [0] 162.41/60.06 [0 0] [1] 162.41/60.06 162.41/60.06 [n__a] = [0] 162.41/60.06 [0] 162.41/60.06 162.41/60.06 [n__g](x1) = [1 0] x1 + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 162.41/60.06 [a] = [0] 162.41/60.06 [0] 162.41/60.06 162.41/60.06 [g](x1) = [1 0] x1 + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 162.41/60.06 [activate](x1) = [1 0] x1 + [0] 162.41/60.06 [0 1] [0] 162.41/60.06 162.41/60.06 The order satisfies the following ordering constraints: 162.41/60.06 162.41/60.06 [f(X)] = [1 1] X + [0] 162.41/60.06 [0 0] [1] 162.41/60.06 >= [1 1] X + [0] 162.41/60.06 [0 0] [1] 162.41/60.06 = [n__f(X)] 162.41/60.06 162.41/60.06 [f(n__f(n__a()))] = [1] 162.41/60.06 [1] 162.41/60.06 > [0] 162.41/60.06 [1] 162.41/60.06 = [f(n__g(n__f(n__a())))] 162.41/60.06 162.41/60.06 [a()] = [0] 162.41/60.06 [0] 162.41/60.06 >= [0] 162.41/60.06 [0] 162.41/60.06 = [n__a()] 162.41/60.06 162.41/60.06 [g(X)] = [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 = [n__g(X)] 162.41/60.06 162.41/60.06 [activate(X)] = [1 0] X + [0] 162.41/60.06 [0 1] [0] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 1] [0] 162.41/60.06 = [X] 162.41/60.06 162.41/60.06 [activate(n__f(X))] = [1 1] X + [0] 162.41/60.06 [0 0] [1] 162.41/60.06 >= [1 1] X + [0] 162.41/60.06 [0 0] [1] 162.41/60.06 = [f(X)] 162.41/60.06 162.41/60.06 [activate(n__a())] = [0] 162.41/60.06 [0] 162.41/60.06 >= [0] 162.41/60.06 [0] 162.41/60.06 = [a()] 162.41/60.06 162.41/60.06 [activate(n__g(X))] = [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 = [g(activate(X))] 162.41/60.06 162.41/60.06 162.41/60.06 We return to the main proof. 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(n^2)). 162.41/60.06 162.41/60.06 Strict Trs: { activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Weak Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , a() -> n__a() 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(n^2)) 162.41/60.06 162.41/60.06 We use the processor 'matrix interpretation of dimension 2' to 162.41/60.06 orient following rules strictly. 162.41/60.06 162.41/60.06 Trs: { activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 162.41/60.06 The induced complexity on above rules (modulo remaining rules) is 162.41/60.06 YES(?,O(n^2)) . These rules are moved into the corresponding weak 162.41/60.06 component(s). 162.41/60.06 162.41/60.06 Sub-proof: 162.41/60.06 ---------- 162.41/60.06 TcT has computed the following triangular matrix interpretation. 162.41/60.06 162.41/60.06 [f](x1) = [1 0] x1 + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 162.41/60.06 [n__f](x1) = [1 0] x1 + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 162.41/60.06 [n__a] = [0] 162.41/60.06 [0] 162.41/60.06 162.41/60.06 [n__g](x1) = [1 0] x1 + [0] 162.41/60.06 [0 1] [1] 162.41/60.06 162.41/60.06 [a] = [0] 162.41/60.06 [0] 162.41/60.06 162.41/60.06 [g](x1) = [1 0] x1 + [0] 162.41/60.06 [0 1] [1] 162.41/60.06 162.41/60.06 [activate](x1) = [1 1] x1 + [0] 162.41/60.06 [0 1] [0] 162.41/60.06 162.41/60.06 The order satisfies the following ordering constraints: 162.41/60.06 162.41/60.06 [f(X)] = [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 = [n__f(X)] 162.41/60.06 162.41/60.06 [f(n__f(n__a()))] = [0] 162.41/60.06 [0] 162.41/60.06 >= [0] 162.41/60.06 [0] 162.41/60.06 = [f(n__g(n__f(n__a())))] 162.41/60.06 162.41/60.06 [a()] = [0] 162.41/60.06 [0] 162.41/60.06 >= [0] 162.41/60.06 [0] 162.41/60.06 = [n__a()] 162.41/60.06 162.41/60.06 [g(X)] = [1 0] X + [0] 162.41/60.06 [0 1] [1] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 1] [1] 162.41/60.06 = [n__g(X)] 162.41/60.06 162.41/60.06 [activate(X)] = [1 1] X + [0] 162.41/60.06 [0 1] [0] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 1] [0] 162.41/60.06 = [X] 162.41/60.06 162.41/60.06 [activate(n__f(X))] = [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 >= [1 0] X + [0] 162.41/60.06 [0 0] [0] 162.41/60.06 = [f(X)] 162.41/60.06 162.41/60.06 [activate(n__a())] = [0] 162.41/60.06 [0] 162.41/60.06 >= [0] 162.41/60.06 [0] 162.41/60.06 = [a()] 162.41/60.06 162.41/60.06 [activate(n__g(X))] = [1 1] X + [1] 162.41/60.06 [0 1] [1] 162.41/60.06 > [1 1] X + [0] 162.41/60.06 [0 1] [1] 162.41/60.06 = [g(activate(X))] 162.41/60.06 162.41/60.06 162.41/60.06 We return to the main proof. 162.41/60.06 162.41/60.06 We are left with following problem, upon which TcT provides the 162.41/60.06 certificate YES(O(1),O(1)). 162.41/60.06 162.41/60.06 Weak Trs: 162.41/60.06 { f(X) -> n__f(X) 162.41/60.06 , f(n__f(n__a())) -> f(n__g(n__f(n__a()))) 162.41/60.06 , a() -> n__a() 162.41/60.06 , g(X) -> n__g(X) 162.41/60.06 , activate(X) -> X 162.41/60.06 , activate(n__f(X)) -> f(X) 162.41/60.06 , activate(n__a()) -> a() 162.41/60.06 , activate(n__g(X)) -> g(activate(X)) } 162.41/60.06 Obligation: 162.41/60.06 derivational complexity 162.41/60.06 Answer: 162.41/60.06 YES(O(1),O(1)) 162.41/60.06 162.41/60.06 Empty rules are trivially bounded 162.41/60.06 162.41/60.06 Hurray, we answered YES(O(1),O(n^2)) 162.41/60.07 EOF