YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(X) -> X } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1] x1 + [0] [g](x1) = [1] x1 + [0] [n__h](x1) = [1] x1 + [0] [n__f](x1) = [1] x1 + [0] [h](x1) = [1] x1 + [0] [activate](x1) = [2] x1 + [1] This order satisfies the following ordering constraints: [f(X)] = [1] X + [0] >= [1] X + [0] = [g(n__h(n__f(X)))] [f(X)] = [1] X + [0] >= [1] X + [0] = [n__f(X)] [h(X)] = [1] X + [0] >= [1] X + [0] = [n__h(X)] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__h(X))] = [2] X + [1] >= [2] X + [1] = [h(activate(X))] [activate(n__f(X))] = [2] X + [1] >= [2] X + [1] = [f(activate(X))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(X)) } Weak Trs: { activate(X) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(n__f(X)) -> f(activate(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1] x1 + [2] [g](x1) = [1] x1 + [0] [n__h](x1) = [1] x1 + [0] [n__f](x1) = [1] x1 + [2] [h](x1) = [1] x1 + [0] [activate](x1) = [2] x1 + [0] This order satisfies the following ordering constraints: [f(X)] = [1] X + [2] >= [1] X + [2] = [g(n__h(n__f(X)))] [f(X)] = [1] X + [2] >= [1] X + [2] = [n__f(X)] [h(X)] = [1] X + [0] >= [1] X + [0] = [n__h(X)] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__h(X))] = [2] X + [0] >= [2] X + [0] = [h(activate(X))] [activate(n__f(X))] = [2] X + [4] > [2] X + [2] = [f(activate(X))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(n__h(X)) -> h(activate(X)) } Weak Trs: { activate(X) -> X , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { f(X) -> n__f(X) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1] x1 + [3] [g](x1) = [1] x1 + [1] [n__h](x1) = [1] x1 + [0] [n__f](x1) = [1] x1 + [2] [h](x1) = [1] x1 + [0] [activate](x1) = [2] x1 + [0] This order satisfies the following ordering constraints: [f(X)] = [1] X + [3] >= [1] X + [3] = [g(n__h(n__f(X)))] [f(X)] = [1] X + [3] > [1] X + [2] = [n__f(X)] [h(X)] = [1] X + [0] >= [1] X + [0] = [n__h(X)] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__h(X))] = [2] X + [0] >= [2] X + [0] = [h(activate(X))] [activate(n__f(X))] = [2] X + [4] > [2] X + [3] = [f(activate(X))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(X) -> g(n__h(n__f(X))) , h(X) -> n__h(X) , activate(n__h(X)) -> h(activate(X)) } Weak Trs: { f(X) -> n__f(X) , activate(X) -> X , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { f(X) -> g(n__h(n__f(X))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1] x1 + [3] [g](x1) = [1] x1 + [0] [n__h](x1) = [1] x1 + [0] [n__f](x1) = [1] x1 + [2] [h](x1) = [1] x1 + [0] [activate](x1) = [2] x1 + [0] This order satisfies the following ordering constraints: [f(X)] = [1] X + [3] > [1] X + [2] = [g(n__h(n__f(X)))] [f(X)] = [1] X + [3] > [1] X + [2] = [n__f(X)] [h(X)] = [1] X + [0] >= [1] X + [0] = [n__h(X)] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__h(X))] = [2] X + [0] >= [2] X + [0] = [h(activate(X))] [activate(n__f(X))] = [2] X + [4] > [2] X + [3] = [f(activate(X))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { h(X) -> n__h(X) , activate(n__h(X)) -> h(activate(X)) } Weak Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , activate(X) -> X , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { h(X) -> n__h(X) , activate(n__h(X)) -> h(activate(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1] x1 + [3] [g](x1) = [1] x1 + [0] [n__h](x1) = [1] x1 + [1] [n__f](x1) = [1] x1 + [2] [h](x1) = [1] x1 + [2] [activate](x1) = [3] x1 + [0] This order satisfies the following ordering constraints: [f(X)] = [1] X + [3] >= [1] X + [3] = [g(n__h(n__f(X)))] [f(X)] = [1] X + [3] > [1] X + [2] = [n__f(X)] [h(X)] = [1] X + [2] > [1] X + [1] = [n__h(X)] [activate(X)] = [3] X + [0] >= [1] X + [0] = [X] [activate(n__h(X))] = [3] X + [3] > [3] X + [2] = [h(activate(X))] [activate(n__f(X))] = [3] X + [6] > [3] X + [3] = [f(activate(X))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))