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) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(true(), X, Y) -> X , if(false(), X, Y) -> activate(Y) , activate(X) -> X , activate(n__f(X)) -> f(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: { if(false(), X, Y) -> activate(Y) } 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) = [2] x1 + [0] [if](x1, x2, x3) = [2] x1 + [3] x2 + [2] x3 + [0] [c] = [0] [n__f](x1) = [1] x1 + [0] [true] = [0] [false] = [2] [activate](x1) = [2] x1 + [0] This order satisfies the following ordering constraints: [f(X)] = [2] X + [0] >= [2] X + [0] = [if(X, c(), n__f(true()))] [f(X)] = [2] X + [0] >= [1] X + [0] = [n__f(X)] [if(true(), X, Y)] = [3] X + [2] Y + [0] >= [1] X + [0] = [X] [if(false(), X, Y)] = [3] X + [2] Y + [4] > [2] Y + [0] = [activate(Y)] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__f(X))] = [2] X + [0] >= [2] X + [0] = [f(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) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(true(), X, Y) -> X , activate(X) -> X , activate(n__f(X)) -> f(X) } Weak Trs: { if(false(), X, Y) -> activate(Y) } 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 , activate(n__f(X)) -> 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) = [2] x1 + [0] [if](x1, x2, x3) = [2] x1 + [3] x2 + [2] x3 + [0] [c] = [0] [n__f](x1) = [1] x1 + [0] [true] = [0] [false] = [2] [activate](x1) = [2] x1 + [1] This order satisfies the following ordering constraints: [f(X)] = [2] X + [0] >= [2] X + [0] = [if(X, c(), n__f(true()))] [f(X)] = [2] X + [0] >= [1] X + [0] = [n__f(X)] [if(true(), X, Y)] = [3] X + [2] Y + [0] >= [1] X + [0] = [X] [if(false(), X, Y)] = [3] X + [2] Y + [4] > [2] Y + [1] = [activate(Y)] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__f(X))] = [2] X + [1] > [2] X + [0] = [f(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) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(true(), X, Y) -> X } Weak Trs: { if(false(), X, Y) -> activate(Y) , activate(X) -> X , activate(n__f(X)) -> f(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) -> if(X, c(), n__f(true())) , 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) = [2] x1 + [1] [if](x1, x2, x3) = [2] x1 + [3] x2 + [2] x3 + [0] [c] = [0] [n__f](x1) = [1] x1 + [0] [true] = [0] [false] = [2] [activate](x1) = [2] x1 + [1] This order satisfies the following ordering constraints: [f(X)] = [2] X + [1] > [2] X + [0] = [if(X, c(), n__f(true()))] [f(X)] = [2] X + [1] > [1] X + [0] = [n__f(X)] [if(true(), X, Y)] = [3] X + [2] Y + [0] >= [1] X + [0] = [X] [if(false(), X, Y)] = [3] X + [2] Y + [4] > [2] Y + [1] = [activate(Y)] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__f(X))] = [2] X + [1] >= [2] X + [1] = [f(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: { if(true(), X, Y) -> X } Weak Trs: { f(X) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(false(), X, Y) -> activate(Y) , activate(X) -> X , activate(n__f(X)) -> f(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: { if(true(), X, Y) -> 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) = [2] x1 + [1] [if](x1, x2, x3) = [2] x1 + [3] x2 + [2] x3 + [1] [c] = [0] [n__f](x1) = [1] x1 + [0] [true] = [0] [false] = [0] [activate](x1) = [2] x1 + [1] This order satisfies the following ordering constraints: [f(X)] = [2] X + [1] >= [2] X + [1] = [if(X, c(), n__f(true()))] [f(X)] = [2] X + [1] > [1] X + [0] = [n__f(X)] [if(true(), X, Y)] = [3] X + [2] Y + [1] > [1] X + [0] = [X] [if(false(), X, Y)] = [3] X + [2] Y + [1] >= [2] Y + [1] = [activate(Y)] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__f(X))] = [2] X + [1] >= [2] X + [1] = [f(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) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(true(), X, Y) -> X , if(false(), X, Y) -> activate(Y) , activate(X) -> X , activate(n__f(X)) -> f(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))