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: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(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: { first(0(), X) -> nil() } 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). [first](x1, x2) = [2] x1 + [2] x2 + [0] [0] = [3] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [0] [from](x1) = [2] x1 + [0] [n__from](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [first(X1, X2)] = [2] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [2] X + [6] > [0] = [nil()] [first(s(X), cons(Y, Z))] = [2] X + [2] Y + [2] Z + [0] >= [1] X + [1] Y + [2] Z + [0] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [0] >= [2] X1 + [2] X2 + [0] = [first(X1, X2)] [activate(n__from(X))] = [2] X + [0] >= [2] X + [0] = [from(X)] [from(X)] = [2] X + [0] >= [2] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [0] >= [1] X + [0] = [n__from(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: { first(X1, X2) -> n__first(X1, X2) , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Weak Trs: { first(0(), X) -> nil() } 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: { first(X1, X2) -> n__first(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(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). [first](x1, x2) = [2] x1 + [2] x2 + [1] [0] = [3] [nil] = [3] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [1] [from](x1) = [2] x1 + [0] [n__from](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [first(X1, X2)] = [2] X1 + [2] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [2] X + [7] > [3] = [nil()] [first(s(X), cons(Y, Z))] = [2] X + [2] Y + [2] Z + [1] >= [1] X + [1] Y + [2] Z + [1] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [1] >= [2] X1 + [2] X2 + [1] = [first(X1, X2)] [activate(n__from(X))] = [2] X + [1] > [2] X + [0] = [from(X)] [from(X)] = [2] X + [0] >= [2] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [0] >= [1] X + [0] = [n__from(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: { first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(n__first(X1, X2)) -> first(X1, X2) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Weak Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , activate(X) -> X , activate(n__from(X)) -> from(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__first(X1, X2)) -> first(X1, X2) , from(X) -> n__from(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). [first](x1, x2) = [1] x1 + [2] x2 + [0] [0] = [3] [nil] = [1] [s](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [1] [from](x1) = [2] x1 + [1] [n__from](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [first(X1, X2)] = [1] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [2] X + [3] > [1] = [nil()] [first(s(X), cons(Y, Z))] = [1] X + [2] Y + [2] Z + [1] >= [1] X + [1] Y + [2] Z + [1] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [1] > [1] X1 + [2] X2 + [0] = [first(X1, X2)] [activate(n__from(X))] = [2] X + [1] >= [2] X + [1] = [from(X)] [from(X)] = [2] X + [1] >= [2] X + [1] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [1] > [1] X + [0] = [n__from(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: { first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , from(X) -> cons(X, n__from(s(X))) } Weak Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> n__from(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: { from(X) -> cons(X, n__from(s(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). [first](x1, x2) = [1] x1 + [2] x2 + [1] [0] = [3] [nil] = [3] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [1] [from](x1) = [2] x1 + [1] [n__from](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [first(X1, X2)] = [1] X1 + [2] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [2] X + [4] > [3] = [nil()] [first(s(X), cons(Y, Z))] = [1] X + [2] Y + [2] Z + [1] >= [1] X + [1] Y + [2] Z + [1] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [1] >= [1] X1 + [2] X2 + [1] = [first(X1, X2)] [activate(n__from(X))] = [2] X + [1] >= [2] X + [1] = [from(X)] [from(X)] = [2] X + [1] > [2] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [1] > [1] X + [0] = [n__from(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: { first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) } Weak Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(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: { first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) } 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). [first](x1, x2) = [1] x1 + [2] x2 + [0] [0] = [3] [nil] = [3] [s](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [0] [from](x1) = [2] x1 + [3] [n__from](x1) = [1] x1 + [2] This order satisfies the following ordering constraints: [first(X1, X2)] = [1] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [2] X + [3] >= [3] = [nil()] [first(s(X), cons(Y, Z))] = [1] X + [2] Y + [2] Z + [1] > [1] X + [1] Y + [2] Z + [0] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [0] >= [1] X1 + [2] X2 + [0] = [first(X1, X2)] [activate(n__from(X))] = [2] X + [4] > [2] X + [3] = [from(X)] [from(X)] = [2] X + [3] >= [2] X + [3] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [3] > [1] X + [2] = [n__from(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: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(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))