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: { 2nd(cons1(X, cons(Y, Z))) -> Y , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) , activate(X) -> X , 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)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(2nd) = {1}, Uargs(cons1) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [2nd](x1) = [1] x1 + [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [1] x1 + [0] [from](x1) = [1] x1 + [5] [n__from](x1) = [1] x1 + [0] [s](x1) = [0] The following symbols are considered usable {2nd, activate, from} The order satisfies the following ordering constraints: [2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [0] >= [1] Y + [0] = [Y] [2nd(cons(X, X1))] = [1] X + [1] X1 + [0] >= [1] X + [1] X1 + [0] = [2nd(cons1(X, activate(X1)))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [0] ? [1] X + [5] = [from(X)] [from(X)] = [1] X + [5] > [1] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [5] > [1] X + [0] = [n__from(X)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { 2nd(cons1(X, cons(Y, Z))) -> Y , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) , activate(X) -> X , activate(n__from(X)) -> from(X) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(2nd) = {1}, Uargs(cons1) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [2nd](x1) = [1] x1 + [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [1] x1 + [4] [from](x1) = [1] x1 + [4] [n__from](x1) = [1] x1 + [0] [s](x1) = [0] The following symbols are considered usable {2nd, activate, from} The order satisfies the following ordering constraints: [2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [0] >= [1] Y + [0] = [Y] [2nd(cons(X, X1))] = [1] X + [1] X1 + [0] ? [1] X + [1] X1 + [4] = [2nd(cons1(X, activate(X1)))] [activate(X)] = [1] X + [4] > [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [4] >= [1] X + [4] = [from(X)] [from(X)] = [1] X + [4] > [1] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [4] > [1] X + [0] = [n__from(X)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { 2nd(cons1(X, cons(Y, Z))) -> Y , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) , activate(n__from(X)) -> from(X) } Weak Trs: { activate(X) -> 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)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(2nd) = {1}, Uargs(cons1) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [2nd](x1) = [1] x1 + [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [1] x1 + [1] [from](x1) = [1] x1 + [0] [n__from](x1) = [1] x1 + [0] [s](x1) = [0] The following symbols are considered usable {2nd, activate, from} The order satisfies the following ordering constraints: [2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [0] >= [1] Y + [0] = [Y] [2nd(cons(X, X1))] = [1] X + [1] X1 + [0] ? [1] X + [1] X1 + [1] = [2nd(cons1(X, activate(X1)))] [activate(X)] = [1] X + [1] > [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [1] > [1] X + [0] = [from(X)] [from(X)] = [1] X + [0] >= [1] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [0] >= [1] X + [0] = [n__from(X)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { 2nd(cons1(X, cons(Y, Z))) -> Y , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) } Weak Trs: { activate(X) -> X , 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)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(2nd) = {1}, Uargs(cons1) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [2nd](x1) = [1] x1 + [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [4] [cons](x1, x2) = [1] x1 + [1] x2 + [4] [activate](x1) = [1] x1 + [4] [from](x1) = [1] x1 + [4] [n__from](x1) = [1] x1 + [0] [s](x1) = [0] The following symbols are considered usable {2nd, activate, from} The order satisfies the following ordering constraints: [2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [8] > [1] Y + [0] = [Y] [2nd(cons(X, X1))] = [1] X + [1] X1 + [4] ? [1] X + [1] X1 + [8] = [2nd(cons1(X, activate(X1)))] [activate(X)] = [1] X + [4] > [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [4] >= [1] X + [4] = [from(X)] [from(X)] = [1] X + [4] >= [1] X + [4] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [4] > [1] X + [0] = [n__from(X)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) } Weak Trs: { 2nd(cons1(X, cons(Y, Z))) -> Y , activate(X) -> X , 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 2' to orient following rules strictly. Trs: { 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) } 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: ---------- The following argument positions are usable: Uargs(2nd) = {1}, Uargs(cons1) = {2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [2nd](x1) = [2 1] x1 + [0] [1 0] [0] [cons1](x1, x2) = [1 0] x2 + [0] [0 0] [1] [cons](x1, x2) = [1 1] x1 + [1 0] x2 + [0] [0 0] [0 0] [4] [activate](x1) = [1 0] x1 + [0] [0 1] [4] [from](x1) = [1 1] x1 + [0] [0 0] [4] [n__from](x1) = [1 1] x1 + [0] [0 0] [0] [s](x1) = [0] [0] The following symbols are considered usable {2nd, activate, from} The order satisfies the following ordering constraints: [2nd(cons1(X, cons(Y, Z)))] = [2 2] Y + [2 0] Z + [1] [1 1] [1 0] [0] > [1 0] Y + [0] [0 1] [0] = [Y] [2nd(cons(X, X1))] = [2 2] X + [2 0] X1 + [4] [1 1] [1 0] [0] > [2 0] X1 + [1] [1 0] [0] = [2nd(cons1(X, activate(X1)))] [activate(X)] = [1 0] X + [0] [0 1] [4] >= [1 0] X + [0] [0 1] [0] = [X] [activate(n__from(X))] = [1 1] X + [0] [0 0] [4] >= [1 1] X + [0] [0 0] [4] = [from(X)] [from(X)] = [1 1] X + [0] [0 0] [4] >= [1 1] X + [0] [0 0] [4] = [cons(X, n__from(s(X)))] [from(X)] = [1 1] X + [0] [0 0] [4] >= [1 1] X + [0] [0 0] [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(1)). Weak Trs: { 2nd(cons1(X, cons(Y, Z))) -> Y , 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) , activate(X) -> X , 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))