We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^4)). Strict Trs: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(2nd(X)) -> a__2nd(mark(X)) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^4)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__2nd](x1) = [1] x1 + [4] [cons](x1, x2) = [1] x1 + [1] [mark](x1) = [0] [a__from](x1) = [1] x1 + [0] [from](x1) = [0] [s](x1) = [1] x1 + [0] [2nd](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [a__2nd(X)] = [1] X + [4] > [1] X + [0] = [2nd(X)] [a__2nd(cons(X, cons(Y, Z)))] = [1] X + [5] > [0] = [mark(Y)] [mark(cons(X1, X2))] = [0] ? [1] = [cons(mark(X1), X2)] [mark(from(X))] = [0] >= [0] = [a__from(mark(X))] [mark(s(X))] = [0] >= [0] = [s(mark(X))] [mark(2nd(X))] = [0] ? [4] = [a__2nd(mark(X))] [a__from(X)] = [1] X + [0] ? [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1] X + [0] >= [0] = [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^4)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(2nd(X)) -> a__2nd(mark(X)) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Weak Trs: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^4)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__2nd](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [0] [mark](x1) = [0] [a__from](x1) = [1] x1 + [4] [from](x1) = [0] [s](x1) = [1] x1 + [0] [2nd](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [a__2nd(X)] = [1] X + [0] >= [1] X + [0] = [2nd(X)] [a__2nd(cons(X, cons(Y, Z)))] = [1] X + [0] >= [0] = [mark(Y)] [mark(cons(X1, X2))] = [0] >= [0] = [cons(mark(X1), X2)] [mark(from(X))] = [0] ? [4] = [a__from(mark(X))] [mark(s(X))] = [0] >= [0] = [s(mark(X))] [mark(2nd(X))] = [0] >= [0] = [a__2nd(mark(X))] [a__from(X)] = [1] X + [4] > [0] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1] X + [4] > [0] = [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^4)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(2nd(X)) -> a__2nd(mark(X)) } Weak Trs: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^4)) We use the processor 'matrix interpretation of dimension 4' to orient following rules strictly. Trs: { mark(2nd(X)) -> a__2nd(mark(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^4)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [1 0 0 0] [1] [a__2nd](x1) = [1 0 0 0] x1 + [1] [1 0 0 0] [0] [0 0 0 1] [1] [1 0 0 0] [0 1 1 0] [0] [cons](x1, x2) = [0 0 0 1] x1 + [0 0 0 0] x2 + [0] [1 0 0 0] [0 1 1 0] [0] [0 0 0 1] [0 1 0 0] [0] [1 0 0 1] [0] [mark](x1) = [1 0 0 1] x1 + [1] [1 0 0 1] [0] [0 0 0 1] [1] [1 0 0 1] [0] [a__from](x1) = [1 0 0 1] x1 + [1] [1 0 0 1] [0] [0 0 0 1] [1] [1 0 0 1] [0] [from](x1) = [0 0 0 0] x1 + [0] [0 0 0 0] [0] [0 0 0 1] [1] [1 0 0 1] [0] [s](x1) = [0 0 0 1] x1 + [1] [0 0 0 1] [0] [0 0 0 1] [1] [1 0 0 0] [1] [2nd](x1) = [0 0 0 0] x1 + [0] [0 0 0 0] [0] [0 0 0 1] [1] The order satisfies the following ordering constraints: [a__2nd(X)] = [1 0 0 0] [1] [1 0 0 0] X + [1] [1 0 0 0] [0] [0 0 0 1] [1] >= [1 0 0 0] [1] [0 0 0 0] X + [0] [0 0 0 0] [0] [0 0 0 1] [1] = [2nd(X)] [a__2nd(cons(X, cons(Y, Z)))] = [1 0 0 0] [1 0 0 1] [0 1 1 0] [1] [1 0 0 0] X + [1 0 0 1] Y + [0 1 1 0] Z + [1] [1 0 0 0] [1 0 0 1] [0 1 1 0] [0] [0 0 0 1] [0 0 0 1] [0 0 0 0] [1] > [1 0 0 1] [0] [1 0 0 1] Y + [1] [1 0 0 1] [0] [0 0 0 1] [1] = [mark(Y)] [mark(cons(X1, X2))] = [1 0 0 1] [0 2 1 0] [0] [1 0 0 1] X1 + [0 2 1 0] X2 + [1] [1 0 0 1] [0 2 1 0] [0] [0 0 0 1] [0 1 0 0] [1] >= [1 0 0 1] [0 1 1 0] [0] [0 0 0 1] X1 + [0 0 0 0] X2 + [1] [1 0 0 1] [0 1 1 0] [0] [0 0 0 1] [0 1 0 0] [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1 0 0 2] [1] [1 0 0 2] X + [2] [1 0 0 2] [1] [0 0 0 1] [2] >= [1 0 0 2] [1] [1 0 0 2] X + [2] [1 0 0 2] [1] [0 0 0 1] [2] = [a__from(mark(X))] [mark(s(X))] = [1 0 0 2] [1] [1 0 0 2] X + [2] [1 0 0 2] [1] [0 0 0 1] [2] >= [1 0 0 2] [1] [0 0 0 1] X + [2] [0 0 0 1] [1] [0 0 0 1] [2] = [s(mark(X))] [mark(2nd(X))] = [1 0 0 1] [2] [1 0 0 1] X + [3] [1 0 0 1] [2] [0 0 0 1] [2] > [1 0 0 1] [1] [1 0 0 1] X + [1] [1 0 0 1] [0] [0 0 0 1] [2] = [a__2nd(mark(X))] [a__from(X)] = [1 0 0 1] [0] [1 0 0 1] X + [1] [1 0 0 1] [0] [0 0 0 1] [1] >= [1 0 0 1] [0] [0 0 0 1] X + [1] [1 0 0 1] [0] [0 0 0 1] [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 0 0 1] [0] [1 0 0 1] X + [1] [1 0 0 1] [0] [0 0 0 1] [1] >= [1 0 0 1] [0] [0 0 0 0] X + [0] [0 0 0 0] [0] [0 0 0 1] [1] = [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^4)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) } Weak Trs: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) , mark(2nd(X)) -> a__2nd(mark(X)) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^4)) We use the processor 'matrix interpretation of dimension 4' to orient following rules strictly. Trs: { mark(s(X)) -> s(mark(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^4)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [1 0 0 1] [0] [a__2nd](x1) = [0 1 0 1] x1 + [0] [0 0 1 0] [1] [0 0 0 1] [1] [1 0 0 0] [0 1 0 0] [0] [cons](x1, x2) = [1 0 0 0] x1 + [0 1 0 0] x2 + [0] [0 0 0 1] [0 0 1 0] [0] [0 0 0 1] [0 0 1 0] [0] [1 0 0 1] [0] [mark](x1) = [1 0 0 1] x1 + [0] [0 0 0 1] [1] [0 0 0 1] [1] [1 0 0 1] [1] [a__from](x1) = [1 0 0 1] x1 + [0] [0 0 0 1] [1] [0 0 0 1] [1] [1 0 0 1] [1] [from](x1) = [0 0 0 0] x1 + [0] [0 0 0 0] [0] [0 0 0 1] [1] [1 0 0 0] [0] [s](x1) = [0 0 0 1] x1 + [0] [0 0 0 0] [1] [0 0 0 1] [1] [1 0 0 1] [0] [2nd](x1) = [0 0 0 1] x1 + [0] [0 0 1 0] [0] [0 0 0 1] [1] The order satisfies the following ordering constraints: [a__2nd(X)] = [1 0 0 1] [0] [0 1 0 1] X + [0] [0 0 1 0] [1] [0 0 0 1] [1] >= [1 0 0 1] [0] [0 0 0 1] X + [0] [0 0 1 0] [0] [0 0 0 1] [1] = [2nd(X)] [a__2nd(cons(X, cons(Y, Z)))] = [1 0 0 1] [1 0 0 1] [0 1 1 0] [0] [1 0 0 1] X + [1 0 0 1] Y + [0 1 1 0] Z + [0] [0 0 0 1] [0 0 0 1] [0 0 1 0] [1] [0 0 0 1] [0 0 0 1] [0 0 1 0] [1] >= [1 0 0 1] [0] [1 0 0 1] Y + [0] [0 0 0 1] [1] [0 0 0 1] [1] = [mark(Y)] [mark(cons(X1, X2))] = [1 0 0 1] [0 1 1 0] [0] [1 0 0 1] X1 + [0 1 1 0] X2 + [0] [0 0 0 1] [0 0 1 0] [1] [0 0 0 1] [0 0 1 0] [1] >= [1 0 0 1] [0 1 0 0] [0] [1 0 0 1] X1 + [0 1 0 0] X2 + [0] [0 0 0 1] [0 0 1 0] [1] [0 0 0 1] [0 0 1 0] [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1 0 0 2] [2] [1 0 0 2] X + [2] [0 0 0 1] [2] [0 0 0 1] [2] >= [1 0 0 2] [2] [1 0 0 2] X + [1] [0 0 0 1] [2] [0 0 0 1] [2] = [a__from(mark(X))] [mark(s(X))] = [1 0 0 1] [1] [1 0 0 1] X + [1] [0 0 0 1] [2] [0 0 0 1] [2] > [1 0 0 1] [0] [0 0 0 1] X + [1] [0 0 0 0] [1] [0 0 0 1] [2] = [s(mark(X))] [mark(2nd(X))] = [1 0 0 2] [1] [1 0 0 2] X + [1] [0 0 0 1] [2] [0 0 0 1] [2] >= [1 0 0 2] [1] [1 0 0 2] X + [1] [0 0 0 1] [2] [0 0 0 1] [2] = [a__2nd(mark(X))] [a__from(X)] = [1 0 0 1] [1] [1 0 0 1] X + [0] [0 0 0 1] [1] [0 0 0 1] [1] > [1 0 0 1] [0] [1 0 0 1] X + [0] [0 0 0 1] [1] [0 0 0 1] [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 0 0 1] [1] [1 0 0 1] X + [0] [0 0 0 1] [1] [0 0 0 1] [1] >= [1 0 0 1] [1] [0 0 0 0] X + [0] [0 0 0 0] [0] [0 0 0 1] [1] = [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^4)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) } Weak Trs: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) , mark(s(X)) -> s(mark(X)) , mark(2nd(X)) -> a__2nd(mark(X)) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^4)) We use the processor 'matrix interpretation of dimension 4' to orient following rules strictly. Trs: { mark(from(X)) -> a__from(mark(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^4)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [1 0 1 0] [0] [a__2nd](x1) = [1 0 1 0] x1 + [0] [0 0 1 0] [0] [0 0 1 1] [0] [1 0 0 0] [0 1 0 1] [0] [cons](x1, x2) = [0 0 1 0] x1 + [0 0 0 0] x2 + [0] [0 0 1 0] [0 1 0 0] [0] [1 0 0 0] [0 0 0 1] [0] [1 0 1 0] [0] [mark](x1) = [1 0 1 0] x1 + [0] [0 0 1 0] [0] [1 0 1 0] [0] [1 0 1 0] [1] [a__from](x1) = [0 0 1 1] x1 + [1] [0 0 1 0] [1] [1 0 1 0] [1] [1 0 1 0] [1] [from](x1) = [0 0 0 0] x1 + [1] [0 0 1 0] [1] [0 0 0 0] [0] [1 0 0 0] [0] [s](x1) = [0 0 0 0] x1 + [0] [0 0 1 0] [1] [0 0 0 0] [0] [1 0 1 0] [0] [2nd](x1) = [0 0 1 0] x1 + [0] [0 0 1 0] [0] [0 0 1 0] [0] The order satisfies the following ordering constraints: [a__2nd(X)] = [1 0 1 0] [0] [1 0 1 0] X + [0] [0 0 1 0] [0] [0 0 1 1] [0] >= [1 0 1 0] [0] [0 0 1 0] X + [0] [0 0 1 0] [0] [0 0 1 0] [0] = [2nd(X)] [a__2nd(cons(X, cons(Y, Z)))] = [1 0 1 0] [1 0 2 0] [0 0 0 1] [0] [1 0 1 0] X + [1 0 2 0] Y + [0 0 0 1] Z + [0] [0 0 1 0] [0 0 1 0] [0 0 0 0] [0] [1 0 1 0] [1 0 1 0] [0 0 0 1] [0] >= [1 0 1 0] [0] [1 0 1 0] Y + [0] [0 0 1 0] [0] [1 0 1 0] [0] = [mark(Y)] [mark(cons(X1, X2))] = [1 0 1 0] [0 2 0 1] [0] [1 0 1 0] X1 + [0 2 0 1] X2 + [0] [0 0 1 0] [0 1 0 0] [0] [1 0 1 0] [0 2 0 1] [0] >= [1 0 1 0] [0 1 0 1] [0] [0 0 1 0] X1 + [0 0 0 0] X2 + [0] [0 0 1 0] [0 1 0 0] [0] [1 0 1 0] [0 0 0 1] [0] = [cons(mark(X1), X2)] [mark(from(X))] = [1 0 2 0] [2] [1 0 2 0] X + [2] [0 0 1 0] [1] [1 0 2 0] [2] > [1 0 2 0] [1] [1 0 2 0] X + [1] [0 0 1 0] [1] [1 0 2 0] [1] = [a__from(mark(X))] [mark(s(X))] = [1 0 1 0] [1] [1 0 1 0] X + [1] [0 0 1 0] [1] [1 0 1 0] [1] > [1 0 1 0] [0] [0 0 0 0] X + [0] [0 0 1 0] [1] [0 0 0 0] [0] = [s(mark(X))] [mark(2nd(X))] = [1 0 2 0] [0] [1 0 2 0] X + [0] [0 0 1 0] [0] [1 0 2 0] [0] >= [1 0 2 0] [0] [1 0 2 0] X + [0] [0 0 1 0] [0] [1 0 2 0] [0] = [a__2nd(mark(X))] [a__from(X)] = [1 0 1 0] [1] [0 0 1 1] X + [1] [0 0 1 0] [1] [1 0 1 0] [1] >= [1 0 1 0] [1] [0 0 1 0] X + [0] [0 0 1 0] [1] [1 0 1 0] [0] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 0 1 0] [1] [0 0 1 1] X + [1] [0 0 1 0] [1] [1 0 1 0] [1] >= [1 0 1 0] [1] [0 0 0 0] X + [1] [0 0 1 0] [1] [0 0 0 0] [0] = [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^4)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) } Weak Trs: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(2nd(X)) -> a__2nd(mark(X)) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^4)) We use the processor 'matrix interpretation of dimension 4' to orient following rules strictly. Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^4)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [1 0 0 1] [0] [a__2nd](x1) = [0 1 0 0] x1 + [0] [1 1 0 0] [0] [0 0 0 1] [0] [1 0 0 0] [0 0 1 0] [0] [cons](x1, x2) = [0 0 0 1] x1 + [0 1 0 0] x2 + [0] [1 0 0 0] [0 0 0 0] [1] [0 0 0 1] [0 1 0 0] [1] [1 0 0 1] [0] [mark](x1) = [0 0 0 1] x1 + [0] [1 0 0 1] [1] [0 0 0 1] [0] [1 0 0 1] [1] [a__from](x1) = [0 0 0 1] x1 + [1] [1 0 0 1] [1] [0 0 0 1] [1] [1 0 0 1] [0] [from](x1) = [0 0 0 0] x1 + [0] [0 0 0 0] [0] [0 0 0 1] [1] [1 0 0 1] [0] [s](x1) = [0 0 0 0] x1 + [0] [0 0 1 1] [0] [0 0 0 1] [1] [1 0 0 1] [0] [2nd](x1) = [0 1 0 0] x1 + [0] [0 0 0 0] [0] [0 0 0 1] [0] The order satisfies the following ordering constraints: [a__2nd(X)] = [1 0 0 1] [0] [0 1 0 0] X + [0] [1 1 0 0] [0] [0 0 0 1] [0] >= [1 0 0 1] [0] [0 1 0 0] X + [0] [0 0 0 0] [0] [0 0 0 1] [0] = [2nd(X)] [a__2nd(cons(X, cons(Y, Z)))] = [1 0 0 1] [1 0 0 1] [0 1 0 0] [2] [0 0 0 1] X + [0 0 0 1] Y + [0 1 0 0] Z + [0] [1 0 0 1] [1 0 0 1] [0 1 0 0] [1] [0 0 0 1] [0 0 0 1] [0 1 0 0] [1] > [1 0 0 1] [0] [0 0 0 1] Y + [0] [1 0 0 1] [1] [0 0 0 1] [0] = [mark(Y)] [mark(cons(X1, X2))] = [1 0 0 1] [0 1 1 0] [1] [0 0 0 1] X1 + [0 1 0 0] X2 + [1] [1 0 0 1] [0 1 1 0] [2] [0 0 0 1] [0 1 0 0] [1] > [1 0 0 1] [0 0 1 0] [0] [0 0 0 1] X1 + [0 1 0 0] X2 + [0] [1 0 0 1] [0 0 0 0] [1] [0 0 0 1] [0 1 0 0] [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1 0 0 2] [1] [0 0 0 1] X + [1] [1 0 0 2] [2] [0 0 0 1] [1] >= [1 0 0 2] [1] [0 0 0 1] X + [1] [1 0 0 2] [1] [0 0 0 1] [1] = [a__from(mark(X))] [mark(s(X))] = [1 0 0 2] [1] [0 0 0 1] X + [1] [1 0 0 2] [2] [0 0 0 1] [1] > [1 0 0 2] [0] [0 0 0 0] X + [0] [1 0 0 2] [1] [0 0 0 1] [1] = [s(mark(X))] [mark(2nd(X))] = [1 0 0 2] [0] [0 0 0 1] X + [0] [1 0 0 2] [1] [0 0 0 1] [0] >= [1 0 0 2] [0] [0 0 0 1] X + [0] [1 0 0 2] [0] [0 0 0 1] [0] = [a__2nd(mark(X))] [a__from(X)] = [1 0 0 1] [1] [0 0 0 1] X + [1] [1 0 0 1] [1] [0 0 0 1] [1] > [1 0 0 1] [0] [0 0 0 1] X + [0] [1 0 0 1] [1] [0 0 0 1] [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 0 0 1] [1] [0 0 0 1] X + [1] [1 0 0 1] [1] [0 0 0 1] [1] > [1 0 0 1] [0] [0 0 0 0] X + [0] [0 0 0 0] [0] [0 0 0 1] [1] = [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: { a__2nd(X) -> 2nd(X) , a__2nd(cons(X, cons(Y, Z))) -> mark(Y) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(2nd(X)) -> a__2nd(mark(X)) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> 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^4))