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: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , q(s(X)) -> s(p(q(X), d(X))) , s(X) -> ns(X) , p(X, 0()) -> X , p(0(), X) -> X , p(s(X), s(Y)) -> s(s(p(X, Y))) , d(0()) -> 0() , d(s(X)) -> s(s(d(X))) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , f(s(X), cs(Y, Z)) -> cs(Y, nf(X, a(Z))) , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { q(s(X)) -> s(p(q(X), d(X))) , p(s(X), s(Y)) -> s(s(p(X, Y))) , d(s(X)) -> s(s(d(X))) , f(s(X), cs(Y, Z)) -> cs(Y, nf(X, a(Z))) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , p(X, 0()) -> X , p(0(), X) -> X , d(0()) -> 0() , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { t^#(N) -> c_1(q^#(N)) , t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() , a^#(X) -> c_10() , a^#(nt(X)) -> c_11(t^#(a(X))) , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { t^#(N) -> c_1(q^#(N)) , t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() , a^#(X) -> c_10() , a^#(nt(X)) -> c_11(t^#(a(X))) , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } Strict Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , p(X, 0()) -> X , p(0(), X) -> X , d(0()) -> 0() , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { t^#(N) -> c_1(q^#(N)) , t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() , a^#(X) -> c_10() , a^#(nt(X)) -> c_11(t^#(a(X))) , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } Strict Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(t) = {1}, Uargs(cs) = {1}, Uargs(r) = {1}, Uargs(s) = {1}, Uargs(f) = {1, 2}, Uargs(t^#) = {1}, Uargs(c_1) = {1}, Uargs(s^#) = {1}, Uargs(f^#) = {1, 2}, Uargs(c_11) = {1}, Uargs(c_12) = {1}, Uargs(c_13) = {1} TcT has computed the following constructor-restricted matrix interpretation. [t](x1) = [1 1] x1 + [1] [0 0] [2] [cs](x1, x2) = [1 0] x1 + [0] [0 0] [1] [r](x1) = [1 0] x1 + [0] [0 0] [2] [q](x1) = [0 1] x1 + [0] [0 0] [1] [nt](x1) = [1 1] x1 + [0] [0 0] [2] [ns](x1) = [1 0] x1 + [1] [0 1] [2] [0] = [0] [1] [s](x1) = [1 0] x1 + [2] [0 1] [2] [f](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [nil] = [1] [1] [nf](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 1] [0 1] [2] [a](x1) = [2 1] x1 + [1] [0 1] [0] [t^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_1](x1) = [1 0] x1 + [1] [0 1] [1] [q^#](x1) = [1 0] x1 + [2] [1 2] [2] [c_2] = [0] [1] [c_3] = [1] [0] [s^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_4] = [0] [1] [p^#](x1, x2) = [2 2] x1 + [1 2] x2 + [2] [1 2] [1 2] [2] [c_5] = [1] [0] [c_6] = [1] [0] [d^#](x1) = [1 2] x1 + [2] [2 2] [2] [c_7] = [1] [0] [f^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 0] [0 0] [2] [c_8] = [1] [1] [c_9] = [1] [1] [a^#](x1) = [2 1] x1 + [0] [0 0] [0] [c_10] = [1] [1] [c_11](x1) = [1 0] x1 + [2] [0 1] [2] [c_12](x1) = [1 0] x1 + [2] [0 1] [2] [c_13](x1) = [1 0] x1 + [2] [0 1] [2] The following symbols are considered usable {t, q, s, f, a, t^#, q^#, s^#, p^#, d^#, f^#, a^#} The order satisfies the following ordering constraints: [t(N)] = [1 1] N + [1] [0 0] [2] > [0 1] N + [0] [0 0] [1] = [cs(r(q(N)), nt(ns(N)))] [t(X)] = [1 1] X + [1] [0 0] [2] > [1 1] X + [0] [0 0] [2] = [nt(X)] [q(0())] = [1] [1] > [0] [1] = [0()] [s(X)] = [1 0] X + [2] [0 1] [2] > [1 0] X + [1] [0 1] [2] = [ns(X)] [f(X1, X2)] = [1 0] X1 + [1 0] X2 + [2] [0 1] [0 1] [2] > [1 0] X1 + [1 0] X2 + [1] [0 1] [0 1] [2] = [nf(X1, X2)] [f(0(), X)] = [1 0] X + [2] [0 1] [3] > [1] [1] = [nil()] [a(X)] = [2 1] X + [1] [0 1] [0] > [1 0] X + [0] [0 1] [0] = [X] [a(nt(X))] = [2 2] X + [3] [0 0] [2] > [2 2] X + [2] [0 0] [2] = [t(a(X))] [a(ns(X))] = [2 1] X + [5] [0 1] [2] > [2 1] X + [3] [0 1] [2] = [s(a(X))] [a(nf(X1, X2))] = [2 1] X1 + [2 1] X2 + [5] [0 1] [0 1] [2] > [2 1] X1 + [2 1] X2 + [4] [0 1] [0 1] [2] = [f(a(X1), a(X2))] [t^#(N)] = [1 0] N + [1] [0 0] [2] ? [1 0] N + [3] [1 2] [3] = [c_1(q^#(N))] [t^#(X)] = [1 0] X + [1] [0 0] [2] > [0] [1] = [c_2()] [q^#(0())] = [2] [4] > [1] [0] = [c_3()] [s^#(X)] = [1 0] X + [1] [0 0] [2] > [0] [1] = [c_4()] [p^#(X, 0())] = [2 2] X + [4] [1 2] [4] > [1] [0] = [c_5()] [p^#(0(), X)] = [1 2] X + [4] [1 2] [4] > [1] [0] = [c_6()] [d^#(0())] = [4] [4] > [1] [0] = [c_7()] [f^#(X1, X2)] = [1 0] X1 + [1 0] X2 + [2] [0 0] [0 0] [2] > [1] [1] = [c_8()] [f^#(0(), X)] = [1 0] X + [2] [0 0] [2] > [1] [1] = [c_9()] [a^#(X)] = [2 1] X + [0] [0 0] [0] ? [1] [1] = [c_10()] [a^#(nt(X))] = [2 2] X + [2] [0 0] [0] ? [2 1] X + [4] [0 0] [4] = [c_11(t^#(a(X)))] [a^#(ns(X))] = [2 1] X + [4] [0 0] [0] ? [2 1] X + [4] [0 0] [4] = [c_12(s^#(a(X)))] [a^#(nf(X1, X2))] = [2 1] X1 + [2 1] X2 + [4] [0 0] [0 0] [0] ? [2 1] X1 + [2 1] X2 + [6] [0 0] [0 0] [4] = [c_13(f^#(a(X1), a(X2)))] 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(1)). Strict DPs: { t^#(N) -> c_1(q^#(N)) , a^#(X) -> c_10() , a^#(nt(X)) -> c_11(t^#(a(X))) , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } Weak DPs: { t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() } Weak Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {1,2,4,5} by applications of Pre({1,2,4,5}) = {3}. Here rules are labeled as follows: DPs: { 1: t^#(N) -> c_1(q^#(N)) , 2: a^#(X) -> c_10() , 3: a^#(nt(X)) -> c_11(t^#(a(X))) , 4: a^#(ns(X)) -> c_12(s^#(a(X))) , 5: a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) , 6: t^#(X) -> c_2() , 7: q^#(0()) -> c_3() , 8: s^#(X) -> c_4() , 9: p^#(X, 0()) -> c_5() , 10: p^#(0(), X) -> c_6() , 11: d^#(0()) -> c_7() , 12: f^#(X1, X2) -> c_8() , 13: f^#(0(), X) -> c_9() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { a^#(nt(X)) -> c_11(t^#(a(X))) } Weak DPs: { t^#(N) -> c_1(q^#(N)) , t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() , a^#(X) -> c_10() , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } Weak Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {1} by applications of Pre({1}) = {}. Here rules are labeled as follows: DPs: { 1: a^#(nt(X)) -> c_11(t^#(a(X))) , 2: t^#(N) -> c_1(q^#(N)) , 3: t^#(X) -> c_2() , 4: q^#(0()) -> c_3() , 5: s^#(X) -> c_4() , 6: p^#(X, 0()) -> c_5() , 7: p^#(0(), X) -> c_6() , 8: d^#(0()) -> c_7() , 9: f^#(X1, X2) -> c_8() , 10: f^#(0(), X) -> c_9() , 11: a^#(X) -> c_10() , 12: a^#(ns(X)) -> c_12(s^#(a(X))) , 13: a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { t^#(N) -> c_1(q^#(N)) , t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() , a^#(X) -> c_10() , a^#(nt(X)) -> c_11(t^#(a(X))) , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } Weak Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { t^#(N) -> c_1(q^#(N)) , t^#(X) -> c_2() , q^#(0()) -> c_3() , s^#(X) -> c_4() , p^#(X, 0()) -> c_5() , p^#(0(), X) -> c_6() , d^#(0()) -> c_7() , f^#(X1, X2) -> c_8() , f^#(0(), X) -> c_9() , a^#(X) -> c_10() , a^#(nt(X)) -> c_11(t^#(a(X))) , a^#(ns(X)) -> c_12(s^#(a(X))) , a^#(nf(X1, X2)) -> c_13(f^#(a(X1), a(X2))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { t(N) -> cs(r(q(N)), nt(ns(N))) , t(X) -> nt(X) , q(0()) -> 0() , s(X) -> ns(X) , f(X1, X2) -> nf(X1, X2) , f(0(), X) -> nil() , a(X) -> X , a(nt(X)) -> t(a(X)) , a(ns(X)) -> s(a(X)) , a(nf(X1, X2)) -> f(a(X1), a(X2)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))