We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { f(x, y, z) -> g(<=(x, y), x, y, z) , g(true(), x, y, z) -> z , g(false(), x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) , p(0()) -> 0() , p(s(x)) -> x } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We add the following dependency tuples: Strict DPs: { f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , g^#(true(), x, y, z) -> c_2() , g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) , p^#(0()) -> c_4() , p^#(s(x)) -> c_5() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , g^#(true(), x, y, z) -> c_2() , g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) , p^#(0()) -> c_4() , p^#(s(x)) -> c_5() } Weak Trs: { f(x, y, z) -> g(<=(x, y), x, y, z) , g(true(), x, y, z) -> z , g(false(), x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) , p(0()) -> 0() , p(s(x)) -> x } 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: f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , 2: g^#(true(), x, y, z) -> c_2() , 3: g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) , 4: p^#(0()) -> c_4() , 5: p^#(s(x)) -> c_5() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) } Weak DPs: { f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , g^#(true(), x, y, z) -> c_2() , p^#(0()) -> c_4() , p^#(s(x)) -> c_5() } Weak Trs: { f(x, y, z) -> g(<=(x, y), x, y, z) , g(true(), x, y, z) -> z , g(false(), x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) , p(0()) -> 0() , p(s(x)) -> x } 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: g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) , 2: f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , 3: g^#(true(), x, y, z) -> c_2() , 4: p^#(0()) -> c_4() , 5: p^#(s(x)) -> c_5() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , g^#(true(), x, y, z) -> c_2() , g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) , p^#(0()) -> c_4() , p^#(s(x)) -> c_5() } Weak Trs: { f(x, y, z) -> g(<=(x, y), x, y, z) , g(true(), x, y, z) -> z , g(false(), x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) , p(0()) -> 0() , p(s(x)) -> x } 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. { f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z)) , g^#(true(), x, y, z) -> c_2() , g^#(false(), x, y, z) -> c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)), f^#(p(x), y, z), p^#(x), f^#(p(y), z, x), p^#(y), f^#(p(z), x, y), p^#(z)) , p^#(0()) -> c_4() , p^#(s(x)) -> c_5() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { f(x, y, z) -> g(<=(x, y), x, y, z) , g(true(), x, y, z) -> z , g(false(), x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) , p(0()) -> 0() , p(s(x)) -> x } 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(1))