YES(?,O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We add following dependency tuples: Strict DPs: { a__f^#(X1, X2, X3) -> c_1() , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , a__g^#(X) -> c_3() , a__g^#(b()) -> c_4() , a__b^#() -> c_5() , a__b^#() -> c_6() , mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , mark^#(b()) -> c_8(a__b^#()) , mark^#(c()) -> c_9() , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { a__f^#(X1, X2, X3) -> c_1() , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , a__g^#(X) -> c_3() , a__g^#(b()) -> c_4() , a__b^#() -> c_5() , a__b^#() -> c_6() , mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , mark^#(b()) -> c_8(a__b^#()) , mark^#(c()) -> c_9() , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } Weak Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1,3,4,5,6,9} by applications of Pre({1,3,4,5,6,9}) = {2,7,8,10}. Here rules are labeled as follows: DPs: { 1: a__f^#(X1, X2, X3) -> c_1() , 2: a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , 3: a__g^#(X) -> c_3() , 4: a__g^#(b()) -> c_4() , 5: a__b^#() -> c_5() , 6: a__b^#() -> c_6() , 7: mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , 8: mark^#(b()) -> c_8(a__b^#()) , 9: mark^#(c()) -> c_9() , 10: mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , mark^#(b()) -> c_8(a__b^#()) , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } Weak DPs: { a__f^#(X1, X2, X3) -> c_1() , a__g^#(X) -> c_3() , a__g^#(b()) -> c_4() , a__b^#() -> c_5() , a__b^#() -> c_6() , mark^#(c()) -> c_9() } Weak Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1,3} by applications of Pre({1,3}) = {2,4}. Here rules are labeled as follows: DPs: { 1: a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , 2: mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , 3: mark^#(b()) -> c_8(a__b^#()) , 4: mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) , 5: a__f^#(X1, X2, X3) -> c_1() , 6: a__g^#(X) -> c_3() , 7: a__g^#(b()) -> c_4() , 8: a__b^#() -> c_5() , 9: a__b^#() -> c_6() , 10: mark^#(c()) -> c_9() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } Weak DPs: { a__f^#(X1, X2, X3) -> c_1() , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , a__g^#(X) -> c_3() , a__g^#(b()) -> c_4() , a__b^#() -> c_5() , a__b^#() -> c_6() , mark^#(b()) -> c_8(a__b^#()) , mark^#(c()) -> c_9() } Weak Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {2} by applications of Pre({2}) = {1}. Here rules are labeled as follows: DPs: { 1: mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) , 2: mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) , 3: a__f^#(X1, X2, X3) -> c_1() , 4: a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , 5: a__g^#(X) -> c_3() , 6: a__g^#(b()) -> c_4() , 7: a__b^#() -> c_5() , 8: a__b^#() -> c_6() , 9: mark^#(b()) -> c_8(a__b^#()) , 10: mark^#(c()) -> c_9() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) } Weak DPs: { a__f^#(X1, X2, X3) -> c_1() , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , a__g^#(X) -> c_3() , a__g^#(b()) -> c_4() , a__b^#() -> c_5() , a__b^#() -> c_6() , mark^#(b()) -> c_8(a__b^#()) , mark^#(c()) -> c_9() , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } Weak Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { a__f^#(X1, X2, X3) -> c_1() , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y)) , a__g^#(X) -> c_3() , a__g^#(b()) -> c_4() , a__b^#() -> c_5() , a__b^#() -> c_6() , mark^#(b()) -> c_8(a__b^#()) , mark^#(c()) -> c_9() , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) } Weak Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(g(X)) -> c_1(mark^#(X)) } Weak Trs: { a__f(X1, X2, X3) -> f(X1, X2, X3) , a__f(X, g(X), Y) -> a__f(Y, Y, Y) , a__g(X) -> g(X) , a__g(b()) -> c() , a__b() -> b() , a__b() -> c() , mark(g(X)) -> a__g(mark(X)) , mark(b()) -> a__b() , mark(c()) -> c() , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) } Obligation: innermost runtime complexity Answer: YES(?,O(n^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(n^1)). Strict DPs: { mark^#(g(X)) -> c_1(mark^#(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The input was oriented with the instance of 'Small Polynomial Path Order (PS)' as induced by the safe mapping safe(g) = {1}, safe(mark^#) = {}, safe(c_1) = {} and precedence empty . Following symbols are considered recursive: {mark^#} The recursion depth is 1. Further, following argument filtering is employed: pi(g) = [1], pi(mark^#) = [1], pi(c_1) = [1] Usable defined function symbols are a subset of: {mark^#} For your convenience, here are the satisfied ordering constraints: pi(mark^#(g(X))) = mark^#(g(; X);) > c_1(mark^#(X;);) = pi(c_1(mark^#(X))) Hurray, we answered YES(?,O(n^1))