YES(?,O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict Trs: { a__f(X, X) -> a__f(a(), b()) , a__f(X1, X2) -> f(X1, X2) , a__b() -> a() , a__b() -> b() , mark(a()) -> a() , mark(b()) -> a__b() , mark(f(X1, X2)) -> a__f(mark(X1), X2) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We add following dependency tuples: Strict DPs: { a__f^#(X, X) -> c_1(a__f^#(a(), b())) , a__f^#(X1, X2) -> c_2() , a__b^#() -> c_3() , a__b^#() -> c_4() , mark^#(a()) -> c_5() , mark^#(b()) -> c_6(a__b^#()) , mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } 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^#(X, X) -> c_1(a__f^#(a(), b())) , a__f^#(X1, X2) -> c_2() , a__b^#() -> c_3() , a__b^#() -> c_4() , mark^#(a()) -> c_5() , mark^#(b()) -> c_6(a__b^#()) , mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } Weak Trs: { a__f(X, X) -> a__f(a(), b()) , a__f(X1, X2) -> f(X1, X2) , a__b() -> a() , a__b() -> b() , mark(a()) -> a() , mark(b()) -> a__b() , mark(f(X1, X2)) -> a__f(mark(X1), X2) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {2,3,4,5} by applications of Pre({2,3,4,5}) = {1,6,7}. Here rules are labeled as follows: DPs: { 1: a__f^#(X, X) -> c_1(a__f^#(a(), b())) , 2: a__f^#(X1, X2) -> c_2() , 3: a__b^#() -> c_3() , 4: a__b^#() -> c_4() , 5: mark^#(a()) -> c_5() , 6: mark^#(b()) -> c_6(a__b^#()) , 7: mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { a__f^#(X, X) -> c_1(a__f^#(a(), b())) , mark^#(b()) -> c_6(a__b^#()) , mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } Weak DPs: { a__f^#(X1, X2) -> c_2() , a__b^#() -> c_3() , a__b^#() -> c_4() , mark^#(a()) -> c_5() } Weak Trs: { a__f(X, X) -> a__f(a(), b()) , a__f(X1, X2) -> f(X1, X2) , a__b() -> a() , a__b() -> b() , mark(a()) -> a() , mark(b()) -> a__b() , mark(f(X1, X2)) -> a__f(mark(X1), X2) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1,2} by applications of Pre({1,2}) = {3}. Here rules are labeled as follows: DPs: { 1: a__f^#(X, X) -> c_1(a__f^#(a(), b())) , 2: mark^#(b()) -> c_6(a__b^#()) , 3: mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) , 4: a__f^#(X1, X2) -> c_2() , 5: a__b^#() -> c_3() , 6: a__b^#() -> c_4() , 7: mark^#(a()) -> c_5() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } Weak DPs: { a__f^#(X, X) -> c_1(a__f^#(a(), b())) , a__f^#(X1, X2) -> c_2() , a__b^#() -> c_3() , a__b^#() -> c_4() , mark^#(a()) -> c_5() , mark^#(b()) -> c_6(a__b^#()) } Weak Trs: { a__f(X, X) -> a__f(a(), b()) , a__f(X1, X2) -> f(X1, X2) , a__b() -> a() , a__b() -> b() , mark(a()) -> a() , mark(b()) -> a__b() , mark(f(X1, X2)) -> a__f(mark(X1), X2) } 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^#(X, X) -> c_1(a__f^#(a(), b())) , a__f^#(X1, X2) -> c_2() , a__b^#() -> c_3() , a__b^#() -> c_4() , mark^#(a()) -> c_5() , mark^#(b()) -> c_6(a__b^#()) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } Weak Trs: { a__f(X, X) -> a__f(a(), b()) , a__f(X1, X2) -> f(X1, X2) , a__b() -> a() , a__b() -> b() , mark(a()) -> a() , mark(b()) -> a__b() , mark(f(X1, X2)) -> a__f(mark(X1), X2) } 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^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(f(X1, X2)) -> c_1(mark^#(X1)) } Weak Trs: { a__f(X, X) -> a__f(a(), b()) , a__f(X1, X2) -> f(X1, X2) , a__b() -> a() , a__b() -> b() , mark(a()) -> a() , mark(b()) -> a__b() , mark(f(X1, X2)) -> a__f(mark(X1), X2) } 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^#(f(X1, X2)) -> c_1(mark^#(X1)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following argument positions are usable: Uargs(c_1) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [f](x1, x2) = [1] x1 + [1] x2 + [3] [mark^#](x1) = [1] x1 + [3] [c_1](x1) = [1] x1 + [2] This order satisfies following ordering constraints: [mark^#(f(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [5] = [c_1(mark^#(X1))] Hurray, we answered YES(?,O(n^1))