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: { 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(1),O(n^1)) We add the 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(1),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(1),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(1),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(1),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(1),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(1),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(1),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(1),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(1),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(1),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(1),O(n^1)). Strict DPs: { mark^#(f(X1, X2)) -> c_1(mark^#(X1)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: mark^#(f(X1, X2)) -> c_1(mark^#(X1)) } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(a__f) = {}, safe(a) = {}, safe(b) = {}, safe(a__b) = {}, safe(mark) = {}, safe(f) = {1, 2}, safe(a__f^#) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(a__b^#) = {}, safe(c_3) = {}, safe(c_4) = {}, safe(mark^#) = {}, safe(c_5) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(c) = {}, safe(c_1) = {} and precedence empty . Following symbols are considered recursive: {mark^#} The recursion depth is 1. Further, following argument filtering is employed: pi(a__f) = [], pi(a) = [], pi(b) = [], pi(a__b) = [], pi(mark) = [], pi(f) = [1, 2], pi(a__f^#) = [], pi(c_1) = [], pi(c_2) = [], pi(a__b^#) = [], pi(c_3) = [], pi(c_4) = [], pi(mark^#) = [1], pi(c_5) = [], pi(c_6) = [], pi(c_7) = [], pi(c) = [], pi(c_1) = [1] Usable defined function symbols are a subset of: {a__f^#, a__b^#, mark^#} For your convenience, here are the satisfied ordering constraints: pi(mark^#(f(X1, X2))) = mark^#(f(; X1, X2);) > c_1(mark^#(X1;);) = pi(c_1(mark^#(X1))) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { mark^#(f(X1, X2)) -> c_1(mark^#(X1)) } 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. { mark^#(f(X1, X2)) -> c_1(mark^#(X1)) } 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))