YES(O(1),O(1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { main(S(x1)) -> x1 , main(0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We add the following innermost weak dependency pairs: Strict DPs: { main^#(S(x1)) -> c_1() , main^#(0()) -> c_2() } 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: { main^#(S(x1)) -> c_1() , main^#(0()) -> c_2() } Strict Trs: { main(S(x1)) -> x1 , main(0()) -> 0() } 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)). Strict DPs: { main^#(S(x1)) -> c_1() , main^#(0()) -> c_2() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: none TcT has computed the following constructor-restricted matrix interpretation. [S](x1) = [2] [1] [0] = [1] [1] [main^#](x1) = [1 1] x1 + [2] [1 1] [1] [c_1] = [0] [0] [c_2] = [1] [1] The following symbols are considered usable {main^#} The order satisfies the following ordering constraints: [main^#(S(x1))] = [5] [4] > [0] [0] = [c_1()] [main^#(0())] = [4] [3] > [1] [1] = [c_2()] 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)). Weak DPs: { main^#(S(x1)) -> c_1() , main^#(0()) -> c_2() } 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. { main^#(S(x1)) -> c_1() , main^#(0()) -> c_2() } 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))