MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) , sum(cons(0(), x), y) -> sum(x, y) , sum(nil(), y) -> y , weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0(), x))) , weight(cons(n, nil())) -> n } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { sum^#(cons(s(n), x), cons(m, y)) -> c_1(sum^#(cons(n, x), cons(s(m), y))) , sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) , sum^#(nil(), y) -> c_3() , weight^#(cons(n, cons(m, x))) -> c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))), sum^#(cons(n, cons(m, x)), cons(0(), x))) , weight^#(cons(n, nil())) -> c_5() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sum^#(cons(s(n), x), cons(m, y)) -> c_1(sum^#(cons(n, x), cons(s(m), y))) , sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) , sum^#(nil(), y) -> c_3() , weight^#(cons(n, cons(m, x))) -> c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))), sum^#(cons(n, cons(m, x)), cons(0(), x))) , weight^#(cons(n, nil())) -> c_5() } Weak Trs: { sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) , sum(cons(0(), x), y) -> sum(x, y) , sum(nil(), y) -> y , weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0(), x))) , weight(cons(n, nil())) -> n } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {3,5} by applications of Pre({3,5}) = {2,4}. Here rules are labeled as follows: DPs: { 1: sum^#(cons(s(n), x), cons(m, y)) -> c_1(sum^#(cons(n, x), cons(s(m), y))) , 2: sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) , 3: sum^#(nil(), y) -> c_3() , 4: weight^#(cons(n, cons(m, x))) -> c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))), sum^#(cons(n, cons(m, x)), cons(0(), x))) , 5: weight^#(cons(n, nil())) -> c_5() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sum^#(cons(s(n), x), cons(m, y)) -> c_1(sum^#(cons(n, x), cons(s(m), y))) , sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) , weight^#(cons(n, cons(m, x))) -> c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))), sum^#(cons(n, cons(m, x)), cons(0(), x))) } Weak DPs: { sum^#(nil(), y) -> c_3() , weight^#(cons(n, nil())) -> c_5() } Weak Trs: { sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) , sum(cons(0(), x), y) -> sum(x, y) , sum(nil(), y) -> y , weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0(), x))) , weight(cons(n, nil())) -> n } Obligation: innermost runtime complexity Answer: MAYBE The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { sum^#(nil(), y) -> c_3() , weight^#(cons(n, nil())) -> c_5() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sum^#(cons(s(n), x), cons(m, y)) -> c_1(sum^#(cons(n, x), cons(s(m), y))) , sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) , weight^#(cons(n, cons(m, x))) -> c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))), sum^#(cons(n, cons(m, x)), cons(0(), x))) } Weak Trs: { sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) , sum(cons(0(), x), y) -> sum(x, y) , sum(nil(), y) -> y , weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0(), x))) , weight(cons(n, nil())) -> n } Obligation: innermost runtime complexity Answer: MAYBE We replace rewrite rules by usable rules: Weak Usable Rules: { sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) , sum(cons(0(), x), y) -> sum(x, y) , sum(nil(), y) -> y } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sum^#(cons(s(n), x), cons(m, y)) -> c_1(sum^#(cons(n, x), cons(s(m), y))) , sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) , weight^#(cons(n, cons(m, x))) -> c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))), sum^#(cons(n, cons(m, x)), cons(0(), x))) } Weak Trs: { sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) , sum(cons(0(), x), y) -> sum(x, y) , sum(nil(), y) -> y } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..