We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) , lt(x, 0()) -> false() , lt(0(), s(y)) -> true() , lt(s(x), s(y)) -> lt(x, y) , if(false(), x, y) -> y , if(true(), x, y) -> x , div(x, 0()) -> 0() , div(0(), y) -> 0() , div(s(x), s(y)) -> if(lt(x, y), 0(), s(div(-(x, y), s(y)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We add the following dependency tuples: Strict DPs: { -^#(x, 0()) -> c_1() , -^#(0(), s(y)) -> c_2() , -^#(s(x), s(y)) -> c_3(-^#(x, y)) , lt^#(x, 0()) -> c_4() , lt^#(0(), s(y)) -> c_5() , lt^#(s(x), s(y)) -> c_6(lt^#(x, y)) , if^#(false(), x, y) -> c_7() , if^#(true(), x, y) -> c_8() , div^#(x, 0()) -> c_9() , div^#(0(), y) -> c_10() , div^#(s(x), s(y)) -> c_11(if^#(lt(x, y), 0(), s(div(-(x, y), s(y)))), lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict DPs: { -^#(x, 0()) -> c_1() , -^#(0(), s(y)) -> c_2() , -^#(s(x), s(y)) -> c_3(-^#(x, y)) , lt^#(x, 0()) -> c_4() , lt^#(0(), s(y)) -> c_5() , lt^#(s(x), s(y)) -> c_6(lt^#(x, y)) , if^#(false(), x, y) -> c_7() , if^#(true(), x, y) -> c_8() , div^#(x, 0()) -> c_9() , div^#(0(), y) -> c_10() , div^#(s(x), s(y)) -> c_11(if^#(lt(x, y), 0(), s(div(-(x, y), s(y)))), lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) , lt(x, 0()) -> false() , lt(0(), s(y)) -> true() , lt(s(x), s(y)) -> lt(x, y) , if(false(), x, y) -> y , if(true(), x, y) -> x , div(x, 0()) -> 0() , div(0(), y) -> 0() , div(s(x), s(y)) -> if(lt(x, y), 0(), s(div(-(x, y), s(y)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We estimate the number of application of {1,2,4,5,7,8,9,10} by applications of Pre({1,2,4,5,7,8,9,10}) = {3,6,11}. Here rules are labeled as follows: DPs: { 1: -^#(x, 0()) -> c_1() , 2: -^#(0(), s(y)) -> c_2() , 3: -^#(s(x), s(y)) -> c_3(-^#(x, y)) , 4: lt^#(x, 0()) -> c_4() , 5: lt^#(0(), s(y)) -> c_5() , 6: lt^#(s(x), s(y)) -> c_6(lt^#(x, y)) , 7: if^#(false(), x, y) -> c_7() , 8: if^#(true(), x, y) -> c_8() , 9: div^#(x, 0()) -> c_9() , 10: div^#(0(), y) -> c_10() , 11: div^#(s(x), s(y)) -> c_11(if^#(lt(x, y), 0(), s(div(-(x, y), s(y)))), lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict DPs: { -^#(s(x), s(y)) -> c_3(-^#(x, y)) , lt^#(s(x), s(y)) -> c_6(lt^#(x, y)) , div^#(s(x), s(y)) -> c_11(if^#(lt(x, y), 0(), s(div(-(x, y), s(y)))), lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Weak DPs: { -^#(x, 0()) -> c_1() , -^#(0(), s(y)) -> c_2() , lt^#(x, 0()) -> c_4() , lt^#(0(), s(y)) -> c_5() , if^#(false(), x, y) -> c_7() , if^#(true(), x, y) -> c_8() , div^#(x, 0()) -> c_9() , div^#(0(), y) -> c_10() } Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) , lt(x, 0()) -> false() , lt(0(), s(y)) -> true() , lt(s(x), s(y)) -> lt(x, y) , if(false(), x, y) -> y , if(true(), x, y) -> x , div(x, 0()) -> 0() , div(0(), y) -> 0() , div(s(x), s(y)) -> if(lt(x, y), 0(), s(div(-(x, y), s(y)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { -^#(x, 0()) -> c_1() , -^#(0(), s(y)) -> c_2() , lt^#(x, 0()) -> c_4() , lt^#(0(), s(y)) -> c_5() , if^#(false(), x, y) -> c_7() , if^#(true(), x, y) -> c_8() , div^#(x, 0()) -> c_9() , div^#(0(), y) -> c_10() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict DPs: { -^#(s(x), s(y)) -> c_3(-^#(x, y)) , lt^#(s(x), s(y)) -> c_6(lt^#(x, y)) , div^#(s(x), s(y)) -> c_11(if^#(lt(x, y), 0(), s(div(-(x, y), s(y)))), lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) , lt(x, 0()) -> false() , lt(0(), s(y)) -> true() , lt(s(x), s(y)) -> lt(x, y) , if(false(), x, y) -> y , if(true(), x, y) -> x , div(x, 0()) -> 0() , div(0(), y) -> 0() , div(s(x), s(y)) -> if(lt(x, y), 0(), s(div(-(x, y), s(y)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { div^#(s(x), s(y)) -> c_11(if^#(lt(x, y), 0(), s(div(-(x, y), s(y)))), lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict DPs: { -^#(s(x), s(y)) -> c_1(-^#(x, y)) , lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) , div^#(s(x), s(y)) -> c_3(lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) , lt(x, 0()) -> false() , lt(0(), s(y)) -> true() , lt(s(x), s(y)) -> lt(x, y) , if(false(), x, y) -> y , if(true(), x, y) -> x , div(x, 0()) -> 0() , div(0(), y) -> 0() , div(s(x), s(y)) -> if(lt(x, y), 0(), s(div(-(x, y), s(y)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We replace rewrite rules by usable rules: Weak Usable Rules: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict DPs: { -^#(s(x), s(y)) -> c_1(-^#(x, y)) , lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) , div^#(s(x), s(y)) -> c_3(lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'Small Polynomial Path Order (PS,2-bounded)' to orient following rules strictly. DPs: { 1: -^#(s(x), s(y)) -> c_1(-^#(x, y)) , 2: lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) , 3: div^#(s(x), s(y)) -> c_3(lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Trs: { -(s(x), s(y)) -> -(x, y) } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,2-bounded)' as induced by the safe mapping safe(-) = {1}, safe(0) = {}, safe(s) = {1}, safe(-^#) = {2}, safe(lt^#) = {2}, safe(div^#) = {2}, safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {} and precedence div^# > -^#, div^# > lt^# . Following symbols are considered recursive: {-^#, lt^#, div^#} The recursion depth is 2. Further, following argument filtering is employed: pi(-) = 1, pi(0) = [], pi(s) = [1], pi(-^#) = [1], pi(lt^#) = [1], pi(div^#) = [1, 2], pi(c_1) = [1], pi(c_2) = [1], pi(c_3) = [1, 2, 3] Usable defined function symbols are a subset of: {-, -^#, lt^#, div^#} For your convenience, here are the satisfied ordering constraints: pi(-^#(s(x), s(y))) = -^#(s(; x);) > c_1(-^#(x;);) = pi(c_1(-^#(x, y))) pi(lt^#(s(x), s(y))) = lt^#(s(; x);) > c_2(lt^#(x;);) = pi(c_2(lt^#(x, y))) pi(div^#(s(x), s(y))) = div^#(s(; x); s(; y)) > c_3(lt^#(x;), div^#(x; s(; y)), -^#(x;);) = pi(c_3(lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y))) pi(-(x, 0())) = x >= x = pi(x) pi(-(0(), s(y))) = 0() >= 0() = pi(0()) pi(-(s(x), s(y))) = s(; x) > x = pi(-(x, y)) 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: { -^#(s(x), s(y)) -> c_1(-^#(x, y)) , lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) , div^#(s(x), s(y)) -> c_3(lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) } 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. { -^#(s(x), s(y)) -> c_1(-^#(x, y)) , lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) , div^#(s(x), s(y)) -> c_3(lt^#(x, y), div^#(-(x, y), s(y)), -^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { -(x, 0()) -> x , -(0(), s(y)) -> 0() , -(s(x), s(y)) -> -(x, y) } 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)). 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^2))