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: { gcd(x, y) -> gcd[Ite][False][Ite][False][Ite](equal0(x, y), x, y) , monus(S(x'), S(x)) -> monus(x', x) , equal0(a, b) -> equal0[Ite](<(a, b), a, b) } Weak Trs: { equal0[Ite][True][Ite](True(), a, b) -> True() , equal0[Ite][True][Ite](False(), a, b) -> False() , <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() , equal0[Ite](True(), a, b) -> equal0[Ite][True][Ite](<(b, a), a, b) , equal0[Ite](False(), a, b) -> False() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { gcd^#(x, y) -> c_1(equal0^#(x, y)) , equal0^#(a, b) -> c_3(equal0[Ite]^#(<(a, b), a, b)) , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Weak DPs: { equal0[Ite]^#(True(), a, b) -> c_9(equal0[Ite][True][Ite]^#(<(b, a), a, b)) , equal0[Ite]^#(False(), a, b) -> c_10() , equal0[Ite][True][Ite]^#(True(), a, b) -> c_4() , equal0[Ite][True][Ite]^#(False(), a, b) -> c_5() , <^#(x, 0()) -> c_6() , <^#(S(x), S(y)) -> c_7(<^#(x, y)) , <^#(0(), S(y)) -> c_8() } 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: { gcd^#(x, y) -> c_1(equal0^#(x, y)) , equal0^#(a, b) -> c_3(equal0[Ite]^#(<(a, b), a, b)) , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Strict Trs: { gcd(x, y) -> gcd[Ite][False][Ite][False][Ite](equal0(x, y), x, y) , monus(S(x'), S(x)) -> monus(x', x) , equal0(a, b) -> equal0[Ite](<(a, b), a, b) } Weak DPs: { equal0[Ite]^#(True(), a, b) -> c_9(equal0[Ite][True][Ite]^#(<(b, a), a, b)) , equal0[Ite]^#(False(), a, b) -> c_10() , equal0[Ite][True][Ite]^#(True(), a, b) -> c_4() , equal0[Ite][True][Ite]^#(False(), a, b) -> c_5() , <^#(x, 0()) -> c_6() , <^#(S(x), S(y)) -> c_7(<^#(x, y)) , <^#(0(), S(y)) -> c_8() } Weak Trs: { equal0[Ite][True][Ite](True(), a, b) -> True() , equal0[Ite][True][Ite](False(), a, b) -> False() , <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() , equal0[Ite](True(), a, b) -> equal0[Ite][True][Ite](<(b, a), a, b) , equal0[Ite](False(), a, b) -> False() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Weak Usable Rules: { <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { gcd^#(x, y) -> c_1(equal0^#(x, y)) , equal0^#(a, b) -> c_3(equal0[Ite]^#(<(a, b), a, b)) , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Weak DPs: { equal0[Ite]^#(True(), a, b) -> c_9(equal0[Ite][True][Ite]^#(<(b, a), a, b)) , equal0[Ite]^#(False(), a, b) -> c_10() , equal0[Ite][True][Ite]^#(True(), a, b) -> c_4() , equal0[Ite][True][Ite]^#(False(), a, b) -> c_5() , <^#(x, 0()) -> c_6() , <^#(S(x), S(y)) -> c_7(<^#(x, y)) , <^#(0(), S(y)) -> c_8() } Weak Trs: { <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(equal0[Ite]^#) = {1}, Uargs(equal0[Ite][True][Ite]^#) = {1}, Uargs(c_7) = {1}, Uargs(c_9) = {1} TcT has computed the following constructor-restricted matrix interpretation. [True] = [0] [0] [S](x1) = [1 0] x1 + [0] [0 0] [0] [<](x1, x2) = [0] [0] [0] = [0] [0] [False] = [0] [0] [gcd^#](x1, x2) = [2 2] x1 + [2 2] x2 + [2] [2 2] [1 2] [2] [c_1](x1) = [1 0] x1 + [0] [0 1] [0] [equal0^#](x1, x2) = [1 2] x1 + [1 2] x2 + [1] [2 1] [1 1] [2] [monus^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2] [1 2] [1 2] [2] [c_2](x1) = [1 0] x1 + [1] [0 1] [1] [c_3](x1) = [1 0] x1 + [2] [0 1] [2] [equal0[Ite]^#](x1, x2, x3) = [2 0] x1 + [0] [0 0] [0] [equal0[Ite][True][Ite]^#](x1, x2, x3) = [2 0] x1 + [0] [0 0] [0] [c_4] = [0] [0] [c_5] = [0] [0] [<^#](x1, x2) = [1] [0] [c_6] = [1] [0] [c_7](x1) = [1 0] x1 + [0] [0 1] [0] [c_8] = [1] [0] [c_9](x1) = [1 0] x1 + [0] [0 1] [0] [c_10] = [0] [0] The following symbols are considered usable {<, gcd^#, equal0^#, monus^#, equal0[Ite]^#, equal0[Ite][True][Ite]^#, <^#} The order satisfies the following ordering constraints: [<(x, 0())] = [0] [0] >= [0] [0] = [False()] [<(S(x), S(y))] = [0] [0] >= [0] [0] = [<(x, y)] [<(0(), S(y))] = [0] [0] >= [0] [0] = [True()] [gcd^#(x, y)] = [2 2] x + [2 2] y + [2] [2 2] [1 2] [2] > [1 2] x + [1 2] y + [1] [2 1] [1 1] [2] = [c_1(equal0^#(x, y))] [equal0^#(a, b)] = [1 2] a + [1 2] b + [1] [2 1] [1 1] [2] ? [2] [2] = [c_3(equal0[Ite]^#(<(a, b), a, b))] [monus^#(S(x'), S(x))] = [0 0] x' + [0 0] x + [2] [1 0] [1 0] [2] ? [0 0] x' + [0 0] x + [3] [1 2] [1 2] [3] = [c_2(monus^#(x', x))] [equal0[Ite]^#(True(), a, b)] = [0] [0] >= [0] [0] = [c_9(equal0[Ite][True][Ite]^#(<(b, a), a, b))] [equal0[Ite]^#(False(), a, b)] = [0] [0] >= [0] [0] = [c_10()] [equal0[Ite][True][Ite]^#(True(), a, b)] = [0] [0] >= [0] [0] = [c_4()] [equal0[Ite][True][Ite]^#(False(), a, b)] = [0] [0] >= [0] [0] = [c_5()] [<^#(x, 0())] = [1] [0] >= [1] [0] = [c_6()] [<^#(S(x), S(y))] = [1] [0] >= [1] [0] = [c_7(<^#(x, y))] [<^#(0(), S(y))] = [1] [0] >= [1] [0] = [c_8()] 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(n^1)). Strict DPs: { equal0^#(a, b) -> c_3(equal0[Ite]^#(<(a, b), a, b)) , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Weak DPs: { gcd^#(x, y) -> c_1(equal0^#(x, y)) , equal0[Ite]^#(True(), a, b) -> c_9(equal0[Ite][True][Ite]^#(<(b, a), a, b)) , equal0[Ite]^#(False(), a, b) -> c_10() , equal0[Ite][True][Ite]^#(True(), a, b) -> c_4() , equal0[Ite][True][Ite]^#(False(), a, b) -> c_5() , <^#(x, 0()) -> c_6() , <^#(S(x), S(y)) -> c_7(<^#(x, y)) , <^#(0(), S(y)) -> c_8() } Weak Trs: { <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() } 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. { equal0[Ite]^#(True(), a, b) -> c_9(equal0[Ite][True][Ite]^#(<(b, a), a, b)) , equal0[Ite]^#(False(), a, b) -> c_10() , equal0[Ite][True][Ite]^#(True(), a, b) -> c_4() , equal0[Ite][True][Ite]^#(False(), a, b) -> c_5() , <^#(x, 0()) -> c_6() , <^#(S(x), S(y)) -> c_7(<^#(x, y)) , <^#(0(), S(y)) -> c_8() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { equal0^#(a, b) -> c_3(equal0[Ite]^#(<(a, b), a, b)) , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Weak DPs: { gcd^#(x, y) -> c_1(equal0^#(x, y)) } Weak Trs: { <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() } 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: { equal0^#(a, b) -> c_3(equal0[Ite]^#(<(a, b), a, b)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { equal0^#(a, b) -> c_1() , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Weak DPs: { gcd^#(x, y) -> c_3(equal0^#(x, y)) } Weak Trs: { <(x, 0()) -> False() , <(S(x), S(y)) -> <(x, y) , <(0(), S(y)) -> True() } 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: { equal0^#(a, b) -> c_1() , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Weak DPs: { gcd^#(x, y) -> c_3(equal0^#(x, y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph 1: equal0^#(a, b) -> c_1() 2: monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) -->_1 monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) :2 3: gcd^#(x, y) -> c_3(equal0^#(x, y)) -->_1 equal0^#(a, b) -> c_1() :1 Following roots of the dependency graph are removed, as the considered set of starting terms is closed under reduction with respect to these rules (modulo compound contexts). { gcd^#(x, y) -> c_3(equal0^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { equal0^#(a, b) -> c_1() , monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph 1: equal0^#(a, b) -> c_1() 2: monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) -->_1 monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) :2 Following roots of the dependency graph are removed, as the considered set of starting terms is closed under reduction with respect to these rules (modulo compound contexts). { equal0^#(a, b) -> c_1() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } 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: monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(equal0[Ite][True][Ite]) = {}, safe(True) = {}, safe(S) = {1}, safe(gcd[Ite][False][Ite][False][Ite]) = {1, 2, 3}, safe(<) = {}, safe(gcd) = {}, safe(0) = {}, safe(monus) = {}, safe(equal0) = {}, safe(equal0[Ite]) = {}, safe(False) = {}, safe(gcd^#) = {}, safe(c_1) = {}, safe(equal0^#) = {}, safe(monus^#) = {2}, safe(c_2) = {}, safe(c_3) = {}, safe(equal0[Ite]^#) = {}, safe(equal0[Ite][True][Ite]^#) = {}, safe(c_4) = {}, safe(c_5) = {}, safe(<^#) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(c_8) = {}, safe(c_9) = {}, safe(c_10) = {}, safe(c) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {} and precedence empty . Following symbols are considered recursive: {monus^#} The recursion depth is 1. Further, following argument filtering is employed: pi(equal0[Ite][True][Ite]) = [], pi(True) = [], pi(S) = [1], pi(gcd[Ite][False][Ite][False][Ite]) = [], pi(<) = [], pi(gcd) = [], pi(0) = [], pi(monus) = [], pi(equal0) = [], pi(equal0[Ite]) = [], pi(False) = [], pi(gcd^#) = [], pi(c_1) = [], pi(equal0^#) = [], pi(monus^#) = [1, 2], pi(c_2) = [], pi(c_3) = [], pi(equal0[Ite]^#) = [], pi(equal0[Ite][True][Ite]^#) = [], pi(c_4) = [], pi(c_5) = [], pi(<^#) = [], pi(c_6) = [], pi(c_7) = [], pi(c_8) = [], pi(c_9) = [], pi(c_10) = [], pi(c) = [], pi(c_1) = [], pi(c_2) = [1], pi(c_3) = [] Usable defined function symbols are a subset of: {gcd^#, equal0^#, monus^#, equal0[Ite]^#, equal0[Ite][True][Ite]^#, <^#} For your convenience, here are the satisfied ordering constraints: pi(monus^#(S(x'), S(x))) = monus^#(S(; x'); S(; x)) > c_2(monus^#(x'; x);) = pi(c_2(monus^#(x', x))) 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: { monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } 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. { monus^#(S(x'), S(x)) -> c_2(monus^#(x', x)) } 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))