YES(O(1),O(1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { gcd(x, 0()) -> x , gcd(0(), y) -> y , gcd(s(x), s(y)) -> if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y))) } Obligation: runtime complexity Answer: YES(O(1),O(1)) We add the following weak dependency pairs: Strict DPs: { gcd^#(x, 0()) -> c_1(x) , gcd^#(0(), y) -> c_2(y) , gcd^#(s(x), s(y)) -> c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y))) } 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: { gcd^#(x, 0()) -> c_1(x) , gcd^#(0(), y) -> c_2(y) , gcd^#(s(x), s(y)) -> c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y))) } Strict Trs: { gcd(x, 0()) -> x , gcd(0(), y) -> y , gcd(s(x), s(y)) -> if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y))) } Obligation: 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: { gcd^#(x, 0()) -> c_1(x) , gcd^#(0(), y) -> c_2(y) , gcd^#(s(x), s(y)) -> c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y))) } Obligation: 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. [0] = [0] [0] [s](x1) = [0] [2] [-](x1, x2) = [0] [0] [gcd^#](x1, x2) = [0 2] x1 + [0] [0 0] [0] [c_1](x1) = [0 1] x1 + [1] [0 0] [0] [c_2](x1) = [1] [0] [c_3](x1, x2, x3, x4) = [1] [0] The following symbols are considered usable {gcd^#} The order satisfies the following ordering constraints: [gcd^#(x, 0())] = [0 2] x + [0] [0 0] [0] ? [0 1] x + [1] [0 0] [0] = [c_1(x)] [gcd^#(0(), y)] = [0] [0] ? [1] [0] = [c_2(y)] [gcd^#(s(x), s(y))] = [4] [0] > [1] [0] = [c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y)))] 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)). Strict DPs: { gcd^#(x, 0()) -> c_1(x) , gcd^#(0(), y) -> c_2(y) } Weak DPs: { gcd^#(s(x), s(y)) -> c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y))) } Obligation: runtime complexity Answer: YES(O(1),O(1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: gcd^#(x, 0()) -> c_1(x) , 2: gcd^#(0(), y) -> c_2(y) } Sub-proof: ---------- The following argument positions are usable: none TcT has computed the following constructor-restricted matrix interpretation. Note that the diagonal of the component-wise maxima of interpretation-entries (of constructors) contains no more than 0 non-zero entries. [gcd](x1, x2) = [0] [0] = [0] [s](x1) = [0] [if](x1, x2, x3) = [0] [<](x1, x2) = [0] [-](x1, x2) = [0] [gcd^#](x1, x2) = [1] [c_1](x1) = [0] [c_2](x1) = [0] [c_3](x1, x2, x3, x4) = [1] The following symbols are considered usable {gcd^#} The order satisfies the following ordering constraints: [gcd^#(x, 0())] = [1] > [0] = [c_1(x)] [gcd^#(0(), y)] = [1] > [0] = [c_2(y)] [gcd^#(s(x), s(y))] = [1] >= [1] = [c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(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: { gcd^#(x, 0()) -> c_1(x) , gcd^#(0(), y) -> c_2(y) , gcd^#(s(x), s(y)) -> c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y))) } Obligation: 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. { gcd^#(x, 0()) -> c_1(x) , gcd^#(0(), y) -> c_2(y) , gcd^#(s(x), s(y)) -> c_3(x, y, gcd^#(s(x), -(y, x)), gcd^#(-(x, y), s(y))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(1))