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: { f(c(a(), z, x)) -> b(a(), z) , b(y, z) -> z , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) , b^#(y, z) -> c_2(z) , b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, 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^1)). Strict DPs: { f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) , b^#(y, z) -> c_2(z) , b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } Strict Trs: { f(c(a(), z, x)) -> b(a(), z) , b(y, z) -> z , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) } Obligation: 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(f) = {1}, Uargs(c) = {1, 2, 3}, Uargs(b) = {1, 2}, Uargs(f^#) = {1}, Uargs(c_1) = {1}, Uargs(b^#) = {2}, Uargs(c_2) = {1}, Uargs(c_3) = {1} TcT has computed the following constructor-restricted matrix interpretation. [f](x1) = [1 0] x1 + [0] [2 0] [0] [c](x1, x2, x3) = [1 0] x1 + [1 2] x2 + [1 0] x3 + [2] [0 0] [0 0] [0 0] [0] [a] = [0] [0] [b](x1, x2) = [1 0] x1 + [1 1] x2 + [1] [2 2] [1 2] [2] [f^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_1](x1) = [1 0] x1 + [2] [0 1] [1] [b^#](x1, x2) = [2 2] x1 + [1 1] x2 + [2] [1 2] [1 2] [1] [c_2](x1) = [1 0] x1 + [1] [0 1] [1] [c_3](x1) = [1 0] x1 + [2] [0 1] [2] The following symbols are considered usable {f, b, f^#, b^#} The order satisfies the following ordering constraints: [f(c(a(), z, x))] = [1 2] z + [1 0] x + [2] [2 4] [2 0] [4] > [1 1] z + [1] [1 2] [2] = [b(a(), z)] [b(y, z)] = [1 1] z + [1 0] y + [1] [1 2] [2 2] [2] > [1 0] z + [0] [0 1] [0] = [z] [b(x, b(z, y))] = [3 2] z + [1 0] x + [2 3] y + [4] [5 4] [2 2] [3 5] [7] > [2 2] z + [1 0] x + [1 0] y + [3] [4 4] [2 0] [2 0] [6] = [f(b(f(f(z)), c(x, z, y)))] [f^#(c(a(), z, x))] = [1 2] z + [1 0] x + [3] [0 0] [0 0] [2] ? [1 1] z + [4] [1 2] [2] = [c_1(b^#(a(), z))] [b^#(y, z)] = [1 1] z + [2 2] y + [2] [1 2] [1 2] [1] > [1 0] z + [1] [0 1] [1] = [c_2(z)] [b^#(x, b(z, y))] = [3 2] z + [2 2] x + [2 3] y + [5] [5 4] [1 2] [3 5] [6] ? [2 2] z + [1 0] x + [1 0] y + [6] [0 0] [0 0] [0 0] [4] = [c_3(f^#(b(f(f(z)), c(x, z, 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(n^1)). Strict DPs: { f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) , b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } Weak DPs: { b^#(y, z) -> c_2(z) } Weak Trs: { f(c(a(), z, x)) -> b(a(), z) , b(y, z) -> z , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. DPs: { 2: b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } Trs: { b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [f](x1) = [1 0] x1 + [0] [1 0] [4] [c](x1, x2, x3) = [1 2] x2 + [0] [0 0] [0] [a] = [0] [0] [b](x1, x2) = [0 0] x1 + [1 2] x2 + [0] [2 4] [0 2] [4] [f^#](x1) = [2 0] x1 + [0] [0 0] [0] [c_1](x1) = [1 0] x1 + [0] [0 0] [0] [b^#](x1, x2) = [0 4] x1 + [2 2] x2 + [0] [4 0] [0 0] [0] [c_2](x1) = [2 1] x1 + [0] [0 0] [0] [c_3](x1) = [2 0] x1 + [5] [0 0] [0] The following symbols are considered usable {f, b, f^#, b^#} The order satisfies the following ordering constraints: [f(c(a(), z, x))] = [1 2] z + [0] [1 2] [4] >= [1 2] z + [0] [0 2] [4] = [b(a(), z)] [b(y, z)] = [1 2] z + [0 0] y + [0] [0 2] [2 4] [4] >= [1 0] z + [0] [0 1] [0] = [z] [b(x, b(z, y))] = [4 8] z + [0 0] x + [1 6] y + [8] [4 8] [2 4] [0 4] [12] > [1 2] z + [0] [1 2] [4] = [f(b(f(f(z)), c(x, z, y)))] [f^#(c(a(), z, x))] = [2 4] z + [0] [0 0] [0] >= [2 2] z + [0] [0 0] [0] = [c_1(b^#(a(), z))] [b^#(y, z)] = [2 2] z + [0 4] y + [0] [0 0] [4 0] [0] >= [2 1] z + [0] [0 0] [0] = [c_2(z)] [b^#(x, b(z, y))] = [4 8] z + [0 4] x + [2 8] y + [8] [0 0] [4 0] [0 0] [0] > [4 8] z + [5] [0 0] [0] = [c_3(f^#(b(f(f(z)), c(x, z, 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(n^1)). Strict DPs: { f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) } Weak DPs: { b^#(y, z) -> c_2(z) , b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } Weak Trs: { f(c(a(), z, x)) -> b(a(), z) , b(y, z) -> z , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. DPs: { 1: f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) , 3: b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } Trs: { f(c(a(), z, x)) -> b(a(), z) , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [f](x1) = [2 0] x1 + [4] [1 0] [1] [c](x1, x2, x3) = [1 2] x2 + [1] [0 0] [0] [a] = [0] [0] [b](x1, x2) = [0 0] x1 + [1 4] x2 + [0] [2 1] [0 2] [2] [f^#](x1) = [1 0] x1 + [0] [0 0] [0] [c_1](x1) = [1 0] x1 + [0] [0 0] [0] [b^#](x1, x2) = [0 0] x1 + [1 2] x2 + [0] [4 4] [0 4] [0] [c_2](x1) = [1 1] x1 + [0] [0 4] [0] [c_3](x1) = [1 0] x1 + [1] [0 0] [3] The following symbols are considered usable {f, b, f^#, b^#} The order satisfies the following ordering constraints: [f(c(a(), z, x))] = [2 4] z + [6] [1 2] [2] > [1 4] z + [0] [0 2] [2] = [b(a(), z)] [b(y, z)] = [1 4] z + [0 0] y + [0] [0 2] [2 1] [2] >= [1 0] z + [0] [0 1] [0] = [z] [b(x, b(z, y))] = [8 4] z + [0 0] x + [1 12] y + [8] [4 2] [2 1] [0 4] [6] > [2 4] z + [6] [1 2] [2] = [f(b(f(f(z)), c(x, z, y)))] [f^#(c(a(), z, x))] = [1 2] z + [1] [0 0] [0] > [1 2] z + [0] [0 0] [0] = [c_1(b^#(a(), z))] [b^#(y, z)] = [1 2] z + [0 0] y + [0] [0 4] [4 4] [0] >= [1 1] z + [0] [0 4] [0] = [c_2(z)] [b^#(x, b(z, y))] = [4 2] z + [0 0] x + [1 8] y + [4] [8 4] [4 4] [0 8] [8] > [1 2] z + [2] [0 0] [3] = [c_3(f^#(b(f(f(z)), c(x, z, 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: { f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) , b^#(y, z) -> c_2(z) , b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } Weak Trs: { f(c(a(), z, x)) -> b(a(), z) , b(y, z) -> z , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, 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. { f^#(c(a(), z, x)) -> c_1(b^#(a(), z)) , b^#(y, z) -> c_2(z) , b^#(x, b(z, y)) -> c_3(f^#(b(f(f(z)), c(x, z, y)))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { f(c(a(), z, x)) -> b(a(), z) , b(y, z) -> z , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, 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)). Rules: Empty Obligation: runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))