We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , bits(0()) -> 0() , bits(s(x)) -> s(bits(half(s(x)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(bits) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [4] [bits](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [half(0())] = [0] >= [0] = [0()] [half(s(0()))] = [4] > [0] = [0()] [half(s(s(x)))] = [1] x + [8] > [1] x + [4] = [s(half(x))] [bits(0())] = [0] >= [0] = [0()] [bits(s(x))] = [1] x + [4] ? [1] x + [8] = [s(bits(half(s(x))))] 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 Trs: { half(0()) -> 0() , bits(0()) -> 0() , bits(s(x)) -> s(bits(half(s(x)))) } Weak Trs: { half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(bits) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] x1 + [4] [0] = [4] [s](x1) = [1] x1 + [4] [bits](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [half(0())] = [8] > [4] = [0()] [half(s(0()))] = [12] > [4] = [0()] [half(s(s(x)))] = [1] x + [12] > [1] x + [8] = [s(half(x))] [bits(0())] = [4] >= [4] = [0()] [bits(s(x))] = [1] x + [4] ? [1] x + [12] = [s(bits(half(s(x))))] 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 Trs: { bits(0()) -> 0() , bits(s(x)) -> s(bits(half(s(x)))) } Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(bits) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [7] [0] = [0] [s](x1) = [1] x1 + [0] [bits](x1) = [1] x1 + [1] The order satisfies the following ordering constraints: [half(0())] = [7] > [0] = [0()] [half(s(0()))] = [7] > [0] = [0()] [half(s(s(x)))] = [7] >= [7] = [s(half(x))] [bits(0())] = [1] > [0] = [0()] [bits(s(x))] = [1] x + [1] ? [8] = [s(bits(half(s(x))))] 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 Trs: { bits(s(x)) -> s(bits(half(s(x)))) } Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , bits(0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { bits(s(x)) -> s(bits(half(s(x)))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(s) = {1}, Uargs(bits) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1 0] x1 + [0] [1 0] [0] [0] = [0] [0] [s](x1) = [1 0] x1 + [2] [1 0] [4] [bits](x1) = [1 3] x1 + [1] [3 1] [3] The order satisfies the following ordering constraints: [half(0())] = [0] [0] >= [0] [0] = [0()] [half(s(0()))] = [2] [2] > [0] [0] = [0()] [half(s(s(x)))] = [1 0] x + [4] [1 0] [4] > [1 0] x + [2] [1 0] [4] = [s(half(x))] [bits(0())] = [1] [3] > [0] [0] = [0()] [bits(s(x))] = [4 0] x + [15] [4 0] [13] > [4 0] x + [11] [4 0] [13] = [s(bits(half(s(x))))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(x))) -> s(half(x)) , bits(0()) -> 0() , bits(s(x)) -> s(bits(half(s(x)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))