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(s(x))) -> s(half(x)) , log(s(0())) -> 0() , log(s(s(x))) -> s(log(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(log) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] x1 + [0] [0] = [4] [s](x1) = [1] x1 + [4] [log](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [half(0())] = [4] >= [4] = [0()] [half(s(s(x)))] = [1] x + [8] > [1] x + [4] = [s(half(x))] [log(s(0()))] = [8] > [4] = [0()] [log(s(s(x)))] = [1] x + [8] >= [1] x + [8] = [s(log(s(half(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() , log(s(s(x))) -> s(log(s(half(x)))) } Weak Trs: { half(s(s(x))) -> s(half(x)) , log(s(0())) -> 0() } 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(log) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [half](x1) = [1] x1 + [4] [0] = [0] [s](x1) = [1] x1 + [2] [log](x1) = [1] x1 + [5] The order satisfies the following ordering constraints: [half(0())] = [4] > [0] = [0()] [half(s(s(x)))] = [1] x + [8] > [1] x + [6] = [s(half(x))] [log(s(0()))] = [7] > [0] = [0()] [log(s(s(x)))] = [1] x + [9] ? [1] x + [13] = [s(log(s(half(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: { log(s(s(x))) -> s(log(s(half(x)))) } Weak Trs: { half(0()) -> 0() , half(s(s(x))) -> s(half(x)) , log(s(0())) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { log(s(s(x))) -> s(log(s(half(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(log) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [half](x1) = [1] x1 + [0] [0] = [1] [s](x1) = [1] x1 + [1] [log](x1) = [4] x1 + [5] The order satisfies the following ordering constraints: [half(0())] = [1] >= [1] = [0()] [half(s(s(x)))] = [1] x + [2] > [1] x + [1] = [s(half(x))] [log(s(0()))] = [13] > [1] = [0()] [log(s(s(x)))] = [4] x + [13] > [4] x + [10] = [s(log(s(half(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(s(x))) -> s(half(x)) , log(s(0())) -> 0() , log(s(s(x))) -> s(log(s(half(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))