We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { average(x, s(s(s(y)))) -> s(average(s(x), y)) , average(s(x), y) -> average(x, s(y)) , average(0(), s(s(0()))) -> s(0()) , average(0(), s(0())) -> 0() , average(0(), 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: { average(0(), s(s(0()))) -> s(0()) , average(0(), s(0())) -> 0() , average(0(), 0()) -> 0() } 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} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [average](x1, x2) = [5] x1 + [4] x2 + [0] [s](x1) = [1] x1 + [0] [0] = [1] The order satisfies the following ordering constraints: [average(x, s(s(s(y))))] = [5] x + [4] y + [0] >= [5] x + [4] y + [0] = [s(average(s(x), y))] [average(s(x), y)] = [5] x + [4] y + [0] >= [5] x + [4] y + [0] = [average(x, s(y))] [average(0(), s(s(0())))] = [9] > [1] = [s(0())] [average(0(), s(0()))] = [9] > [1] = [0()] [average(0(), 0())] = [9] > [1] = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { average(x, s(s(s(y)))) -> s(average(s(x), y)) , average(s(x), y) -> average(x, s(y)) } Weak Trs: { average(0(), s(s(0()))) -> s(0()) , average(0(), s(0())) -> 0() , average(0(), 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: { average(x, s(s(s(y)))) -> s(average(s(x), y)) } 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} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [average](x1, x2) = [2] x1 + [2] x2 + [0] [s](x1) = [1] x1 + [2] [0] = [0] The order satisfies the following ordering constraints: [average(x, s(s(s(y))))] = [2] x + [2] y + [12] > [2] x + [2] y + [6] = [s(average(s(x), y))] [average(s(x), y)] = [2] x + [2] y + [4] >= [2] x + [2] y + [4] = [average(x, s(y))] [average(0(), s(s(0())))] = [8] > [2] = [s(0())] [average(0(), s(0()))] = [4] > [0] = [0()] [average(0(), 0())] = [0] >= [0] = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { average(s(x), y) -> average(x, s(y)) } Weak Trs: { average(x, s(s(s(y)))) -> s(average(s(x), y)) , average(0(), s(s(0()))) -> s(0()) , average(0(), s(0())) -> 0() , average(0(), 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: { average(s(x), y) -> average(x, s(y)) } 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} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [average](x1, x2) = [4] x1 + [3] x2 + [4] [s](x1) = [1] x1 + [1] [0] = [0] The order satisfies the following ordering constraints: [average(x, s(s(s(y))))] = [4] x + [3] y + [13] > [4] x + [3] y + [9] = [s(average(s(x), y))] [average(s(x), y)] = [4] x + [3] y + [8] > [4] x + [3] y + [7] = [average(x, s(y))] [average(0(), s(s(0())))] = [10] > [1] = [s(0())] [average(0(), s(0()))] = [7] > [0] = [0()] [average(0(), 0())] = [4] > [0] = [0()] 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: { average(x, s(s(s(y)))) -> s(average(s(x), y)) , average(s(x), y) -> average(x, s(y)) , average(0(), s(s(0()))) -> s(0()) , average(0(), s(0())) -> 0() , average(0(), 0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))