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: { a(Z(), y) -> Z() , a(C(x1, x2), y) -> C(a(x1, y), a(x2, C(x1, x2))) , second(C(x1, x2)) -> x2 , eqZList(Z(), Z()) -> True() , eqZList(Z(), C(y1, y2)) -> False() , eqZList(C(x1, x2), Z()) -> False() , eqZList(C(x1, x2), C(y1, y2)) -> and(eqZList(x1, y1), eqZList(x2, y2)) , first(C(x1, x2)) -> x1 } Weak Trs: { and(True(), True()) -> True() , and(True(), False()) -> False() , and(False(), True()) -> False() , and(False(), False()) -> False() } 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(C) = {1, 2}, Uargs(and) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a](x1, x2) = [1] x1 + [3] [second](x1) = [1] x1 + [7] [eqZList](x1, x2) = [1] x1 + [1] x2 + [1] [Z] = [7] [True] = [4] [C](x1, x2) = [1] x1 + [1] x2 + [7] [and](x1, x2) = [1] x1 + [1] x2 + [0] [first](x1) = [1] x1 + [7] [False] = [4] The following symbols are considered usable {a, second, eqZList, and, first} The order satisfies the following ordering constraints: [a(Z(), y)] = [10] > [7] = [Z()] [a(C(x1, x2), y)] = [1] x1 + [1] x2 + [10] ? [1] x1 + [1] x2 + [13] = [C(a(x1, y), a(x2, C(x1, x2)))] [second(C(x1, x2))] = [1] x1 + [1] x2 + [14] > [1] x2 + [0] = [x2] [eqZList(Z(), Z())] = [15] > [4] = [True()] [eqZList(Z(), C(y1, y2))] = [1] y1 + [1] y2 + [15] > [4] = [False()] [eqZList(C(x1, x2), Z())] = [1] x1 + [1] x2 + [15] > [4] = [False()] [eqZList(C(x1, x2), C(y1, y2))] = [1] x1 + [1] x2 + [1] y1 + [1] y2 + [15] > [1] x1 + [1] x2 + [1] y1 + [1] y2 + [2] = [and(eqZList(x1, y1), eqZList(x2, y2))] [and(True(), True())] = [8] > [4] = [True()] [and(True(), False())] = [8] > [4] = [False()] [and(False(), True())] = [8] > [4] = [False()] [and(False(), False())] = [8] > [4] = [False()] [first(C(x1, x2))] = [1] x1 + [1] x2 + [14] > [1] x1 + [0] = [x1] 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: { a(C(x1, x2), y) -> C(a(x1, y), a(x2, C(x1, x2))) } Weak Trs: { a(Z(), y) -> Z() , second(C(x1, x2)) -> x2 , eqZList(Z(), Z()) -> True() , eqZList(Z(), C(y1, y2)) -> False() , eqZList(C(x1, x2), Z()) -> False() , eqZList(C(x1, x2), C(y1, y2)) -> and(eqZList(x1, y1), eqZList(x2, y2)) , and(True(), True()) -> True() , and(True(), False()) -> False() , and(False(), True()) -> False() , and(False(), False()) -> False() , first(C(x1, x2)) -> x1 } 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: { a(C(x1, x2), y) -> C(a(x1, y), a(x2, C(x1, x2))) } 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(C) = {1, 2}, Uargs(and) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [a](x1, x2) = [2] x1 + [0] [second](x1) = [3] x1 + [3] [eqZList](x1, x2) = [0] [Z] = [4] [True] = [0] [C](x1, x2) = [1] x1 + [1] x2 + [4] [and](x1, x2) = [2] x1 + [1] x2 + [0] [first](x1) = [3] x1 + [3] [False] = [0] The following symbols are considered usable {a, second, eqZList, and, first} The order satisfies the following ordering constraints: [a(Z(), y)] = [8] > [4] = [Z()] [a(C(x1, x2), y)] = [2] x1 + [2] x2 + [8] > [2] x1 + [2] x2 + [4] = [C(a(x1, y), a(x2, C(x1, x2)))] [second(C(x1, x2))] = [3] x1 + [3] x2 + [15] > [1] x2 + [0] = [x2] [eqZList(Z(), Z())] = [0] >= [0] = [True()] [eqZList(Z(), C(y1, y2))] = [0] >= [0] = [False()] [eqZList(C(x1, x2), Z())] = [0] >= [0] = [False()] [eqZList(C(x1, x2), C(y1, y2))] = [0] >= [0] = [and(eqZList(x1, y1), eqZList(x2, y2))] [and(True(), True())] = [0] >= [0] = [True()] [and(True(), False())] = [0] >= [0] = [False()] [and(False(), True())] = [0] >= [0] = [False()] [and(False(), False())] = [0] >= [0] = [False()] [first(C(x1, x2))] = [3] x1 + [3] x2 + [15] > [1] x1 + [0] = [x1] 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: { a(Z(), y) -> Z() , a(C(x1, x2), y) -> C(a(x1, y), a(x2, C(x1, x2))) , second(C(x1, x2)) -> x2 , eqZList(Z(), Z()) -> True() , eqZList(Z(), C(y1, y2)) -> False() , eqZList(C(x1, x2), Z()) -> False() , eqZList(C(x1, x2), C(y1, y2)) -> and(eqZList(x1, y1), eqZList(x2, y2)) , and(True(), True()) -> True() , and(True(), False()) -> False() , and(False(), True()) -> False() , and(False(), False()) -> False() , first(C(x1, x2)) -> x1 } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))