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: { norm(nil()) -> 0() , norm(g(x, y)) -> s(norm(x)) , f(x, nil()) -> g(nil(), x) , f(x, g(y, z)) -> g(f(x, y), z) , rem(nil(), y) -> nil() , rem(g(x, y), 0()) -> g(x, y) , rem(g(x, y), s(z)) -> rem(x, z) } 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: { rem(nil(), y) -> nil() , rem(g(x, y), 0()) -> g(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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [norm](x1) = [2] x1 + [0] [nil] = [0] [0] = [0] [g](x1, x2) = [1] x1 + [1] x2 + [0] [s](x1) = [1] x1 + [0] [f](x1, x2) = [3] x1 + [2] x2 + [0] [rem](x1, x2) = [2] x1 + [1] x2 + [1] This order satisfies the following ordering constraints: [norm(nil())] = [0] >= [0] = [0()] [norm(g(x, y))] = [2] x + [2] y + [0] >= [2] x + [0] = [s(norm(x))] [f(x, nil())] = [3] x + [0] >= [1] x + [0] = [g(nil(), x)] [f(x, g(y, z))] = [3] x + [2] y + [2] z + [0] >= [3] x + [2] y + [1] z + [0] = [g(f(x, y), z)] [rem(nil(), y)] = [1] y + [1] > [0] = [nil()] [rem(g(x, y), 0())] = [2] x + [2] y + [1] > [1] x + [1] y + [0] = [g(x, y)] [rem(g(x, y), s(z))] = [2] x + [2] y + [1] z + [1] >= [2] x + [1] z + [1] = [rem(x, z)] 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: { norm(nil()) -> 0() , norm(g(x, y)) -> s(norm(x)) , f(x, nil()) -> g(nil(), x) , f(x, g(y, z)) -> g(f(x, y), z) , rem(g(x, y), s(z)) -> rem(x, z) } Weak Trs: { rem(nil(), y) -> nil() , rem(g(x, y), 0()) -> g(x, y) } 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: { norm(nil()) -> 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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [norm](x1) = [1] x1 + [0] [nil] = [1] [0] = [0] [g](x1, x2) = [1] x1 + [1] x2 + [0] [s](x1) = [1] x1 + [0] [f](x1, x2) = [3] x1 + [1] x2 + [0] [rem](x1, x2) = [1] x1 + [2] x2 + [0] This order satisfies the following ordering constraints: [norm(nil())] = [1] > [0] = [0()] [norm(g(x, y))] = [1] x + [1] y + [0] >= [1] x + [0] = [s(norm(x))] [f(x, nil())] = [3] x + [1] >= [1] x + [1] = [g(nil(), x)] [f(x, g(y, z))] = [3] x + [1] y + [1] z + [0] >= [3] x + [1] y + [1] z + [0] = [g(f(x, y), z)] [rem(nil(), y)] = [2] y + [1] >= [1] = [nil()] [rem(g(x, y), 0())] = [1] x + [1] y + [0] >= [1] x + [1] y + [0] = [g(x, y)] [rem(g(x, y), s(z))] = [1] x + [1] y + [2] z + [0] >= [1] x + [2] z + [0] = [rem(x, z)] 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: { norm(g(x, y)) -> s(norm(x)) , f(x, nil()) -> g(nil(), x) , f(x, g(y, z)) -> g(f(x, y), z) , rem(g(x, y), s(z)) -> rem(x, z) } Weak Trs: { norm(nil()) -> 0() , rem(nil(), y) -> nil() , rem(g(x, y), 0()) -> g(x, y) } 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: { f(x, nil()) -> g(nil(), 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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [norm](x1) = [1] x1 + [0] [nil] = [0] [0] = [0] [g](x1, x2) = [1] x1 + [1] x2 + [0] [s](x1) = [1] x1 + [0] [f](x1, x2) = [3] x1 + [1] x2 + [2] [rem](x1, x2) = [2] x1 + [2] x2 + [0] This order satisfies the following ordering constraints: [norm(nil())] = [0] >= [0] = [0()] [norm(g(x, y))] = [1] x + [1] y + [0] >= [1] x + [0] = [s(norm(x))] [f(x, nil())] = [3] x + [2] > [1] x + [0] = [g(nil(), x)] [f(x, g(y, z))] = [3] x + [1] y + [1] z + [2] >= [3] x + [1] y + [1] z + [2] = [g(f(x, y), z)] [rem(nil(), y)] = [2] y + [0] >= [0] = [nil()] [rem(g(x, y), 0())] = [2] x + [2] y + [0] >= [1] x + [1] y + [0] = [g(x, y)] [rem(g(x, y), s(z))] = [2] x + [2] y + [2] z + [0] >= [2] x + [2] z + [0] = [rem(x, z)] 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: { norm(g(x, y)) -> s(norm(x)) , f(x, g(y, z)) -> g(f(x, y), z) , rem(g(x, y), s(z)) -> rem(x, z) } Weak Trs: { norm(nil()) -> 0() , f(x, nil()) -> g(nil(), x) , rem(nil(), y) -> nil() , rem(g(x, y), 0()) -> g(x, y) } 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: { norm(g(x, y)) -> s(norm(x)) , f(x, g(y, z)) -> g(f(x, y), z) , rem(g(x, y), s(z)) -> rem(x, z) } 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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [norm](x1) = [1] x1 + [0] [nil] = [0] [0] = [0] [g](x1, x2) = [1] x1 + [1] x2 + [1] [s](x1) = [1] x1 + [0] [f](x1, x2) = [3] x1 + [3] x2 + [1] [rem](x1, x2) = [1] x1 + [1] x2 + [0] This order satisfies the following ordering constraints: [norm(nil())] = [0] >= [0] = [0()] [norm(g(x, y))] = [1] x + [1] y + [1] > [1] x + [0] = [s(norm(x))] [f(x, nil())] = [3] x + [1] >= [1] x + [1] = [g(nil(), x)] [f(x, g(y, z))] = [3] x + [3] y + [3] z + [4] > [3] x + [3] y + [1] z + [2] = [g(f(x, y), z)] [rem(nil(), y)] = [1] y + [0] >= [0] = [nil()] [rem(g(x, y), 0())] = [1] x + [1] y + [1] >= [1] x + [1] y + [1] = [g(x, y)] [rem(g(x, y), s(z))] = [1] x + [1] y + [1] z + [1] > [1] x + [1] z + [0] = [rem(x, z)] 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: { norm(nil()) -> 0() , norm(g(x, y)) -> s(norm(x)) , f(x, nil()) -> g(nil(), x) , f(x, g(y, z)) -> g(f(x, y), z) , rem(nil(), y) -> nil() , rem(g(x, y), 0()) -> g(x, y) , rem(g(x, y), s(z)) -> rem(x, z) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))