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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) , map(Nil()) -> Nil() , f(x) -> *(x, x) , +Full(S(x), y) -> +Full(x, S(y)) , +Full(0(), y) -> y , goal(xs) -> map(xs) } Weak Trs: { *(x, S(S(y))) -> +(x, *(x, S(y))) , *(x, S(0())) -> x , *(x, 0()) -> 0() , *(0(), y) -> 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(+) = {2}, Uargs(Cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [map](x1) = [1] x1 + [7] [+](x1, x2) = [1] x2 + [0] [S](x1) = [1] x1 + [7] [f](x1) = [1] x1 + [7] [Cons](x1, x2) = [1] x1 + [1] x2 + [1] [Nil] = [7] [+Full](x1, x2) = [1] x1 + [1] x2 + [7] [0] = [7] [goal](x1) = [1] x1 + [7] [*](x1, x2) = [1] x1 + [7] The following symbols are considered usable {map, f, +Full, goal, *} The order satisfies the following ordering constraints: [map(Cons(x, xs))] = [1] x + [1] xs + [8] ? [1] x + [1] xs + [15] = [Cons(f(x), map(xs))] [map(Nil())] = [14] > [7] = [Nil()] [f(x)] = [1] x + [7] >= [1] x + [7] = [*(x, x)] [+Full(S(x), y)] = [1] x + [1] y + [14] >= [1] x + [1] y + [14] = [+Full(x, S(y))] [+Full(0(), y)] = [1] y + [14] > [1] y + [0] = [y] [goal(xs)] = [1] xs + [7] >= [1] xs + [7] = [map(xs)] [*(x, S(S(y)))] = [1] x + [7] >= [1] x + [7] = [+(x, *(x, S(y)))] [*(x, S(0()))] = [1] x + [7] > [1] x + [0] = [x] [*(x, 0())] = [1] x + [7] >= [7] = [0()] [*(0(), y)] = [14] > [7] = [0()] 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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) , f(x) -> *(x, x) , +Full(S(x), y) -> +Full(x, S(y)) , goal(xs) -> map(xs) } Weak Trs: { map(Nil()) -> Nil() , +Full(0(), y) -> y , *(x, S(S(y))) -> +(x, *(x, S(y))) , *(x, S(0())) -> x , *(x, 0()) -> 0() , *(0(), y) -> 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(+) = {2}, Uargs(Cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [map](x1) = [1] x1 + [0] [+](x1, x2) = [1] x2 + [0] [S](x1) = [1] x1 + [0] [f](x1) = [1] x1 + [0] [Cons](x1, x2) = [1] x1 + [1] x2 + [0] [Nil] = [7] [+Full](x1, x2) = [1] x1 + [1] x2 + [0] [0] = [7] [goal](x1) = [1] x1 + [7] [*](x1, x2) = [1] x1 + [7] The following symbols are considered usable {map, f, +Full, goal, *} The order satisfies the following ordering constraints: [map(Cons(x, xs))] = [1] x + [1] xs + [0] >= [1] x + [1] xs + [0] = [Cons(f(x), map(xs))] [map(Nil())] = [7] >= [7] = [Nil()] [f(x)] = [1] x + [0] ? [1] x + [7] = [*(x, x)] [+Full(S(x), y)] = [1] x + [1] y + [0] >= [1] x + [1] y + [0] = [+Full(x, S(y))] [+Full(0(), y)] = [1] y + [7] > [1] y + [0] = [y] [goal(xs)] = [1] xs + [7] > [1] xs + [0] = [map(xs)] [*(x, S(S(y)))] = [1] x + [7] >= [1] x + [7] = [+(x, *(x, S(y)))] [*(x, S(0()))] = [1] x + [7] > [1] x + [0] = [x] [*(x, 0())] = [1] x + [7] >= [7] = [0()] [*(0(), y)] = [14] > [7] = [0()] 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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) , f(x) -> *(x, x) , +Full(S(x), y) -> +Full(x, S(y)) } Weak Trs: { map(Nil()) -> Nil() , +Full(0(), y) -> y , goal(xs) -> map(xs) , *(x, S(S(y))) -> +(x, *(x, S(y))) , *(x, S(0())) -> x , *(x, 0()) -> 0() , *(0(), y) -> 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(+) = {2}, Uargs(Cons) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [map](x1) = [1] x1 + [0] [+](x1, x2) = [1] x2 + [0] [S](x1) = [1] x1 + [0] [f](x1) = [1] x1 + [1] [Cons](x1, x2) = [1] x1 + [1] x2 + [0] [Nil] = [7] [+Full](x1, x2) = [1] x1 + [1] x2 + [0] [0] = [0] [goal](x1) = [1] x1 + [7] [*](x1, x2) = [1] x1 + [0] The following symbols are considered usable {map, f, +Full, goal, *} The order satisfies the following ordering constraints: [map(Cons(x, xs))] = [1] x + [1] xs + [0] ? [1] x + [1] xs + [1] = [Cons(f(x), map(xs))] [map(Nil())] = [7] >= [7] = [Nil()] [f(x)] = [1] x + [1] > [1] x + [0] = [*(x, x)] [+Full(S(x), y)] = [1] x + [1] y + [0] >= [1] x + [1] y + [0] = [+Full(x, S(y))] [+Full(0(), y)] = [1] y + [0] >= [1] y + [0] = [y] [goal(xs)] = [1] xs + [7] > [1] xs + [0] = [map(xs)] [*(x, S(S(y)))] = [1] x + [0] >= [1] x + [0] = [+(x, *(x, S(y)))] [*(x, S(0()))] = [1] x + [0] >= [1] x + [0] = [x] [*(x, 0())] = [1] x + [0] >= [0] = [0()] [*(0(), y)] = [0] >= [0] = [0()] 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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) , +Full(S(x), y) -> +Full(x, S(y)) } Weak Trs: { map(Nil()) -> Nil() , f(x) -> *(x, x) , +Full(0(), y) -> y , goal(xs) -> map(xs) , *(x, S(S(y))) -> +(x, *(x, S(y))) , *(x, S(0())) -> x , *(x, 0()) -> 0() , *(0(), y) -> 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: { +Full(S(x), y) -> +Full(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(+) = {2}, Uargs(Cons) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [map](x1) = [4] x1 + [0] [+](x1, x2) = [1] x2 + [3] [S](x1) = [1] x1 + [4] [f](x1) = [4] x1 + [0] [Cons](x1, x2) = [1] x1 + [1] x2 + [0] [Nil] = [2] [+Full](x1, x2) = [2] x1 + [1] x2 + [0] [0] = [0] [goal](x1) = [7] x1 + [7] [*](x1, x2) = [3] x1 + [1] x2 + [0] The following symbols are considered usable {map, f, +Full, goal, *} The order satisfies the following ordering constraints: [map(Cons(x, xs))] = [4] x + [4] xs + [0] >= [4] x + [4] xs + [0] = [Cons(f(x), map(xs))] [map(Nil())] = [8] > [2] = [Nil()] [f(x)] = [4] x + [0] >= [4] x + [0] = [*(x, x)] [+Full(S(x), y)] = [2] x + [1] y + [8] > [2] x + [1] y + [4] = [+Full(x, S(y))] [+Full(0(), y)] = [1] y + [0] >= [1] y + [0] = [y] [goal(xs)] = [7] xs + [7] > [4] xs + [0] = [map(xs)] [*(x, S(S(y)))] = [3] x + [1] y + [8] > [3] x + [1] y + [7] = [+(x, *(x, S(y)))] [*(x, S(0()))] = [3] x + [4] > [1] x + [0] = [x] [*(x, 0())] = [3] x + [0] >= [0] = [0()] [*(0(), y)] = [1] y + [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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) } Weak Trs: { map(Nil()) -> Nil() , f(x) -> *(x, x) , +Full(S(x), y) -> +Full(x, S(y)) , +Full(0(), y) -> y , goal(xs) -> map(xs) , *(x, S(S(y))) -> +(x, *(x, S(y))) , *(x, S(0())) -> x , *(x, 0()) -> 0() , *(0(), y) -> 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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) } 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(+) = {2}, Uargs(Cons) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [map](x1) = [3] x1 + [0] [+](x1, x2) = [1] x2 + [0] [S](x1) = [1] x1 + [3] [f](x1) = [3] x1 + [4] [Cons](x1, x2) = [1] x1 + [1] x2 + [4] [Nil] = [3] [+Full](x1, x2) = [3] x1 + [1] x2 + [0] [0] = [0] [goal](x1) = [7] x1 + [7] [*](x1, x2) = [2] x1 + [1] x2 + [4] The following symbols are considered usable {map, f, +Full, goal, *} The order satisfies the following ordering constraints: [map(Cons(x, xs))] = [3] x + [3] xs + [12] > [3] x + [3] xs + [8] = [Cons(f(x), map(xs))] [map(Nil())] = [9] > [3] = [Nil()] [f(x)] = [3] x + [4] >= [3] x + [4] = [*(x, x)] [+Full(S(x), y)] = [3] x + [1] y + [9] > [3] x + [1] y + [3] = [+Full(x, S(y))] [+Full(0(), y)] = [1] y + [0] >= [1] y + [0] = [y] [goal(xs)] = [7] xs + [7] > [3] xs + [0] = [map(xs)] [*(x, S(S(y)))] = [2] x + [1] y + [10] > [2] x + [1] y + [7] = [+(x, *(x, S(y)))] [*(x, S(0()))] = [2] x + [7] > [1] x + [0] = [x] [*(x, 0())] = [2] x + [4] > [0] = [0()] [*(0(), y)] = [1] y + [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: { map(Cons(x, xs)) -> Cons(f(x), map(xs)) , map(Nil()) -> Nil() , f(x) -> *(x, x) , +Full(S(x), y) -> +Full(x, S(y)) , +Full(0(), y) -> y , goal(xs) -> map(xs) , *(x, S(S(y))) -> +(x, *(x, S(y))) , *(x, S(0())) -> x , *(x, 0()) -> 0() , *(0(), y) -> 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))