interpretations
YES(?,O(n^3))
We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).
Strict Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, mark(nil()) -> nil()
, mark(0()) -> 0()
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, a__length(X) -> length(X)
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length(nil()) -> 0()
, a__length1(X) -> a__length(X)
, a__length1(X) -> length1(X) }
Obligation:
innermost runtime complexity
Answer:
YES(?,O(n^3))
The following argument positions are usable:
Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(mark) = {},
Uargs(from) = {}, Uargs(s) = {1}, Uargs(a__length) = {},
Uargs(a__length1) = {}, Uargs(length) = {}, Uargs(length1) = {}
TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).
[1 2 0] [3]
[a__from](x1) = [0 1 0] x1 + [2]
[0 0 1] [2]
[1 0 0] [0 0 1] [0]
[cons](x1, x2) = [0 1 0] x1 + [0 0 0] x2 + [1]
[0 0 0] [0 0 1] [2]
[1 2 0] [1]
[mark](x1) = [0 1 0] x1 + [0]
[0 1 1] [3]
[1 2 0] [2]
[from](x1) = [0 1 0] x1 + [2]
[0 0 1] [0]
[1 0 0] [0]
[s](x1) = [0 1 0] x1 + [1]
[0 0 0] [0]
[0 0 1] [2]
[a__length](x1) = [0 0 1] x1 + [2]
[0 0 0] [2]
[2]
[nil] = [0]
[0]
[0]
[0] = [0]
[0]
[0 0 1] [3]
[a__length1](x1) = [0 0 1] x1 + [2]
[0 0 0] [2]
[0 0 0] [0]
[length](x1) = [0 0 1] x1 + [2]
[0 0 0] [2]
[0 0 0] [1]
[length1](x1) = [0 0 1] x1 + [2]
[0 0 0] [0]
This order satisfies following ordering constraints
[a__from(X)] = [1 2 0] [3]
[0 1 0] X + [2]
[0 0 1] [2]
> [1 2 0] [1]
[0 1 0] X + [1]
[0 0 0] [2]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 2 0] [3]
[0 1 0] X + [2]
[0 0 1] [2]
> [1 2 0] [2]
[0 1 0] X + [2]
[0 0 1] [0]
= [from(X)]
[mark(cons(X1, X2))] = [1 2 0] [0 0 1] [3]
[0 1 0] X1 + [0 0 0] X2 + [1]
[0 1 0] [0 0 1] [6]
> [1 2 0] [0 0 1] [1]
[0 1 0] X1 + [0 0 0] X2 + [1]
[0 0 0] [0 0 1] [2]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 4 0] [7]
[0 1 0] X + [2]
[0 1 1] [5]
> [1 4 0] [4]
[0 1 0] X + [2]
[0 1 1] [5]
= [a__from(mark(X))]
[mark(s(X))] = [1 2 0] [3]
[0 1 0] X + [1]
[0 1 0] [4]
> [1 2 0] [1]
[0 1 0] X + [1]
[0 0 0] [0]
= [s(mark(X))]
[mark(nil())] = [3]
[0]
[3]
> [2]
[0]
[0]
= [nil()]
[mark(0())] = [1]
[0]
[3]
> [0]
[0]
[0]
= [0()]
[mark(length(X))] = [0 0 2] [5]
[0 0 1] X + [2]
[0 0 1] [7]
> [0 0 1] [2]
[0 0 1] X + [2]
[0 0 0] [2]
= [a__length(X)]
[mark(length1(X))] = [0 0 2] [6]
[0 0 1] X + [2]
[0 0 1] [5]
> [0 0 1] [3]
[0 0 1] X + [2]
[0 0 0] [2]
= [a__length1(X)]
[a__length(X)] = [0 0 1] [2]
[0 0 1] X + [2]
[0 0 0] [2]
> [0 0 0] [0]
[0 0 1] X + [2]
[0 0 0] [2]
= [length(X)]
[a__length(cons(X, Y))] = [0 0 1] [4]
[0 0 1] Y + [4]
[0 0 0] [2]
> [0 0 1] [3]
[0 0 1] Y + [3]
[0 0 0] [0]
= [s(a__length1(Y))]
[a__length(nil())] = [2]
[2]
[2]
> [0]
[0]
[0]
= [0()]
[a__length1(X)] = [0 0 1] [3]
[0 0 1] X + [2]
[0 0 0] [2]
> [0 0 1] [2]
[0 0 1] X + [2]
[0 0 0] [2]
= [a__length(X)]
[a__length1(X)] = [0 0 1] [3]
[0 0 1] X + [2]
[0 0 0] [2]
> [0 0 0] [1]
[0 0 1] X + [2]
[0 0 0] [0]
= [length1(X)]
Hurray, we answered YES(?,O(n^3))
lmpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__length(nil()) -> 0()
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length1(X) -> a__length(X)
, mark(from(X)) -> a__from(mark(X))
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(s(X)) -> s(mark(X))
, mark(nil()) -> nil()
, mark(0()) -> 0()
, a__from(X) -> from(X)
, a__length(X) -> length(X)
, a__length1(X) -> length1(X) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
mpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__length(nil()) -> 0()
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length1(X) -> a__length(X)
, mark(from(X)) -> a__from(mark(X))
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(s(X)) -> s(mark(X))
, mark(nil()) -> nil()
, mark(0()) -> 0()
, a__from(X) -> from(X)
, a__length(X) -> length(X)
, a__length1(X) -> length1(X) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__length(nil()) -> 0()
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length1(X) -> a__length(X)
, mark(from(X)) -> a__from(mark(X))
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(s(X)) -> s(mark(X))
, mark(nil()) -> nil()
, mark(0()) -> 0()
, a__from(X) -> from(X)
, a__length(X) -> length(X)
, a__length1(X) -> length1(X) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar-ps
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__length(nil()) -> 0()
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length1(X) -> a__length(X)
, mark(from(X)) -> a__from(mark(X))
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(s(X)) -> s(mark(X))
, mark(nil()) -> nil()
, mark(0()) -> 0()
, a__from(X) -> from(X)
, a__length(X) -> length(X)
, a__length1(X) -> length1(X) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..