interpretations
YES(?,O(n^3))
We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).
Strict Trs:
{ a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(?,O(n^3))
The following argument positions are usable:
Uargs(a__first) = {1, 2}, Uargs(s) = {1}, Uargs(cons) = {1},
Uargs(mark) = {}, Uargs(first) = {}, Uargs(a__from) = {1},
Uargs(from) = {}
TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).
[1 0 2] [1 0 2] [3]
[a__first](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [0]
[0 0 1] [0 3 1] [3]
[0]
[0] = [0]
[1]
[0]
[nil] = [0]
[1]
[1 0 0] [3]
[s](x1) = [0 0 0] x1 + [0]
[0 0 1] [2]
[1 0 1] [0]
[cons](x1, x2) = [0 0 0] x1 + [0]
[0 0 1] [1]
[1 0 1] [0]
[mark](x1) = [0 0 0] x1 + [0]
[0 0 1] [0]
[1 0 2] [1 0 2] [2]
[first](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [0]
[0 0 1] [0 0 1] [3]
[1 0 3] [3]
[a__from](x1) = [0 2 0] x1 + [0]
[0 0 1] [3]
[1 0 3] [2]
[from](x1) = [0 1 0] x1 + [0]
[0 0 1] [3]
This order satisfies following ordering constraints
[a__first(X1, X2)] = [1 0 2] [1 0 2] [3]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 3 1] [3]
> [1 0 2] [1 0 2] [2]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 0 1] [3]
= [first(X1, X2)]
[a__first(0(), X)] = [1 0 2] [5]
[0 1 0] X + [0]
[0 3 1] [4]
> [0]
[0]
[1]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1 0 2] [1 0 3] [12]
[0 0 0] X + [0 0 0] Y + [0]
[0 0 1] [0 0 1] [6]
> [1 0 2] [0]
[0 0 0] Y + [0]
[0 0 1] [1]
= [cons(mark(Y), first(X, Z))]
[mark(0())] = [1]
[0]
[1]
> [0]
[0]
[1]
= [0()]
[mark(nil())] = [1]
[0]
[1]
> [0]
[0]
[1]
= [nil()]
[mark(s(X))] = [1 0 1] [5]
[0 0 0] X + [0]
[0 0 1] [2]
> [1 0 1] [3]
[0 0 0] X + [0]
[0 0 1] [2]
= [s(mark(X))]
[mark(cons(X1, X2))] = [1 0 2] [1]
[0 0 0] X1 + [0]
[0 0 1] [1]
> [1 0 2] [0]
[0 0 0] X1 + [0]
[0 0 1] [1]
= [cons(mark(X1), X2)]
[mark(first(X1, X2))] = [1 0 3] [1 0 3] [5]
[0 0 0] X1 + [0 0 0] X2 + [0]
[0 0 1] [0 0 1] [3]
> [1 0 3] [1 0 3] [3]
[0 0 0] X1 + [0 0 0] X2 + [0]
[0 0 1] [0 0 1] [3]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [1 0 4] [5]
[0 0 0] X + [0]
[0 0 1] [3]
> [1 0 4] [3]
[0 0 0] X + [0]
[0 0 1] [3]
= [a__from(mark(X))]
[a__from(X)] = [1 0 3] [3]
[0 2 0] X + [0]
[0 0 1] [3]
> [1 0 2] [0]
[0 0 0] X + [0]
[0 0 1] [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 0 3] [3]
[0 2 0] X + [0]
[0 0 1] [3]
> [1 0 3] [2]
[0 1 0] X + [0]
[0 0 1] [3]
= [from(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__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(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__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(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__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(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__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(s(X)) -> s(mark(X))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..