interpretations
YES(?,O(n^3))
We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).
Strict Trs:
{ a__fst(X1, X2) -> fst(X1, X2)
, a__fst(0(), Z) -> nil()
, a__fst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
, mark(0()) -> 0()
, mark(nil()) -> nil()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(fst(X1, X2)) -> a__fst(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(len(X)) -> a__len(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__len(X) -> len(X)
, a__len(nil()) -> 0()
, a__len(cons(X, Z)) -> s(len(Z)) }
Obligation:
innermost runtime complexity
Answer:
YES(?,O(n^3))
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(s) = {}, Uargs(cons) = {1},
Uargs(mark) = {}, Uargs(fst) = {}, Uargs(a__from) = {1},
Uargs(from) = {}, Uargs(a__add) = {1, 2}, Uargs(add) = {},
Uargs(a__len) = {1}, Uargs(len) = {}
TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).
[1 0 0] [1 1 1] [2]
[a__fst](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [3]
[0 0 1] [0 1 0] [3]
[0]
[0] = [0]
[1]
[0]
[nil] = [1]
[2]
[1]
[s](x1) = [0]
[1]
[1 2 1] [0]
[cons](x1, x2) = [0 1 0] x1 + [2]
[0 1 0] [3]
[1 0 1] [0]
[mark](x1) = [0 1 0] x1 + [0]
[0 1 0] [1]
[1 0 0] [1 1 1] [1]
[fst](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [3]
[0 0 1] [0 1 0] [3]
[1 3 1] [3]
[a__from](x1) = [0 1 0] x1 + [2]
[0 1 0] [3]
[1 3 1] [2]
[from](x1) = [0 1 0] x1 + [2]
[0 1 0] [3]
[1 3 0] [1 2 1] [2]
[a__add](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [3]
[0 0 1] [0 1 0] [3]
[1 3 0] [1 2 1] [1]
[add](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [3]
[0 0 1] [0 1 0] [3]
[1 0 1] [2]
[a__len](x1) = [0 1 0] x1 + [3]
[0 1 0] [3]
[1 0 1] [1]
[len](x1) = [0 1 0] x1 + [3]
[0 1 0] [3]
This order satisfies following ordering constraints
[a__fst(X1, X2)] = [1 0 0] [1 1 1] [2]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 0 1] [0 1 0] [3]
> [1 0 0] [1 1 1] [1]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 0 1] [0 1 0] [3]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1 1 1] [2]
[0 1 0] Z + [3]
[0 1 0] [4]
> [0]
[1]
[2]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1 4 1] [8]
[0 1 0] Y + [5]
[0 1 0] [6]
> [1 3 1] [1]
[0 1 0] Y + [2]
[0 1 0] [3]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [1]
[0]
[1]
> [0]
[0]
[1]
= [0()]
[mark(nil())] = [2]
[1]
[2]
> [0]
[1]
[2]
= [nil()]
[mark(s(X))] = [2]
[0]
[1]
> [1]
[0]
[1]
= [s(X)]
[mark(cons(X1, X2))] = [1 3 1] [3]
[0 1 0] X1 + [2]
[0 1 0] [3]
> [1 3 1] [1]
[0 1 0] X1 + [2]
[0 1 0] [3]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [1 0 1] [1 2 1] [4]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 1 0] [4]
> [1 0 1] [1 2 1] [3]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 1 0] [4]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [1 4 1] [5]
[0 1 0] X + [2]
[0 1 0] [3]
> [1 4 1] [4]
[0 1 0] X + [2]
[0 1 0] [3]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [1 3 1] [1 3 1] [4]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 1 0] [4]
> [1 3 1] [1 3 1] [3]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 1 0] [4]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [1 1 1] [4]
[0 1 0] X + [3]
[0 1 0] [4]
> [1 1 1] [3]
[0 1 0] X + [3]
[0 1 0] [3]
= [a__len(mark(X))]
[a__from(X)] = [1 3 1] [3]
[0 1 0] X + [2]
[0 1 0] [3]
> [1 3 1] [1]
[0 1 0] X + [2]
[0 1 0] [3]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 3 1] [3]
[0 1 0] X + [2]
[0 1 0] [3]
> [1 3 1] [2]
[0 1 0] X + [2]
[0 1 0] [3]
= [from(X)]
[a__add(X1, X2)] = [1 3 0] [1 2 1] [2]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 0 1] [0 1 0] [3]
> [1 3 0] [1 2 1] [1]
[0 1 0] X1 + [0 1 0] X2 + [3]
[0 0 1] [0 1 0] [3]
= [add(X1, X2)]
[a__add(0(), X)] = [1 2 1] [2]
[0 1 0] X + [3]
[0 1 0] [4]
> [1 0 1] [0]
[0 1 0] X + [0]
[0 1 0] [1]
= [mark(X)]
[a__add(s(X), Y)] = [1 2 1] [3]
[0 1 0] Y + [3]
[0 1 0] [4]
> [1]
[0]
[1]
= [s(add(X, Y))]
[a__len(X)] = [1 0 1] [2]
[0 1 0] X + [3]
[0 1 0] [3]
> [1 0 1] [1]
[0 1 0] X + [3]
[0 1 0] [3]
= [len(X)]
[a__len(nil())] = [4]
[4]
[4]
> [0]
[0]
[1]
= [0()]
[a__len(cons(X, Z))] = [1 3 1] [5]
[0 1 0] X + [5]
[0 1 0] [5]
> [1]
[0]
[1]
= [s(len(Z))]
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__fst(0(), Z) -> nil()
, a__fst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__len(nil()) -> 0()
, a__len(cons(X, Z)) -> s(len(Z))
, mark(fst(X1, X2)) -> a__fst(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(len(X)) -> a__len(mark(X))
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__fst(X1, X2) -> fst(X1, X2)
, a__from(X) -> from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__len(X) -> len(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__fst(0(), Z) -> nil()
, a__fst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__len(nil()) -> 0()
, a__len(cons(X, Z)) -> s(len(Z))
, mark(fst(X1, X2)) -> a__fst(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(len(X)) -> a__len(mark(X))
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__fst(X1, X2) -> fst(X1, X2)
, a__from(X) -> from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__len(X) -> len(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__fst(0(), Z) -> nil()
, a__fst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__len(nil()) -> 0()
, a__len(cons(X, Z)) -> s(len(Z))
, mark(fst(X1, X2)) -> a__fst(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(len(X)) -> a__len(mark(X))
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__fst(X1, X2) -> fst(X1, X2)
, a__from(X) -> from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__len(X) -> len(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__fst(0(), Z) -> nil()
, a__fst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__len(nil()) -> 0()
, a__len(cons(X, Z)) -> s(len(Z))
, mark(fst(X1, X2)) -> a__fst(mark(X1), mark(X2))
, mark(from(X)) -> a__from(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(len(X)) -> a__len(mark(X))
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, a__fst(X1, X2) -> fst(X1, X2)
, a__from(X) -> from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__len(X) -> len(X) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..