We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [4]
[0] = [0]
[nil] = [0]
[s](x1) = [4]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [0]
[fst](x1, x2) = [0]
[a__from](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [0]
[add](x1, x2) = [1] x2 + [0]
[a__len](x1) = [1] x1 + [0]
[len](x1) = [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [4]
> [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [4]
> [0]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [8]
> [0]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(s(X))] = [0]
? [4]
= [s(X)]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [0]
? [4]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [0]
>= [0]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [0]
>= [0]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [0]
>= [0]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [0]
>= [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [0]
>= [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [0]
>= [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [4]
>= [4]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [0]
>= [0]
= [len(X)]
[a__len(nil())] = [0]
>= [0]
= [0()]
[a__len(cons(X, Z))] = [1] X + [0]
? [4]
= [s(len(Z))]
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^2)).
Strict Trs:
{ 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)) }
Weak 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)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [4]
[0] = [4]
[nil] = [5]
[s](x1) = [4]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1] x1 + [0]
[fst](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__len](x1) = [1] x1 + [0]
[len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [8]
> [5]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [8]
> [1] Y + [0]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [4]
>= [4]
= [0()]
[mark(nil())] = [5]
>= [5]
= [nil()]
[mark(s(X))] = [4]
>= [4]
= [s(X)]
[mark(cons(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [4]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [1] X + [0]
>= [1] X + [0]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [1] X + [0]
>= [1] X + [0]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [0]
>= [1] X + [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [0]
>= [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [4]
>= [4]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [0]
>= [1] X + [0]
= [len(X)]
[a__len(nil())] = [5]
> [4]
= [0()]
[a__len(cons(X, Z))] = [1] X + [0]
? [4]
= [s(len(Z))]
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^2)).
Strict Trs:
{ 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(s(X), Y) -> s(add(X, Y))
, a__len(X) -> len(X)
, a__len(cons(X, Z)) -> s(len(Z)) }
Weak 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))
, a__add(0(), X) -> mark(X)
, a__len(nil()) -> 0() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [7]
[nil] = [7]
[s](x1) = [7]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [0]
[fst](x1, x2) = [0]
[a__from](x1) = [1] x1 + [1]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [1]
[add](x1, x2) = [1] x2 + [0]
[a__len](x1) = [1] x1 + [1]
[len](x1) = [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [1]
> [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [8]
> [7]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [8]
> [0]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [0]
? [7]
= [0()]
[mark(nil())] = [0]
? [7]
= [nil()]
[mark(s(X))] = [0]
? [7]
= [s(X)]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [0]
? [1]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [0]
? [1]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [0]
? [1]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [0]
? [1]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [1]
> [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [1]
> [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [8]
> [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [8]
> [7]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [1]
> [0]
= [len(X)]
[a__len(nil())] = [8]
> [7]
= [0()]
[a__len(cons(X, Z))] = [1] X + [1]
? [7]
= [s(len(Z))]
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^2)).
Strict Trs:
{ 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__len(cons(X, Z)) -> s(len(Z)) }
Weak 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))
, 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() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [7]
[nil] = [7]
[s](x1) = [0]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [0]
[fst](x1, x2) = [0]
[a__from](x1) = [1] x1 + [5]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [5]
[add](x1, x2) = [1] x2 + [0]
[a__len](x1) = [1] x1 + [1]
[len](x1) = [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [1]
> [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [8]
> [7]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [1]
> [0]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [0]
? [7]
= [0()]
[mark(nil())] = [0]
? [7]
= [nil()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [0]
? [1]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [0]
? [5]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [0]
? [5]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [0]
? [1]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [5]
> [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [5]
> [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [5]
> [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [12]
> [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [5]
> [0]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [1]
> [0]
= [len(X)]
[a__len(nil())] = [8]
> [7]
= [0()]
[a__len(cons(X, Z))] = [1] X + [1]
> [0]
= [s(len(Z))]
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^2)).
Strict Trs:
{ 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)) }
Weak 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))
, 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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [6]
[0] = [2]
[nil] = [4]
[s](x1) = [0]
[cons](x1, x2) = [1] x1 + [3]
[mark](x1) = [1]
[fst](x1, x2) = [0]
[a__from](x1) = [1] x1 + [7]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [6]
[add](x1, x2) = [1] x2 + [0]
[a__len](x1) = [1] x1 + [0]
[len](x1) = [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [6]
> [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [8]
> [4]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [9]
> [4]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [1]
? [2]
= [0()]
[mark(nil())] = [1]
? [4]
= [nil()]
[mark(s(X))] = [1]
> [0]
= [s(X)]
[mark(cons(X1, X2))] = [1]
? [4]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [1]
? [8]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [1]
? [8]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [1]
? [8]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [1]
>= [1]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [7]
> [4]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [7]
> [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [8]
> [1]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [6]
> [0]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [0]
>= [0]
= [len(X)]
[a__len(nil())] = [4]
> [2]
= [0()]
[a__len(cons(X, Z))] = [1] X + [3]
> [0]
= [s(len(Z))]
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^2)).
Strict Trs:
{ mark(0()) -> 0()
, mark(nil()) -> nil()
, 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)) }
Weak 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(s(X)) -> s(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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [7]
[0] = [1]
[nil] = [0]
[s](x1) = [1]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1]
[fst](x1, x2) = [0]
[a__from](x1) = [1] x1 + [7]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [7]
[add](x1, x2) = [1] x2 + [0]
[a__len](x1) = [1] x1 + [4]
[len](x1) = [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [7]
> [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [8]
> [0]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [8]
> [1]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [1]
>= [1]
= [0()]
[mark(nil())] = [1]
> [0]
= [nil()]
[mark(s(X))] = [1]
>= [1]
= [s(X)]
[mark(cons(X1, X2))] = [1]
>= [1]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [1]
? [9]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [1]
? [8]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [1]
? [9]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [1]
? [5]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [7]
> [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [7]
> [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [7]
> [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [8]
> [1]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [8]
> [1]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [4]
> [0]
= [len(X)]
[a__len(nil())] = [4]
> [1]
= [0()]
[a__len(cons(X, Z))] = [1] X + [4]
> [1]
= [s(len(Z))]
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^2)).
Strict Trs:
{ mark(0()) -> 0()
, 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)) }
Weak 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(nil()) -> nil()
, mark(s(X)) -> s(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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__fst](x1, x2) = [1] x1 + [1] x2 + [7]
[0] = [0]
[nil] = [0]
[s](x1) = [0]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1]
[fst](x1, x2) = [0]
[a__from](x1) = [1] x1 + [7]
[from](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [6]
[add](x1, x2) = [1] x2 + [0]
[a__len](x1) = [1] x1 + [0]
[len](x1) = [0]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1] X1 + [1] X2 + [7]
> [0]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1] Z + [7]
> [0]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1] Y + [7]
> [1]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [1]
> [0]
= [0()]
[mark(nil())] = [1]
> [0]
= [nil()]
[mark(s(X))] = [1]
> [0]
= [s(X)]
[mark(cons(X1, X2))] = [1]
>= [1]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [1]
? [9]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [1]
? [8]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [1]
? [8]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [1]
>= [1]
= [a__len(mark(X))]
[a__from(X)] = [1] X + [7]
> [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [7]
> [1] X + [0]
= [from(X)]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [1] X + [6]
> [1]
= [mark(X)]
[a__add(s(X), Y)] = [1] Y + [6]
> [0]
= [s(add(X, Y))]
[a__len(X)] = [1] X + [0]
>= [0]
= [len(X)]
[a__len(nil())] = [0]
>= [0]
= [0()]
[a__len(cons(X, Z))] = [1] X + [0]
>= [0]
= [s(len(Z))]
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^2)).
Strict Trs:
{ 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)) }
Weak 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)
, 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(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ 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)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__fst) = {1, 2}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(a__add) = {1, 2}, Uargs(a__len) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__fst](x1, x2) = [1 2] x1 + [1 6] x2 + [6]
[0 1] [0 1] [3]
[0] = [2]
[2]
[nil] = [0]
[4]
[s](x1) = [1]
[0]
[cons](x1, x2) = [1 0] x1 + [0]
[0 1] [1]
[mark](x1) = [1 3] x1 + [3]
[0 1] [0]
[fst](x1, x2) = [1 2] x1 + [1 6] x2 + [1]
[0 1] [0 1] [3]
[a__from](x1) = [1 6] x1 + [4]
[0 1] [1]
[from](x1) = [1 6] x1 + [2]
[0 1] [1]
[a__add](x1, x2) = [1 4] x1 + [1 3] x2 + [2]
[0 1] [0 1] [4]
[add](x1, x2) = [1 4] x1 + [1 3] x2 + [0]
[0 1] [0 1] [4]
[a__len](x1) = [1 1] x1 + [1]
[0 1] [2]
[len](x1) = [1 1] x1 + [0]
[0 1] [2]
The order satisfies the following ordering constraints:
[a__fst(X1, X2)] = [1 2] X1 + [1 6] X2 + [6]
[0 1] [0 1] [3]
> [1 2] X1 + [1 6] X2 + [1]
[0 1] [0 1] [3]
= [fst(X1, X2)]
[a__fst(0(), Z)] = [1 6] Z + [12]
[0 1] [5]
> [0]
[4]
= [nil()]
[a__fst(s(X), cons(Y, Z))] = [1 6] Y + [13]
[0 1] [4]
> [1 3] Y + [3]
[0 1] [1]
= [cons(mark(Y), fst(X, Z))]
[mark(0())] = [11]
[2]
> [2]
[2]
= [0()]
[mark(nil())] = [15]
[4]
> [0]
[4]
= [nil()]
[mark(s(X))] = [4]
[0]
> [1]
[0]
= [s(X)]
[mark(cons(X1, X2))] = [1 3] X1 + [6]
[0 1] [1]
> [1 3] X1 + [3]
[0 1] [1]
= [cons(mark(X1), X2)]
[mark(fst(X1, X2))] = [1 5] X1 + [1 9] X2 + [13]
[0 1] [0 1] [3]
> [1 5] X1 + [1 9] X2 + [12]
[0 1] [0 1] [3]
= [a__fst(mark(X1), mark(X2))]
[mark(from(X))] = [1 9] X + [8]
[0 1] [1]
> [1 9] X + [7]
[0 1] [1]
= [a__from(mark(X))]
[mark(add(X1, X2))] = [1 7] X1 + [1 6] X2 + [15]
[0 1] [0 1] [4]
> [1 7] X1 + [1 6] X2 + [8]
[0 1] [0 1] [4]
= [a__add(mark(X1), mark(X2))]
[mark(len(X))] = [1 4] X + [9]
[0 1] [2]
> [1 4] X + [4]
[0 1] [2]
= [a__len(mark(X))]
[a__from(X)] = [1 6] X + [4]
[0 1] [1]
> [1 3] X + [3]
[0 1] [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 6] X + [4]
[0 1] [1]
> [1 6] X + [2]
[0 1] [1]
= [from(X)]
[a__add(X1, X2)] = [1 4] X1 + [1 3] X2 + [2]
[0 1] [0 1] [4]
> [1 4] X1 + [1 3] X2 + [0]
[0 1] [0 1] [4]
= [add(X1, X2)]
[a__add(0(), X)] = [1 3] X + [12]
[0 1] [6]
> [1 3] X + [3]
[0 1] [0]
= [mark(X)]
[a__add(s(X), Y)] = [1 3] Y + [3]
[0 1] [4]
> [1]
[0]
= [s(add(X, Y))]
[a__len(X)] = [1 1] X + [1]
[0 1] [2]
> [1 1] X + [0]
[0 1] [2]
= [len(X)]
[a__len(nil())] = [5]
[6]
> [2]
[2]
= [0()]
[a__len(cons(X, Z))] = [1 1] X + [2]
[0 1] [3]
> [1]
[0]
= [s(len(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:
{ 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(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))