We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 0()
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [0]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[inf](x1) = [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [4]
[take](x1, x2) = [0]
[a__length](x1) = [1] x1 + [0]
[length](x1) = [0]
[mark](x1) = [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [0]
>= [0]
= [false()]
[a__eq(X1, X2)] = [0]
>= [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [0]
>= [0]
= [true()]
[a__eq(s(X), s(Y))] = [0]
>= [0]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [0]
>= [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [0]
>= [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [0]
? [4]
= [nil()]
[a__take(s(X), cons(Y, L))] = [0]
>= [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [0]
>= [0]
= [length(X)]
[a__length(cons(X, L))] = [0]
>= [0]
= [s(length(L))]
[a__length(nil())] = [4]
> [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [0]
>= [0]
= [a__inf(mark(X))]
[mark(nil())] = [0]
? [4]
= [nil()]
[mark(take(X1, X2))] = [0]
>= [0]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [0]
>= [0]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
>= [0]
= [a__eq(X1, X2)]
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:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Weak Trs: { a__length(nil()) -> 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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [4]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[inf](x1) = [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [4]
[take](x1, x2) = [0]
[a__length](x1) = [1] x1 + [0]
[length](x1) = [0]
[mark](x1) = [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [4]
> [0]
= [false()]
[a__eq(X1, X2)] = [4]
> [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [4]
> [0]
= [true()]
[a__eq(s(X), s(Y))] = [4]
>= [4]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [0]
>= [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [0]
>= [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [0]
? [4]
= [nil()]
[a__take(s(X), cons(Y, L))] = [0]
>= [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [0]
>= [0]
= [length(X)]
[a__length(cons(X, L))] = [0]
>= [0]
= [s(length(L))]
[a__length(nil())] = [4]
> [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [0]
>= [0]
= [a__inf(mark(X))]
[mark(nil())] = [0]
? [4]
= [nil()]
[mark(take(X1, X2))] = [0]
>= [0]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [0]
>= [0]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
? [4]
= [a__eq(X1, X2)]
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:
{ a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__length(nil()) -> 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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [4]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [1]
[cons](x1, x2) = [0]
[inf](x1) = [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [4]
[take](x1, x2) = [0]
[a__length](x1) = [1] x1 + [0]
[length](x1) = [0]
[mark](x1) = [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [4]
> [0]
= [false()]
[a__eq(X1, X2)] = [4]
> [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [4]
> [0]
= [true()]
[a__eq(s(X), s(Y))] = [4]
>= [4]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [1]
> [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [1]
> [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [0]
? [4]
= [nil()]
[a__take(s(X), cons(Y, L))] = [0]
>= [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [0]
>= [0]
= [length(X)]
[a__length(cons(X, L))] = [0]
>= [0]
= [s(length(L))]
[a__length(nil())] = [4]
> [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [0]
? [1]
= [a__inf(mark(X))]
[mark(nil())] = [0]
? [4]
= [nil()]
[mark(take(X1, X2))] = [0]
>= [0]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [0]
>= [0]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
? [4]
= [a__eq(X1, X2)]
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:
{ a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__length(nil()) -> 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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [4]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[inf](x1) = [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [4]
[nil] = [4]
[take](x1, x2) = [0]
[a__length](x1) = [1] x1 + [0]
[length](x1) = [0]
[mark](x1) = [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [4]
> [0]
= [false()]
[a__eq(X1, X2)] = [4]
> [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [4]
> [0]
= [true()]
[a__eq(s(X), s(Y))] = [4]
>= [4]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [0]
>= [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [0]
>= [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [4]
> [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [4]
>= [4]
= [nil()]
[a__take(s(X), cons(Y, L))] = [4]
> [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [0]
>= [0]
= [length(X)]
[a__length(cons(X, L))] = [0]
>= [0]
= [s(length(L))]
[a__length(nil())] = [4]
> [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [0]
>= [0]
= [a__inf(mark(X))]
[mark(nil())] = [0]
? [4]
= [nil()]
[mark(take(X1, X2))] = [0]
? [4]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [0]
>= [0]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
? [4]
= [a__eq(X1, X2)]
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:
{ a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__take(0(), X) -> nil()
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(nil()) -> 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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [5]
[0] = [0]
[true] = [0]
[s](x1) = [1] x1 + [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x2 + [4]
[inf](x1) = [1] x1 + [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [4]
[nil] = [0]
[take](x1, x2) = [1] x1 + [1] x2 + [1]
[a__length](x1) = [1] x1 + [1]
[length](x1) = [1] x1 + [0]
[mark](x1) = [1] x1 + [0]
[eq](x1, x2) = [4]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [5]
> [0]
= [false()]
[a__eq(X1, X2)] = [5]
> [4]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [5]
> [0]
= [true()]
[a__eq(s(X), s(Y))] = [5]
>= [5]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [4]
>= [1] X + [4]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [4]
> [1] X + [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [1]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [4]
> [0]
= [nil()]
[a__take(s(X), cons(Y, L))] = [1] X + [1] L + [8]
> [1] X + [1] L + [5]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [1]
> [1] X + [0]
= [length(X)]
[a__length(cons(X, L))] = [1] L + [5]
> [1] L + [0]
= [s(length(L))]
[a__length(nil())] = [1]
> [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [1] X2 + [4]
>= [1] X2 + [4]
= [cons(X1, X2)]
[mark(inf(X))] = [1] X + [0]
? [1] X + [4]
= [a__inf(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(take(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [4]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [1] X + [0]
? [1] X + [1]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [4]
? [5]
= [a__eq(X1, X2)]
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:
{ a__eq(s(X), s(Y)) -> a__eq(X, Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [1] x2 + [6]
[0] = [1]
[true] = [0]
[s](x1) = [1] x1 + [1]
[false] = [0]
[a__inf](x1) = [1] x1 + [5]
[cons](x1, x2) = [1] x2 + [0]
[inf](x1) = [1] x1 + [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [4]
[nil] = [4]
[take](x1, x2) = [1] x1 + [1] x2 + [0]
[a__length](x1) = [1] x1 + [4]
[length](x1) = [1] x1 + [0]
[mark](x1) = [1] x1 + [0]
[eq](x1, x2) = [1] x2 + [3]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [1] Y + [6]
> [0]
= [false()]
[a__eq(X1, X2)] = [1] X2 + [6]
> [1] X2 + [3]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [7]
> [0]
= [true()]
[a__eq(s(X), s(Y))] = [1] Y + [7]
> [1] Y + [6]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [5]
> [1] X + [1]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [5]
> [1] X + [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [5]
> [4]
= [nil()]
[a__take(s(X), cons(Y, L))] = [1] X + [1] L + [5]
> [1] X + [1] L + [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [4]
> [1] X + [0]
= [length(X)]
[a__length(cons(X, L))] = [1] L + [4]
> [1] L + [1]
= [s(length(L))]
[a__length(nil())] = [8]
> [1]
= [0()]
[mark(0())] = [1]
>= [1]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [1] X2 + [0]
>= [1] X2 + [0]
= [cons(X1, X2)]
[mark(inf(X))] = [1] X + [0]
? [1] X + [5]
= [a__inf(mark(X))]
[mark(nil())] = [4]
>= [4]
= [nil()]
[mark(take(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [4]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [1] X + [0]
? [1] X + [4]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [1] X2 + [3]
? [1] X2 + [6]
= [a__eq(X1, X2)]
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:
{ mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__eq](x1, x2) = [7]
[0] = [4]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [7]
[cons](x1, x2) = [0]
[inf](x1) = [1] x1 + [7]
[a__take](x1, x2) = [1] x1 + [1] x2 + [1]
[nil] = [4]
[take](x1, x2) = [1] x1 + [1] x2 + [0]
[a__length](x1) = [1] x1 + [7]
[length](x1) = [1] x1 + [4]
[mark](x1) = [1] x1 + [1]
[eq](x1, x2) = [7]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [7]
> [0]
= [false()]
[a__eq(X1, X2)] = [7]
>= [7]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [7]
> [0]
= [true()]
[a__eq(s(X), s(Y))] = [7]
>= [7]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [7]
> [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [7]
>= [1] X + [7]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [5]
> [4]
= [nil()]
[a__take(s(X), cons(Y, L))] = [1]
> [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [7]
> [1] X + [4]
= [length(X)]
[a__length(cons(X, L))] = [7]
> [0]
= [s(length(L))]
[a__length(nil())] = [11]
> [4]
= [0()]
[mark(0())] = [5]
> [4]
= [0()]
[mark(true())] = [1]
> [0]
= [true()]
[mark(s(X))] = [1]
> [0]
= [s(X)]
[mark(false())] = [1]
> [0]
= [false()]
[mark(cons(X1, X2))] = [1]
> [0]
= [cons(X1, X2)]
[mark(inf(X))] = [1] X + [8]
>= [1] X + [8]
= [a__inf(mark(X))]
[mark(nil())] = [5]
> [4]
= [nil()]
[mark(take(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [3]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [1] X + [5]
? [1] X + [8]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [8]
> [7]
= [a__eq(X1, X2)]
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:
{ mark(inf(X)) -> a__inf(mark(X))
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X)) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 0()
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(nil()) -> nil()
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
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: { mark(length(X)) -> a__length(mark(X)) }
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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__eq](x1, x2) = [0]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[inf](x1) = [1] x1 + [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[take](x1, x2) = [1] x1 + [1] x2 + [0]
[a__length](x1) = [1] x1 + [1]
[length](x1) = [1] x1 + [1]
[mark](x1) = [4] x1 + [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [0]
>= [0]
= [false()]
[a__eq(X1, X2)] = [0]
>= [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [0]
>= [0]
= [true()]
[a__eq(s(X), s(Y))] = [0]
>= [0]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [0]
>= [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [0]
>= [1] X + [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a__take(s(X), cons(Y, L))] = [0]
>= [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [1]
>= [1] X + [1]
= [length(X)]
[a__length(cons(X, L))] = [1]
> [0]
= [s(length(L))]
[a__length(nil())] = [1]
> [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [4] X + [0]
>= [4] X + [0]
= [a__inf(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(take(X1, X2))] = [4] X1 + [4] X2 + [0]
>= [4] X1 + [4] X2 + [0]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [4] X + [4]
> [4] X + [1]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
>= [0]
= [a__eq(X1, X2)]
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:
{ mark(inf(X)) -> a__inf(mark(X))
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 0()
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(nil()) -> nil()
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
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: { mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) }
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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__eq](x1, x2) = [0]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[inf](x1) = [1] x1 + [0]
[a__take](x1, x2) = [1] x1 + [1] x2 + [1]
[nil] = [0]
[take](x1, x2) = [1] x1 + [1] x2 + [1]
[a__length](x1) = [1] x1 + [0]
[length](x1) = [1] x1 + [0]
[mark](x1) = [2] x1 + [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [0]
>= [0]
= [false()]
[a__eq(X1, X2)] = [0]
>= [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [0]
>= [0]
= [true()]
[a__eq(s(X), s(Y))] = [0]
>= [0]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [0]
>= [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [0]
>= [1] X + [0]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [1]
> [0]
= [nil()]
[a__take(s(X), cons(Y, L))] = [1]
> [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [0]
>= [1] X + [0]
= [length(X)]
[a__length(cons(X, L))] = [0]
>= [0]
= [s(length(L))]
[a__length(nil())] = [0]
>= [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [2] X + [0]
>= [2] X + [0]
= [a__inf(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(take(X1, X2))] = [2] X1 + [2] X2 + [2]
> [2] X1 + [2] X2 + [1]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [2] X + [0]
>= [2] X + [0]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
>= [0]
= [a__eq(X1, X2)]
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: { mark(inf(X)) -> a__inf(mark(X)) }
Weak Trs:
{ a__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 0()
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
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: { mark(inf(X)) -> a__inf(mark(X)) }
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(a__inf) = {1}, Uargs(a__take) = {1, 2},
Uargs(a__length) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__eq](x1, x2) = [0]
[0] = [0]
[true] = [0]
[s](x1) = [0]
[false] = [0]
[a__inf](x1) = [1] x1 + [4]
[cons](x1, x2) = [0]
[inf](x1) = [1] x1 + [4]
[a__take](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[take](x1, x2) = [1] x1 + [1] x2 + [0]
[a__length](x1) = [1] x1 + [0]
[length](x1) = [1] x1 + [0]
[mark](x1) = [3] x1 + [0]
[eq](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__eq(X, Y)] = [0]
>= [0]
= [false()]
[a__eq(X1, X2)] = [0]
>= [0]
= [eq(X1, X2)]
[a__eq(0(), 0())] = [0]
>= [0]
= [true()]
[a__eq(s(X), s(Y))] = [0]
>= [0]
= [a__eq(X, Y)]
[a__inf(X)] = [1] X + [4]
> [0]
= [cons(X, inf(s(X)))]
[a__inf(X)] = [1] X + [4]
>= [1] X + [4]
= [inf(X)]
[a__take(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [take(X1, X2)]
[a__take(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[a__take(s(X), cons(Y, L))] = [0]
>= [0]
= [cons(Y, take(X, L))]
[a__length(X)] = [1] X + [0]
>= [1] X + [0]
= [length(X)]
[a__length(cons(X, L))] = [0]
>= [0]
= [s(length(L))]
[a__length(nil())] = [0]
>= [0]
= [0()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(X1, X2)]
[mark(inf(X))] = [3] X + [12]
> [3] X + [4]
= [a__inf(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(take(X1, X2))] = [3] X1 + [3] X2 + [0]
>= [3] X1 + [3] X2 + [0]
= [a__take(mark(X1), mark(X2))]
[mark(length(X))] = [3] X + [0]
>= [3] X + [0]
= [a__length(mark(X))]
[mark(eq(X1, X2))] = [0]
>= [0]
= [a__eq(X1, X2)]
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__eq(X, Y) -> false()
, a__eq(X1, X2) -> eq(X1, X2)
, a__eq(0(), 0()) -> true()
, a__eq(s(X), s(Y)) -> a__eq(X, Y)
, a__inf(X) -> cons(X, inf(s(X)))
, a__inf(X) -> inf(X)
, a__take(X1, X2) -> take(X1, X2)
, a__take(0(), X) -> nil()
, a__take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
, a__length(X) -> length(X)
, a__length(cons(X, L)) -> s(length(L))
, a__length(nil()) -> 0()
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(s(X)) -> s(X)
, mark(false()) -> false()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(inf(X)) -> a__inf(mark(X))
, mark(nil()) -> nil()
, mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
, mark(length(X)) -> a__length(mark(X))
, mark(eq(X1, X2)) -> a__eq(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))