YES(O(1),O(n^2))
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__from](x1) = [1] x1 + [7]
[cons](x1, x2) = [1] x1 + [7]
[mark](x1) = [1] x1 + [7]
[from](x1) = [1] x1 + [7]
[s](x1) = [1] x1 + [7]
[a__length](x1) = [7]
[nil] = [7]
[0] = [3]
[a__length1](x1) = [3]
[length](x1) = [7]
[length1](x1) = [7]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1] X + [7]
? [1] X + [14]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [7]
>= [1] X + [7]
= [from(X)]
[mark(cons(X1, X2))] = [1] X1 + [14]
>= [1] X1 + [14]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1] X + [14]
>= [1] X + [14]
= [a__from(mark(X))]
[mark(s(X))] = [1] X + [14]
>= [1] X + [14]
= [s(mark(X))]
[mark(nil())] = [14]
> [7]
= [nil()]
[mark(0())] = [10]
> [3]
= [0()]
[mark(length(X))] = [14]
> [7]
= [a__length(X)]
[mark(length1(X))] = [14]
> [3]
= [a__length1(X)]
[a__length(X)] = [7]
>= [7]
= [length(X)]
[a__length(cons(X, Y))] = [7]
? [10]
= [s(a__length1(Y))]
[a__length(nil())] = [7]
> [3]
= [0()]
[a__length1(X)] = [3]
? [7]
= [a__length(X)]
[a__length1(X)] = [3]
? [7]
= [length1(X)]
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:
{ 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))
, a__length(X) -> length(X)
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length1(X) -> a__length(X)
, a__length1(X) -> length1(X) }
Weak Trs:
{ mark(nil()) -> nil()
, mark(0()) -> 0()
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, a__length(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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__from](x1) = [1] x1 + [1]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[a__length](x1) = [0]
[nil] = [7]
[0] = [0]
[a__length1](x1) = [0]
[length](x1) = [4]
[length1](x1) = [4]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1] X + [1]
> [1] X + [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [1]
> [1] X + [0]
= [from(X)]
[mark(cons(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1] X + [0]
? [1] X + [1]
= [a__from(mark(X))]
[mark(s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(mark(X))]
[mark(nil())] = [7]
>= [7]
= [nil()]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(length(X))] = [4]
> [0]
= [a__length(X)]
[mark(length1(X))] = [4]
> [0]
= [a__length1(X)]
[a__length(X)] = [0]
? [4]
= [length(X)]
[a__length(cons(X, Y))] = [0]
>= [0]
= [s(a__length1(Y))]
[a__length(nil())] = [0]
>= [0]
= [0()]
[a__length1(X)] = [0]
>= [0]
= [a__length(X)]
[a__length1(X)] = [0]
? [4]
= [length1(X)]
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(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, a__length(X) -> length(X)
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length1(X) -> a__length(X)
, a__length1(X) -> length1(X) }
Weak Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, mark(nil()) -> nil()
, mark(0()) -> 0()
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, a__length(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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__from](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[a__length](x1) = [2]
[nil] = [7]
[0] = [2]
[a__length1](x1) = [0]
[length](x1) = [4]
[length1](x1) = [4]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1] X + [4]
> [1] X + [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [4]
> [1] X + [0]
= [from(X)]
[mark(cons(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1] X + [0]
? [1] X + [4]
= [a__from(mark(X))]
[mark(s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(mark(X))]
[mark(nil())] = [7]
>= [7]
= [nil()]
[mark(0())] = [2]
>= [2]
= [0()]
[mark(length(X))] = [4]
> [2]
= [a__length(X)]
[mark(length1(X))] = [4]
> [0]
= [a__length1(X)]
[a__length(X)] = [2]
? [4]
= [length(X)]
[a__length(cons(X, Y))] = [2]
> [0]
= [s(a__length1(Y))]
[a__length(nil())] = [2]
>= [2]
= [0()]
[a__length1(X)] = [0]
? [2]
= [a__length(X)]
[a__length1(X)] = [0]
? [4]
= [length1(X)]
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(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, a__length(X) -> length(X)
, a__length1(X) -> a__length(X)
, a__length1(X) -> length1(X) }
Weak Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, mark(nil()) -> nil()
, mark(0()) -> 0()
, mark(length(X)) -> a__length(X)
, mark(length1(X)) -> a__length1(X)
, a__length(cons(X, Y)) -> s(a__length1(Y))
, a__length(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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__from](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1] x1 + [1]
[from](x1) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[a__length](x1) = [4]
[nil] = [7]
[0] = [3]
[a__length1](x1) = [0]
[length](x1) = [3]
[length1](x1) = [0]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1] X + [4]
> [1] X + [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [4]
> [1] X + [0]
= [from(X)]
[mark(cons(X1, X2))] = [1] X1 + [1]
>= [1] X1 + [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1] X + [1]
? [1] X + [5]
= [a__from(mark(X))]
[mark(s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(mark(X))]
[mark(nil())] = [8]
> [7]
= [nil()]
[mark(0())] = [4]
> [3]
= [0()]
[mark(length(X))] = [4]
>= [4]
= [a__length(X)]
[mark(length1(X))] = [1]
> [0]
= [a__length1(X)]
[a__length(X)] = [4]
> [3]
= [length(X)]
[a__length(cons(X, Y))] = [4]
> [0]
= [s(a__length1(Y))]
[a__length(nil())] = [4]
> [3]
= [0()]
[a__length1(X)] = [0]
? [4]
= [a__length(X)]
[a__length1(X)] = [0]
>= [0]
= [length1(X)]
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(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, a__length1(X) -> a__length(X)
, a__length1(X) -> length1(X) }
Weak Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(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() }
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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__from](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [1] x1 + [1]
[from](x1) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[a__length](x1) = [4]
[nil] = [7]
[0] = [3]
[a__length1](x1) = [1]
[length](x1) = [4]
[length1](x1) = [0]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1] X + [4]
> [1] X + [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [4]
> [1] X + [0]
= [from(X)]
[mark(cons(X1, X2))] = [1] X1 + [1]
>= [1] X1 + [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1] X + [1]
? [1] X + [5]
= [a__from(mark(X))]
[mark(s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(mark(X))]
[mark(nil())] = [8]
> [7]
= [nil()]
[mark(0())] = [4]
> [3]
= [0()]
[mark(length(X))] = [5]
> [4]
= [a__length(X)]
[mark(length1(X))] = [1]
>= [1]
= [a__length1(X)]
[a__length(X)] = [4]
>= [4]
= [length(X)]
[a__length(cons(X, Y))] = [4]
> [1]
= [s(a__length1(Y))]
[a__length(nil())] = [4]
> [3]
= [0()]
[a__length1(X)] = [1]
? [4]
= [a__length(X)]
[a__length1(X)] = [1]
> [0]
= [length1(X)]
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(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, a__length1(X) -> a__length(X) }
Weak Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(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) -> length1(X) }
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: { a__length1(X) -> a__length(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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[a__from](x1) = [1 0] x1 + [0]
[0 0] [1]
[cons](x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 1] [1]
[mark](x1) = [1 0] x1 + [0]
[0 4] [1]
[from](x1) = [1 0] x1 + [0]
[0 0] [0]
[s](x1) = [1 0] x1 + [0]
[0 0] [0]
[a__length](x1) = [0 1] x1 + [0]
[2 0] [1]
[nil] = [2]
[0]
[0] = [0]
[1]
[a__length1](x1) = [0 1] x1 + [1]
[4 0] [1]
[length](x1) = [0 1] x1 + [0]
[1 0] [1]
[length1](x1) = [0 1] x1 + [1]
[1 0] [1]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1 0] X + [0]
[0 0] [1]
>= [1 0] X + [0]
[0 0] [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 0] X + [0]
[0 0] [1]
>= [1 0] X + [0]
[0 0] [0]
= [from(X)]
[mark(cons(X1, X2))] = [1 0] X1 + [0 0] X2 + [0]
[0 0] [0 4] [5]
>= [1 0] X1 + [0 0] X2 + [0]
[0 0] [0 1] [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 0] X + [0]
[0 0] [1]
>= [1 0] X + [0]
[0 0] [1]
= [a__from(mark(X))]
[mark(s(X))] = [1 0] X + [0]
[0 0] [1]
>= [1 0] X + [0]
[0 0] [0]
= [s(mark(X))]
[mark(nil())] = [2]
[1]
>= [2]
[0]
= [nil()]
[mark(0())] = [0]
[5]
>= [0]
[1]
= [0()]
[mark(length(X))] = [0 1] X + [0]
[4 0] [5]
>= [0 1] X + [0]
[2 0] [1]
= [a__length(X)]
[mark(length1(X))] = [0 1] X + [1]
[4 0] [5]
>= [0 1] X + [1]
[4 0] [1]
= [a__length1(X)]
[a__length(X)] = [0 1] X + [0]
[2 0] [1]
>= [0 1] X + [0]
[1 0] [1]
= [length(X)]
[a__length(cons(X, Y))] = [0 0] X + [0 1] Y + [1]
[2 0] [0 0] [1]
>= [0 1] Y + [1]
[0 0] [0]
= [s(a__length1(Y))]
[a__length(nil())] = [0]
[5]
>= [0]
[1]
= [0()]
[a__length1(X)] = [0 1] X + [1]
[4 0] [1]
> [0 1] X + [0]
[2 0] [1]
= [a__length(X)]
[a__length1(X)] = [0 1] X + [1]
[4 0] [1]
>= [0 1] X + [1]
[1 0] [1]
= [length1(X)]
We return to the main proof.
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(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X)) }
Weak Trs:
{ a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(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(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(from(X)) -> a__from(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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__from](x1) = [1 2] x1 + [0]
[0 1] [4]
[cons](x1, x2) = [1 0] x1 + [0]
[0 1] [4]
[mark](x1) = [1 1] x1 + [0]
[0 1] [0]
[from](x1) = [1 2] x1 + [0]
[0 1] [4]
[s](x1) = [1 0] x1 + [0]
[0 1] [0]
[a__length](x1) = [0]
[0]
[nil] = [0]
[0]
[0] = [0]
[0]
[a__length1](x1) = [0]
[0]
[length](x1) = [0]
[0]
[length1](x1) = [0]
[0]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1 2] X + [0]
[0 1] [4]
>= [1 1] X + [0]
[0 1] [4]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 2] X + [0]
[0 1] [4]
>= [1 2] X + [0]
[0 1] [4]
= [from(X)]
[mark(cons(X1, X2))] = [1 1] X1 + [4]
[0 1] [4]
> [1 1] X1 + [0]
[0 1] [4]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 3] X + [4]
[0 1] [4]
> [1 3] X + [0]
[0 1] [4]
= [a__from(mark(X))]
[mark(s(X))] = [1 1] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [s(mark(X))]
[mark(nil())] = [0]
[0]
>= [0]
[0]
= [nil()]
[mark(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[mark(length(X))] = [0]
[0]
>= [0]
[0]
= [a__length(X)]
[mark(length1(X))] = [0]
[0]
>= [0]
[0]
= [a__length1(X)]
[a__length(X)] = [0]
[0]
>= [0]
[0]
= [length(X)]
[a__length(cons(X, Y))] = [0]
[0]
>= [0]
[0]
= [s(a__length1(Y))]
[a__length(nil())] = [0]
[0]
>= [0]
[0]
= [0()]
[a__length1(X)] = [0]
[0]
>= [0]
[0]
= [a__length(X)]
[a__length1(X)] = [0]
[0]
>= [0]
[0]
= [length1(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs: { mark(s(X)) -> s(mark(X)) }
Weak 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(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(1),O(n^2))
We use the processor 'matrix interpretation of dimension 3' to
orient following rules strictly.
Trs: { mark(s(X)) -> s(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__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(2)).
[1 2 0] [1]
[a__from](x1) = [0 1 0] x1 + [0]
[0 0 0] [4]
[1 0 0] [0 0 1] [1]
[cons](x1, x2) = [0 1 0] x1 + [0 0 0] x2 + [0]
[0 0 0] [0 0 1] [4]
[1 2 0] [0]
[mark](x1) = [0 1 0] x1 + [0]
[2 0 0] [2]
[1 2 0] [1]
[from](x1) = [0 1 0] x1 + [0]
[0 0 0] [0]
[1 0 0] [1]
[s](x1) = [0 1 0] x1 + [4]
[0 0 0] [0]
[0 0 2] [0]
[a__length](x1) = [0 0 1] x1 + [0]
[0 0 0] [0]
[1]
[nil] = [2]
[2]
[0]
[0] = [0]
[0]
[0 0 2] [0]
[a__length1](x1) = [0 0 1] x1 + [0]
[0 0 0] [0]
[0 0 0] [0]
[length](x1) = [0 0 1] x1 + [0]
[0 0 0] [0]
[0 0 0] [0]
[length1](x1) = [0 0 1] x1 + [0]
[0 0 0] [0]
The following symbols are considered usable
{a__from, mark, a__length, a__length1}
The order satisfies the following ordering constraints:
[a__from(X)] = [1 2 0] [1]
[0 1 0] X + [0]
[0 0 0] [4]
>= [1 2 0] [1]
[0 1 0] X + [0]
[0 0 0] [4]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 2 0] [1]
[0 1 0] X + [0]
[0 0 0] [4]
>= [1 2 0] [1]
[0 1 0] X + [0]
[0 0 0] [0]
= [from(X)]
[mark(cons(X1, X2))] = [1 2 0] [0 0 1] [1]
[0 1 0] X1 + [0 0 0] X2 + [0]
[2 0 0] [0 0 2] [4]
>= [1 2 0] [0 0 1] [1]
[0 1 0] X1 + [0 0 0] X2 + [0]
[0 0 0] [0 0 1] [4]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 4 0] [1]
[0 1 0] X + [0]
[2 4 0] [4]
>= [1 4 0] [1]
[0 1 0] X + [0]
[0 0 0] [4]
= [a__from(mark(X))]
[mark(s(X))] = [1 2 0] [9]
[0 1 0] X + [4]
[2 0 0] [4]
> [1 2 0] [1]
[0 1 0] X + [4]
[0 0 0] [0]
= [s(mark(X))]
[mark(nil())] = [5]
[2]
[4]
> [1]
[2]
[2]
= [nil()]
[mark(0())] = [0]
[0]
[2]
>= [0]
[0]
[0]
= [0()]
[mark(length(X))] = [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [2]
>= [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [0]
= [a__length(X)]
[mark(length1(X))] = [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [2]
>= [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [0]
= [a__length1(X)]
[a__length(X)] = [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [0]
>= [0 0 0] [0]
[0 0 1] X + [0]
[0 0 0] [0]
= [length(X)]
[a__length(cons(X, Y))] = [0 0 2] [8]
[0 0 1] Y + [4]
[0 0 0] [0]
> [0 0 2] [1]
[0 0 1] Y + [4]
[0 0 0] [0]
= [s(a__length1(Y))]
[a__length(nil())] = [4]
[2]
[0]
> [0]
[0]
[0]
= [0()]
[a__length1(X)] = [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [0]
>= [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [0]
= [a__length(X)]
[a__length1(X)] = [0 0 2] [0]
[0 0 1] X + [0]
[0 0 0] [0]
>= [0 0 0] [0]
[0 0 1] X + [0]
[0 0 0] [0]
= [length1(X)]
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__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(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))