interpretations
| Execution Time (secs) | - |
| Answer | YES(?,O(n^2)) |
| Input | Der95 11 |
YES(?,O(n^2))
We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).
Strict Trs:
{ D(t()) -> 1()
, D(constant()) -> 0()
, D(+(x, y)) -> +(D(x), D(y))
, D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
, D(-(x, y)) -> -(D(x), D(y))
, D(minus(x)) -> minus(D(x))
, D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2())))
, D(pow(x, y)) ->
+(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y)))
, D(ln(x)) -> div(D(x), x) }
Obligation:
innermost runtime complexity
Answer:
YES(?,O(n^2))
The following argument positions are usable:
Uargs(D) = {}, Uargs(+) = {1, 2}, Uargs(*) = {2},
Uargs(-) = {1, 2}, Uargs(minus) = {1}, Uargs(div) = {1},
Uargs(pow) = {}, Uargs(ln) = {}
TcT has computed following constructor-restricted polynomial
interpretation.
[D](x1) = 1 + 3*x1^2
[t]() = 0
[1]() = 0
[constant]() = 0
[0]() = 0
[+](x1, x2) = 2 + x1 + x2
[*](x1, x2) = 2 + x1 + x2
[-](x1, x2) = 2 + x1 + x2
[minus](x1) = 3 + x1
[div](x1, x2) = 3 + x1 + x2
[pow](x1, x2) = 3 + x1 + x2
[2]() = 2
[ln](x1) = 2 + x1
This order satisfies following ordering constraints
[D(t())] = 1
>
= [1()]
[D(constant())] = 1
>
= [0()]
[D(+(x, y))] = 13 + 12*x + 12*y + 3*x^2 + 3*x*y + 3*y*x + 3*y^2
> 4 + 3*x^2 + 3*y^2
= [+(D(x), D(y))]
[D(*(x, y))] = 13 + 12*x + 12*y + 3*x^2 + 3*x*y + 3*y*x + 3*y^2
> 8 + y + 3*x^2 + x + 3*y^2
= [+(*(y, D(x)), *(x, D(y)))]
[D(-(x, y))] = 13 + 12*x + 12*y + 3*x^2 + 3*x*y + 3*y*x + 3*y^2
> 4 + 3*x^2 + 3*y^2
= [-(D(x), D(y))]
[D(minus(x))] = 28 + 18*x + 3*x^2
> 4 + 3*x^2
= [minus(D(x))]
[D(div(x, y))] = 28 + 18*x + 18*y + 3*x^2 + 3*x*y + 3*y*x + 3*y^2
> 17 + 3*x^2 + 2*y + x + 3*y^2
= [-(div(D(x), y), div(*(x, D(y)), pow(y, 2())))]
[D(pow(x, y))] = 28 + 18*x + 18*y + 3*x^2 + 3*x*y + 3*y*x + 3*y^2
> 22 + 3*y + 3*x + 3*x^2 + 3*y^2
= [+(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y)))]
[D(ln(x))] = 13 + 12*x + 3*x^2
> 4 + 3*x^2 + x
= [div(D(x), x)]
Hurray, we answered YES(?,O(n^2))
lmpo
| Execution Time (secs) | - |
| Answer | YES(?,ELEMENTARY) |
| Input | Der95 11 |
YES(?,ELEMENTARY)
We are left with following problem, upon which TcT provides the
certificate YES(?,ELEMENTARY).
Strict Trs:
{ D(t()) -> 1()
, D(constant()) -> 0()
, D(+(x, y)) -> +(D(x), D(y))
, D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
, D(-(x, y)) -> -(D(x), D(y))
, D(minus(x)) -> minus(D(x))
, D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2())))
, D(ln(x)) -> div(D(x), x)
, D(pow(x, y)) ->
+(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y))) }
Obligation:
innermost runtime complexity
Answer:
YES(?,ELEMENTARY)
The input was oriented with the instance of 'Lightweight Multiset
Path Order' as induced by the safe mapping
safe(D) = {}, safe(t) = {}, safe(1) = {}, safe(constant) = {},
safe(0) = {}, safe(+) = {1, 2}, safe(*) = {1, 2}, safe(-) = {1, 2},
safe(minus) = {1}, safe(div) = {1, 2}, safe(pow) = {1, 2},
safe(2) = {}, safe(ln) = {1}
and precedence
empty .
Following symbols are considered recursive:
{D}
The recursion depth is 1.
For your convenience, here are the oriented rules in predicative
notation, possibly applying argument filtering:
Strict DPs:
Weak DPs :
Strict Trs:
{ D(t();) -> 1()
, D(constant();) -> 0()
, D(+(; x, y);) -> +(; D(x;), D(y;))
, D(*(; x, y);) -> +(; *(; y, D(x;)), *(; x, D(y;)))
, D(-(; x, y);) -> -(; D(x;), D(y;))
, D(minus(; x);) -> minus(; D(x;))
, D(div(; x, y);) ->
-(; div(; D(x;), y), div(; *(; x, D(y;)), pow(; y, 2())))
, D(ln(; x);) -> div(; D(x;), x)
, D(pow(; x, y);) ->
+(; *(; *(; y, pow(; x, -(; y, 1()))), D(x;)),
*(; *(; pow(; x, y), ln(; x)), D(y;))) }
Weak Trs :
Hurray, we answered YES(?,ELEMENTARY)
mpo
| Execution Time (secs) | - |
| Answer | YES(?,PRIMREC) |
| Input | Der95 11 |
YES(?,PRIMREC)
We are left with following problem, upon which TcT provides the
certificate YES(?,PRIMREC).
Strict Trs:
{ D(t()) -> 1()
, D(constant()) -> 0()
, D(+(x, y)) -> +(D(x), D(y))
, D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
, D(-(x, y)) -> -(D(x), D(y))
, D(minus(x)) -> minus(D(x))
, D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2())))
, D(ln(x)) -> div(D(x), x)
, D(pow(x, y)) ->
+(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y))) }
Obligation:
innermost runtime complexity
Answer:
YES(?,PRIMREC)
The input was oriented with the instance of'multiset path orders'
as induced by the precedence
D > 1, D > +, D > *, D > -, D > minus, D > div, D > pow, D > 2,
D > ln, constant > 0 .
Hurray, we answered YES(?,PRIMREC)
popstar
| Execution Time (secs) | 0.297 |
| Answer | MAYBE |
| Input | Der95 11 |
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ D(t()) -> 1()
, D(constant()) -> 0()
, D(+(x, y)) -> +(D(x), D(y))
, D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
, D(-(x, y)) -> -(D(x), D(y))
, D(minus(x)) -> minus(D(x))
, D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2())))
, D(ln(x)) -> div(D(x), x)
, D(pow(x, y)) ->
+(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar-ps
| Execution Time (secs) | 0.384 |
| Answer | MAYBE |
| Input | Der95 11 |
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ D(t()) -> 1()
, D(constant()) -> 0()
, D(+(x, y)) -> +(D(x), D(y))
, D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
, D(-(x, y)) -> -(D(x), D(y))
, D(minus(x)) -> minus(D(x))
, D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2())))
, D(ln(x)) -> div(D(x), x)
, D(pow(x, y)) ->
+(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..