YES(O(1),O(n^1))
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ primes() -> sieve(from(s(s(0()))))
, sieve(X) -> n__sieve(X)
, sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y))))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, head(cons(X, Y)) -> X
, tail(cons(X, Y)) -> activate(Y)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, filter(X1, X2) -> n__filter(X1, X2)
, filter(s(s(X)), cons(Y, Z)) ->
if(divides(s(s(X)), Y),
n__filter(n__s(n__s(X)), activate(Z)),
n__cons(Y, n__filter(X, n__sieve(Y)))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y))))
, head(cons(X, Y)) -> X
, tail(cons(X, Y)) -> activate(Y)
, filter(s(s(X)), cons(Y, Z)) ->
if(divides(s(s(X)), Y),
n__filter(n__s(n__s(X)), activate(Z)),
n__cons(Y, n__filter(X, n__sieve(Y)))) }
All above mentioned rules can be savely removed.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ primes() -> sieve(from(s(s(0()))))
, sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following innermost weak dependency pairs:
Strict DPs:
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, sieve^#(X) -> c_2()
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X)))
, filter^#(X1, X2) -> c_15()
, if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, sieve^#(X) -> c_2()
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X)))
, filter^#(X1, X2) -> c_15()
, if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
Strict Trs:
{ primes() -> sieve(from(s(s(0()))))
, sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y)
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, sieve^#(X) -> c_2()
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X)))
, filter^#(X1, X2) -> c_15()
, if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
Strict Trs:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(sieve) = {1}, Uargs(from) = {1}, Uargs(s) = {1},
Uargs(cons) = {1}, Uargs(filter) = {1, 2}, Uargs(c_1) = {1},
Uargs(sieve^#) = {1}, Uargs(from^#) = {1}, Uargs(c_3) = {1},
Uargs(cons^#) = {1}, Uargs(s^#) = {1}, Uargs(c_8) = {1},
Uargs(c_9) = {1}, Uargs(c_10) = {1}, Uargs(filter^#) = {1, 2},
Uargs(c_11) = {1}, Uargs(c_12) = {1}, Uargs(c_13) = {1},
Uargs(c_14) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[sieve](x1) = [1 0] x1 + [2]
[0 1] [2]
[from](x1) = [1 0] x1 + [2]
[0 1] [2]
[s](x1) = [1 0] x1 + [1]
[0 1] [1]
[0] = [0]
[0]
[cons](x1, x2) = [1 0] x1 + [1]
[0 1] [2]
[n__from](x1) = [1 0] x1 + [0]
[0 1] [2]
[n__s](x1) = [1 0] x1 + [0]
[0 1] [1]
[activate](x1) = [1 2] x1 + [1]
[0 2] [0]
[true] = [1]
[2]
[false] = [1]
[1]
[filter](x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 1] [0 1] [2]
[n__filter](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
[n__cons](x1, x2) = [1 0] x1 + [0]
[0 1] [2]
[n__sieve](x1) = [1 0] x1 + [0]
[0 1] [2]
[primes^#] = [1]
[1]
[c_1](x1) = [1 0] x1 + [1]
[0 1] [2]
[sieve^#](x1) = [1 0] x1 + [1]
[0 0] [2]
[c_2] = [1]
[1]
[from^#](x1) = [1 0] x1 + [1]
[0 0] [2]
[c_3](x1) = [1 0] x1 + [1]
[0 1] [2]
[cons^#](x1, x2) = [1 0] x1 + [1]
[0 0] [2]
[c_4] = [1]
[1]
[s^#](x1) = [1 0] x1 + [1]
[0 0] [2]
[c_5] = [1]
[1]
[c_6] = [1]
[1]
[activate^#](x1) = [1 2] x1 + [0]
[0 0] [0]
[c_7] = [1]
[1]
[c_8](x1) = [1 0] x1 + [2]
[0 1] [2]
[c_9](x1) = [1 0] x1 + [2]
[0 1] [2]
[c_10](x1) = [1 0] x1 + [2]
[0 1] [2]
[filter^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 0] [0 0] [2]
[c_11](x1) = [1 0] x1 + [2]
[0 1] [2]
[c_12](x1) = [1 0] x1 + [2]
[0 1] [2]
[if^#](x1, x2, x3) = [1 1] x1 + [1 2] x2 + [1 2] x3 + [1]
[1 1] [2 2] [1 1] [1]
[c_13](x1) = [1 0] x1 + [2]
[0 1] [2]
[c_14](x1) = [1 0] x1 + [2]
[0 1] [2]
[c_15] = [2]
[1]
The following symbols are considered usable
{sieve, from, s, cons, activate, filter, primes^#, sieve^#, from^#,
cons^#, s^#, activate^#, filter^#, if^#}
The order satisfies the following ordering constraints:
[sieve(X)] = [1 0] X + [2]
[0 1] [2]
> [1 0] X + [0]
[0 1] [2]
= [n__sieve(X)]
[from(X)] = [1 0] X + [2]
[0 1] [2]
> [1 0] X + [1]
[0 1] [2]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1 0] X + [2]
[0 1] [2]
> [1 0] X + [0]
[0 1] [2]
= [n__from(X)]
[s(X)] = [1 0] X + [1]
[0 1] [1]
> [1 0] X + [0]
[0 1] [1]
= [n__s(X)]
[cons(X1, X2)] = [1 0] X1 + [1]
[0 1] [2]
> [1 0] X1 + [0]
[0 1] [2]
= [n__cons(X1, X2)]
[activate(X)] = [1 2] X + [1]
[0 2] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__from(X))] = [1 2] X + [5]
[0 2] [4]
> [1 2] X + [3]
[0 2] [2]
= [from(activate(X))]
[activate(n__s(X))] = [1 2] X + [3]
[0 2] [2]
> [1 2] X + [2]
[0 2] [1]
= [s(activate(X))]
[activate(n__filter(X1, X2))] = [1 2] X1 + [1 2] X2 + [5]
[0 2] [0 2] [4]
> [1 2] X1 + [1 2] X2 + [3]
[0 2] [0 2] [2]
= [filter(activate(X1), activate(X2))]
[activate(n__cons(X1, X2))] = [1 2] X1 + [5]
[0 2] [4]
> [1 2] X1 + [2]
[0 2] [2]
= [cons(activate(X1), X2)]
[activate(n__sieve(X))] = [1 2] X + [5]
[0 2] [4]
> [1 2] X + [3]
[0 2] [2]
= [sieve(activate(X))]
[filter(X1, X2)] = [1 0] X1 + [1 0] X2 + [1]
[0 1] [0 1] [2]
> [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [2]
= [n__filter(X1, X2)]
[primes^#()] = [1]
[1]
? [6]
[4]
= [c_1(sieve^#(from(s(s(0())))))]
[sieve^#(X)] = [1 0] X + [1]
[0 0] [2]
>= [1]
[1]
= [c_2()]
[from^#(X)] = [1 0] X + [1]
[0 0] [2]
? [1 0] X + [2]
[0 0] [4]
= [c_3(cons^#(X, n__from(n__s(X))))]
[from^#(X)] = [1 0] X + [1]
[0 0] [2]
>= [1]
[1]
= [c_4()]
[cons^#(X1, X2)] = [1 0] X1 + [1]
[0 0] [2]
>= [1]
[1]
= [c_6()]
[s^#(X)] = [1 0] X + [1]
[0 0] [2]
>= [1]
[1]
= [c_5()]
[activate^#(X)] = [1 2] X + [0]
[0 0] [0]
? [1]
[1]
= [c_7()]
[activate^#(n__from(X))] = [1 2] X + [4]
[0 0] [0]
? [1 2] X + [4]
[0 0] [4]
= [c_8(from^#(activate(X)))]
[activate^#(n__s(X))] = [1 2] X + [2]
[0 0] [0]
? [1 2] X + [4]
[0 0] [4]
= [c_9(s^#(activate(X)))]
[activate^#(n__filter(X1, X2))] = [1 2] X1 + [1 2] X2 + [4]
[0 0] [0 0] [0]
? [1 2] X1 + [1 2] X2 + [6]
[0 0] [0 0] [4]
= [c_10(filter^#(activate(X1), activate(X2)))]
[activate^#(n__cons(X1, X2))] = [1 2] X1 + [4]
[0 0] [0]
? [1 2] X1 + [4]
[0 0] [4]
= [c_11(cons^#(activate(X1), X2))]
[activate^#(n__sieve(X))] = [1 2] X + [4]
[0 0] [0]
? [1 2] X + [4]
[0 0] [4]
= [c_12(sieve^#(activate(X)))]
[filter^#(X1, X2)] = [1 0] X1 + [1 0] X2 + [2]
[0 0] [0 0] [2]
>= [2]
[1]
= [c_15()]
[if^#(true(), X, Y)] = [1 2] X + [1 2] Y + [4]
[2 2] [1 1] [4]
> [1 2] X + [2]
[0 0] [2]
= [c_13(activate^#(X))]
[if^#(false(), X, Y)] = [1 2] X + [1 2] Y + [3]
[2 2] [1 1] [3]
> [1 2] Y + [2]
[0 0] [2]
= [c_14(activate^#(Y))]
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(1)).
Strict DPs:
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, sieve^#(X) -> c_2()
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X)))
, filter^#(X1, X2) -> c_15() }
Weak DPs:
{ if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
Weak Trs:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
We estimate the number of application of {2,4,5,6,13} by
applications of Pre({2,4,5,6,13}) = {1,3,8,9,10,11,12}. Here rules
are labeled as follows:
DPs:
{ 1: primes^#() -> c_1(sieve^#(from(s(s(0())))))
, 2: sieve^#(X) -> c_2()
, 3: from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, 4: from^#(X) -> c_4()
, 5: cons^#(X1, X2) -> c_6()
, 6: s^#(X) -> c_5()
, 7: activate^#(X) -> c_7()
, 8: activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, 9: activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, 10: activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, 11: activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, 12: activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X)))
, 13: filter^#(X1, X2) -> c_15()
, 14: if^#(true(), X, Y) -> c_13(activate^#(X))
, 15: if^#(false(), X, Y) -> c_14(activate^#(Y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) }
Weak DPs:
{ sieve^#(X) -> c_2()
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, filter^#(X1, X2) -> c_15()
, if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
Weak Trs:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
We estimate the number of application of {1,2} by applications of
Pre({1,2}) = {4}. Here rules are labeled as follows:
DPs:
{ 1: primes^#() -> c_1(sieve^#(from(s(s(0())))))
, 2: from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, 3: activate^#(X) -> c_7()
, 4: activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, 5: activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, 6: activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, 7: activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, 8: activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X)))
, 9: sieve^#(X) -> c_2()
, 10: from^#(X) -> c_4()
, 11: cons^#(X1, X2) -> c_6()
, 12: s^#(X) -> c_5()
, 13: filter^#(X1, X2) -> c_15()
, 14: if^#(true(), X, Y) -> c_13(activate^#(X))
, 15: if^#(false(), X, Y) -> c_14(activate^#(Y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) }
Weak DPs:
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, sieve^#(X) -> c_2()
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, filter^#(X1, X2) -> c_15()
, if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
Weak Trs:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ primes^#() -> c_1(sieve^#(from(s(s(0())))))
, sieve^#(X) -> c_2()
, from^#(X) -> c_3(cons^#(X, n__from(n__s(X))))
, from^#(X) -> c_4()
, cons^#(X1, X2) -> c_6()
, s^#(X) -> c_5()
, filter^#(X1, X2) -> c_15() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ activate^#(X) -> c_7()
, activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) }
Weak DPs:
{ if^#(true(), X, Y) -> c_13(activate^#(X))
, if^#(false(), X, Y) -> c_14(activate^#(Y)) }
Weak Trs:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:
{ activate^#(n__from(X)) -> c_8(from^#(activate(X)))
, activate^#(n__s(X)) -> c_9(s^#(activate(X)))
, activate^#(n__filter(X1, X2)) ->
c_10(filter^#(activate(X1), activate(X2)))
, activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2))
, activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ activate^#(X) -> c_1()
, activate^#(n__from(X)) -> c_2()
, activate^#(n__s(X)) -> c_3()
, activate^#(n__filter(X1, X2)) -> c_4()
, activate^#(n__cons(X1, X2)) -> c_5()
, activate^#(n__sieve(X)) -> c_6() }
Weak DPs:
{ if^#(true(), X, Y) -> c_7(activate^#(X))
, if^#(false(), X, Y) -> c_8(activate^#(Y)) }
Weak Trs:
{ sieve(X) -> n__sieve(X)
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, cons(X1, X2) -> n__cons(X1, X2)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
, activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
, activate(n__sieve(X)) -> sieve(activate(X))
, filter(X1, X2) -> n__filter(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
No rule is usable, rules are removed from the input problem.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ activate^#(X) -> c_1()
, activate^#(n__from(X)) -> c_2()
, activate^#(n__s(X)) -> c_3()
, activate^#(n__filter(X1, X2)) -> c_4()
, activate^#(n__cons(X1, X2)) -> c_5()
, activate^#(n__sieve(X)) -> c_6() }
Weak DPs:
{ if^#(true(), X, Y) -> c_7(activate^#(X))
, if^#(false(), X, Y) -> c_8(activate^#(Y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Consider the dependency graph
1: activate^#(X) -> c_1()
2: activate^#(n__from(X)) -> c_2()
3: activate^#(n__s(X)) -> c_3()
4: activate^#(n__filter(X1, X2)) -> c_4()
5: activate^#(n__cons(X1, X2)) -> c_5()
6: activate^#(n__sieve(X)) -> c_6()
7: if^#(true(), X, Y) -> c_7(activate^#(X))
-->_1 activate^#(n__sieve(X)) -> c_6() :6
-->_1 activate^#(n__cons(X1, X2)) -> c_5() :5
-->_1 activate^#(n__filter(X1, X2)) -> c_4() :4
-->_1 activate^#(n__s(X)) -> c_3() :3
-->_1 activate^#(n__from(X)) -> c_2() :2
-->_1 activate^#(X) -> c_1() :1
8: if^#(false(), X, Y) -> c_8(activate^#(Y))
-->_1 activate^#(n__sieve(X)) -> c_6() :6
-->_1 activate^#(n__cons(X1, X2)) -> c_5() :5
-->_1 activate^#(n__filter(X1, X2)) -> c_4() :4
-->_1 activate^#(n__s(X)) -> c_3() :3
-->_1 activate^#(n__from(X)) -> c_2() :2
-->_1 activate^#(X) -> c_1() :1
Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).
{ if^#(true(), X, Y) -> c_7(activate^#(X))
, if^#(false(), X, Y) -> c_8(activate^#(Y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ activate^#(X) -> c_1()
, activate^#(n__from(X)) -> c_2()
, activate^#(n__s(X)) -> c_3()
, activate^#(n__filter(X1, X2)) -> c_4()
, activate^#(n__cons(X1, X2)) -> c_5()
, activate^#(n__sieve(X)) -> c_6() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Consider the dependency graph
1: activate^#(X) -> c_1()
2: activate^#(n__from(X)) -> c_2()
3: activate^#(n__s(X)) -> c_3()
4: activate^#(n__filter(X1, X2)) -> c_4()
5: activate^#(n__cons(X1, X2)) -> c_5()
6: activate^#(n__sieve(X)) -> c_6()
Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).
{ activate^#(X) -> c_1()
, activate^#(n__from(X)) -> c_2()
, activate^#(n__s(X)) -> c_3()
, activate^#(n__filter(X1, X2)) -> c_4()
, activate^#(n__cons(X1, X2)) -> c_5()
, activate^#(n__sieve(X)) -> c_6() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))