MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__primes() -> a__sieve(a__from(s(s(0()))))
, a__primes() -> primes()
, a__sieve(X) -> sieve(X)
, a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, mark(s(X)) -> s(mark(X))
, mark(0()) -> 0()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
, mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
, mark(sieve(X)) -> a__sieve(mark(X))
, mark(primes()) -> a__primes()
, mark(head(X)) -> a__head(mark(X))
, mark(tail(X)) -> a__tail(mark(X))
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__head(X) -> head(X)
, a__head(cons(X, Y)) -> mark(X)
, a__tail(X) -> tail(X)
, a__tail(cons(X, Y)) -> mark(Y)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__filter(X1, X2) -> filter(X1, X2)
, a__filter(s(s(X)), cons(Y, Z)) ->
a__if(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))) }
Obligation:
runtime complexity
Answer:
MAYBE
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 60 seconds)' failed due to the
following reason:
Computation stopped due to timeout after 60.0 seconds.
2) 'Best' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)'
failed due to the following reason:
Computation stopped due to timeout after 30.0 seconds.
2) 'Best' failed due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due
to the following reason:
The processor is inapplicable, reason:
Processor only applicable for innermost runtime complexity analysis
2) 'bsearch-popstar (timeout of 60 seconds)' failed due to the
following reason:
The processor is inapplicable, reason:
Processor only applicable for innermost runtime complexity analysis
3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed
due to the following reason:
None of the processors succeeded.
Details of failed attempt(s):
-----------------------------
1) 'Bounds with perSymbol-enrichment and initial automaton 'match''
failed due to the following reason:
match-boundness of the problem could not be verified.
2) 'Bounds with minimal-enrichment and initial automaton 'match''
failed due to the following reason:
match-boundness of the problem could not be verified.
3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed
due to the following reason:
We add the following weak dependency pairs:
Strict DPs:
{ a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0())))))
, a__primes^#() -> c_2()
, a__sieve^#(X) -> c_3(X)
, a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y)
, mark^#(s(X)) -> c_7(mark^#(X))
, mark^#(0()) -> c_8()
, mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2)
, mark^#(from(X)) -> c_10(a__from^#(mark(X)))
, mark^#(true()) -> c_11()
, mark^#(false()) -> c_12()
, mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
, mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2)))
, mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)))
, mark^#(primes()) -> c_16(a__primes^#())
, mark^#(head(X)) -> c_17(a__head^#(mark(X)))
, mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
, mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3))
, a__from^#(X) -> c_5(mark^#(X), X)
, a__from^#(X) -> c_6(X)
, a__filter^#(X1, X2) -> c_27(X1, X2)
, a__filter^#(s(s(X)), cons(Y, Z)) ->
c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))))
, a__head^#(X) -> c_20(X)
, a__head^#(cons(X, Y)) -> c_21(mark^#(X))
, a__tail^#(X) -> c_22(X)
, a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
, a__if^#(X1, X2, X3) -> c_24(X1, X2, X3)
, a__if^#(true(), X, Y) -> c_25(mark^#(X))
, a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0())))))
, a__primes^#() -> c_2()
, a__sieve^#(X) -> c_3(X)
, a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y)
, mark^#(s(X)) -> c_7(mark^#(X))
, mark^#(0()) -> c_8()
, mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2)
, mark^#(from(X)) -> c_10(a__from^#(mark(X)))
, mark^#(true()) -> c_11()
, mark^#(false()) -> c_12()
, mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
, mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2)))
, mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)))
, mark^#(primes()) -> c_16(a__primes^#())
, mark^#(head(X)) -> c_17(a__head^#(mark(X)))
, mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
, mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3))
, a__from^#(X) -> c_5(mark^#(X), X)
, a__from^#(X) -> c_6(X)
, a__filter^#(X1, X2) -> c_27(X1, X2)
, a__filter^#(s(s(X)), cons(Y, Z)) ->
c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))))
, a__head^#(X) -> c_20(X)
, a__head^#(cons(X, Y)) -> c_21(mark^#(X))
, a__tail^#(X) -> c_22(X)
, a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
, a__if^#(X1, X2, X3) -> c_24(X1, X2, X3)
, a__if^#(true(), X, Y) -> c_25(mark^#(X))
, a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
Strict Trs:
{ a__primes() -> a__sieve(a__from(s(s(0()))))
, a__primes() -> primes()
, a__sieve(X) -> sieve(X)
, a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, mark(s(X)) -> s(mark(X))
, mark(0()) -> 0()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
, mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
, mark(sieve(X)) -> a__sieve(mark(X))
, mark(primes()) -> a__primes()
, mark(head(X)) -> a__head(mark(X))
, mark(tail(X)) -> a__tail(mark(X))
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__head(X) -> head(X)
, a__head(cons(X, Y)) -> mark(X)
, a__tail(X) -> tail(X)
, a__tail(cons(X, Y)) -> mark(Y)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__filter(X1, X2) -> filter(X1, X2)
, a__filter(s(s(X)), cons(Y, Z)) ->
a__if(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))) }
Obligation:
runtime complexity
Answer:
MAYBE
We estimate the number of application of {2,6,9,10} by applications
of Pre({2,6,9,10}) = {3,4,5,7,11,14,18,19,20,22,23,24,25,26,27,28}.
Here rules are labeled as follows:
DPs:
{ 1: a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0())))))
, 2: a__primes^#() -> c_2()
, 3: a__sieve^#(X) -> c_3(X)
, 4: a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y)
, 5: mark^#(s(X)) -> c_7(mark^#(X))
, 6: mark^#(0()) -> c_8()
, 7: mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2)
, 8: mark^#(from(X)) -> c_10(a__from^#(mark(X)))
, 9: mark^#(true()) -> c_11()
, 10: mark^#(false()) -> c_12()
, 11: mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
, 12: mark^#(filter(X1, X2)) ->
c_14(a__filter^#(mark(X1), mark(X2)))
, 13: mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)))
, 14: mark^#(primes()) -> c_16(a__primes^#())
, 15: mark^#(head(X)) -> c_17(a__head^#(mark(X)))
, 16: mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
, 17: mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3))
, 18: a__from^#(X) -> c_5(mark^#(X), X)
, 19: a__from^#(X) -> c_6(X)
, 20: a__filter^#(X1, X2) -> c_27(X1, X2)
, 21: a__filter^#(s(s(X)), cons(Y, Z)) ->
c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))))
, 22: a__head^#(X) -> c_20(X)
, 23: a__head^#(cons(X, Y)) -> c_21(mark^#(X))
, 24: a__tail^#(X) -> c_22(X)
, 25: a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
, 26: a__if^#(X1, X2, X3) -> c_24(X1, X2, X3)
, 27: a__if^#(true(), X, Y) -> c_25(mark^#(X))
, 28: a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict DPs:
{ a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0())))))
, a__sieve^#(X) -> c_3(X)
, a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y)
, mark^#(s(X)) -> c_7(mark^#(X))
, mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2)
, mark^#(from(X)) -> c_10(a__from^#(mark(X)))
, mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
, mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2)))
, mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)))
, mark^#(primes()) -> c_16(a__primes^#())
, mark^#(head(X)) -> c_17(a__head^#(mark(X)))
, mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
, mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3))
, a__from^#(X) -> c_5(mark^#(X), X)
, a__from^#(X) -> c_6(X)
, a__filter^#(X1, X2) -> c_27(X1, X2)
, a__filter^#(s(s(X)), cons(Y, Z)) ->
c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))))
, a__head^#(X) -> c_20(X)
, a__head^#(cons(X, Y)) -> c_21(mark^#(X))
, a__tail^#(X) -> c_22(X)
, a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
, a__if^#(X1, X2, X3) -> c_24(X1, X2, X3)
, a__if^#(true(), X, Y) -> c_25(mark^#(X))
, a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
Strict Trs:
{ a__primes() -> a__sieve(a__from(s(s(0()))))
, a__primes() -> primes()
, a__sieve(X) -> sieve(X)
, a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X)
, mark(s(X)) -> s(mark(X))
, mark(0()) -> 0()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
, mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
, mark(sieve(X)) -> a__sieve(mark(X))
, mark(primes()) -> a__primes()
, mark(head(X)) -> a__head(mark(X))
, mark(tail(X)) -> a__tail(mark(X))
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__head(X) -> head(X)
, a__head(cons(X, Y)) -> mark(X)
, a__tail(X) -> tail(X)
, a__tail(cons(X, Y)) -> mark(Y)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__filter(X1, X2) -> filter(X1, X2)
, a__filter(s(s(X)), cons(Y, Z)) ->
a__if(divides(s(s(mark(X))), mark(Y)),
filter(s(s(X)), Z),
cons(Y, filter(X, sieve(Y)))) }
Weak DPs:
{ a__primes^#() -> c_2()
, mark^#(0()) -> c_8()
, mark^#(true()) -> c_11()
, mark^#(false()) -> c_12() }
Obligation:
runtime complexity
Answer:
MAYBE
Empty strict component of the problem is NOT empty.
Arrrr..