Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, Y)) -> X
tail1(cons2(X, Y)) -> activate1(Y)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
filter2(s1(s1(X)), cons2(Y, Z)) -> if3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, n__sieve1(activate1(Y))))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
sieve1(X) -> n__sieve1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__filter2(X1, X2)) -> filter2(activate1(X1), activate1(X2))
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__sieve1(X)) -> sieve1(activate1(X))
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, Y)) -> X
tail1(cons2(X, Y)) -> activate1(Y)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
filter2(s1(s1(X)), cons2(Y, Z)) -> if3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, n__sieve1(activate1(Y))))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
sieve1(X) -> n__sieve1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__filter2(X1, X2)) -> filter2(activate1(X1), activate1(X2))
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__sieve1(X)) -> sieve1(activate1(X))
activate1(X) -> X

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

PRIMES -> S1(0)
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
PRIMES -> S1(s1(0))
ACTIVATE1(n__filter2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__filter2(X1, X2)) -> FILTER2(activate1(X1), activate1(X2))
IF3(true, X, Y) -> ACTIVATE1(X)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> IF3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
PRIMES -> SIEVE1(from1(s1(s1(0))))
ACTIVATE1(n__filter2(X1, X2)) -> ACTIVATE1(X1)
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__sieve1(X)) -> ACTIVATE1(X)
PRIMES -> FROM1(s1(s1(0)))
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)
SIEVE1(cons2(X, Y)) -> CONS2(X, n__filter2(X, n__sieve1(activate1(Y))))
IF3(false, X, Y) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__sieve1(X)) -> SIEVE1(activate1(X))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(X1), X2)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
TAIL1(cons2(X, Y)) -> ACTIVATE1(Y)

The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, Y)) -> X
tail1(cons2(X, Y)) -> activate1(Y)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
filter2(s1(s1(X)), cons2(Y, Z)) -> if3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, n__sieve1(activate1(Y))))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
sieve1(X) -> n__sieve1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__filter2(X1, X2)) -> filter2(activate1(X1), activate1(X2))
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__sieve1(X)) -> sieve1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

PRIMES -> S1(0)
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
PRIMES -> S1(s1(0))
ACTIVATE1(n__filter2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__filter2(X1, X2)) -> FILTER2(activate1(X1), activate1(X2))
IF3(true, X, Y) -> ACTIVATE1(X)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> IF3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
PRIMES -> SIEVE1(from1(s1(s1(0))))
ACTIVATE1(n__filter2(X1, X2)) -> ACTIVATE1(X1)
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__sieve1(X)) -> ACTIVATE1(X)
PRIMES -> FROM1(s1(s1(0)))
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)
SIEVE1(cons2(X, Y)) -> CONS2(X, n__filter2(X, n__sieve1(activate1(Y))))
IF3(false, X, Y) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__sieve1(X)) -> SIEVE1(activate1(X))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(X1), X2)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
TAIL1(cons2(X, Y)) -> ACTIVATE1(Y)

The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, Y)) -> X
tail1(cons2(X, Y)) -> activate1(Y)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
filter2(s1(s1(X)), cons2(Y, Z)) -> if3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, n__sieve1(activate1(Y))))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
sieve1(X) -> n__sieve1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__filter2(X1, X2)) -> filter2(activate1(X1), activate1(X2))
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__sieve1(X)) -> sieve1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 13 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__filter2(X1, X2)) -> ACTIVATE1(X2)
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__filter2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__filter2(X1, X2)) -> FILTER2(activate1(X1), activate1(X2))
ACTIVATE1(n__sieve1(X)) -> ACTIVATE1(X)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__sieve1(X)) -> SIEVE1(activate1(X))
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, Y)) -> X
tail1(cons2(X, Y)) -> activate1(Y)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
filter2(s1(s1(X)), cons2(Y, Z)) -> if3(divides2(s1(s1(X)), Y), n__filter2(n__s1(n__s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, n__sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, n__sieve1(activate1(Y))))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
sieve1(X) -> n__sieve1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__filter2(X1, X2)) -> filter2(activate1(X1), activate1(X2))
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__sieve1(X)) -> sieve1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.