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(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
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(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FILTER2(s1(s1(X)), cons2(Y, Z)) -> IF3(divides2(s1(s1(X)), Y), n__filter2(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
SIEVE1(cons2(X, Y)) -> CONS2(X, n__filter2(X, sieve1(activate1(Y))))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> FROM1(X)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> SIEVE1(Y)
SIEVE1(cons2(X, Y)) -> SIEVE1(activate1(Y))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
ACTIVATE1(n__filter2(X1, X2)) -> FILTER2(X1, X2)
PRIMES -> SIEVE1(from1(s1(s1(0))))
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
PRIMES -> FROM1(s1(s1(0)))
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)
IF3(false, X, Y) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(s1(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(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
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:

FILTER2(s1(s1(X)), cons2(Y, Z)) -> IF3(divides2(s1(s1(X)), Y), n__filter2(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
SIEVE1(cons2(X, Y)) -> CONS2(X, n__filter2(X, sieve1(activate1(Y))))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> FROM1(X)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> SIEVE1(Y)
SIEVE1(cons2(X, Y)) -> SIEVE1(activate1(Y))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
ACTIVATE1(n__filter2(X1, X2)) -> FILTER2(X1, X2)
PRIMES -> SIEVE1(from1(s1(s1(0))))
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
PRIMES -> FROM1(s1(s1(0)))
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)
IF3(false, X, Y) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(s1(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(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__filter2(X1, X2)) -> FILTER2(X1, X2)
The remaining pairs can at least be oriented weakly.

FILTER2(s1(s1(X)), cons2(Y, Z)) -> SIEVE1(Y)
SIEVE1(cons2(X, Y)) -> SIEVE1(activate1(Y))
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = 2·x1   
POL(FILTER2(x1, x2)) = 2·x2   
POL(SIEVE1(x1)) = 2·x1   
POL(activate1(x1)) = x1   
POL(cons2(x1, x2)) = 2·x1 + x2   
POL(divides2(x1, x2)) = x2   
POL(filter2(x1, x2)) = 2 + 2·x2   
POL(from1(x1)) = 3·x1 + 3·x12   
POL(if3(x1, x2, x3)) = 1 + 3·x1   
POL(n__cons2(x1, x2)) = 2·x1 + x2   
POL(n__filter2(x1, x2)) = 2 + 2·x2   
POL(n__from1(x1)) = 3·x1 + 3·x12   
POL(s1(x1)) = 0   
POL(sieve1(x1)) = 0   

The following usable rules [14] were oriented:

cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(X) -> X
from1(X) -> n__from1(X)
filter2(s1(s1(X)), cons2(Y, Z)) -> if3(divides2(s1(s1(X)), Y), n__filter2(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
activate1(n__from1(X)) -> from1(X)
from1(X) -> cons2(X, n__from1(s1(X)))
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
filter2(X1, X2) -> n__filter2(X1, X2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ DependencyGraphProof

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

FILTER2(s1(s1(X)), cons2(Y, Z)) -> SIEVE1(Y)
SIEVE1(cons2(X, Y)) -> SIEVE1(activate1(Y))
SIEVE1(cons2(X, Y)) -> ACTIVATE1(Y)
FILTER2(s1(s1(X)), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
QDP

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

SIEVE1(cons2(X, Y)) -> SIEVE1(activate1(Y))

The TRS R consists of the following rules:

primes -> sieve1(from1(s1(s1(0))))
from1(X) -> cons2(X, n__from1(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(s1(s1(X)), activate1(Z)), n__cons2(Y, n__filter2(X, sieve1(Y))))
sieve1(cons2(X, Y)) -> cons2(X, n__filter2(X, sieve1(activate1(Y))))
from1(X) -> n__from1(X)
filter2(X1, X2) -> n__filter2(X1, X2)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__filter2(X1, X2)) -> filter2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

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