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:

a__primes -> a__sieve1(a__from1(s1(s1(0))))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__head1(cons2(X, Y)) -> mark1(X)
a__tail1(cons2(X, Y)) -> mark1(Y)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
a__filter2(s1(s1(X)), cons2(Y, Z)) -> a__if3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
a__sieve1(cons2(X, Y)) -> cons2(mark1(X), filter2(X, sieve1(Y)))
mark1(primes) -> a__primes
mark1(sieve1(X)) -> a__sieve1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(filter2(X1, X2)) -> a__filter2(mark1(X1), mark1(X2))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(true) -> true
mark1(false) -> false
mark1(divides2(X1, X2)) -> divides2(mark1(X1), mark1(X2))
a__primes -> primes
a__sieve1(X) -> sieve1(X)
a__from1(X) -> from1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)
a__filter2(X1, X2) -> filter2(X1, X2)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__primes -> a__sieve1(a__from1(s1(s1(0))))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__head1(cons2(X, Y)) -> mark1(X)
a__tail1(cons2(X, Y)) -> mark1(Y)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
a__filter2(s1(s1(X)), cons2(Y, Z)) -> a__if3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
a__sieve1(cons2(X, Y)) -> cons2(mark1(X), filter2(X, sieve1(Y)))
mark1(primes) -> a__primes
mark1(sieve1(X)) -> a__sieve1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(filter2(X1, X2)) -> a__filter2(mark1(X1), mark1(X2))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(true) -> true
mark1(false) -> false
mark1(divides2(X1, X2)) -> divides2(mark1(X1), mark1(X2))
a__primes -> primes
a__sieve1(X) -> sieve1(X)
a__from1(X) -> from1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)
a__filter2(X1, X2) -> filter2(X1, X2)

Q is empty.

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

A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> A__IF3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
A__TAIL1(cons2(X, Y)) -> MARK1(Y)
MARK1(divides2(X1, X2)) -> MARK1(X1)
A__SIEVE1(cons2(X, Y)) -> MARK1(X)
A__IF3(false, X, Y) -> MARK1(Y)
A__PRIMES -> A__FROM1(s1(s1(0)))
MARK1(divides2(X1, X2)) -> MARK1(X2)
A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> MARK1(Y)
MARK1(head1(X)) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(filter2(X1, X2)) -> A__FILTER2(mark1(X1), mark1(X2))
MARK1(filter2(X1, X2)) -> MARK1(X1)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__PRIMES -> A__SIEVE1(a__from1(s1(s1(0))))
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(sieve1(X)) -> MARK1(X)
MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(filter2(X1, X2)) -> MARK1(X2)
MARK1(head1(X)) -> A__HEAD1(mark1(X))
MARK1(s1(X)) -> MARK1(X)
MARK1(primes) -> A__PRIMES
MARK1(sieve1(X)) -> A__SIEVE1(mark1(X))
MARK1(tail1(X)) -> MARK1(X)
A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> MARK1(X)
A__HEAD1(cons2(X, Y)) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__primes -> a__sieve1(a__from1(s1(s1(0))))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__head1(cons2(X, Y)) -> mark1(X)
a__tail1(cons2(X, Y)) -> mark1(Y)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
a__filter2(s1(s1(X)), cons2(Y, Z)) -> a__if3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
a__sieve1(cons2(X, Y)) -> cons2(mark1(X), filter2(X, sieve1(Y)))
mark1(primes) -> a__primes
mark1(sieve1(X)) -> a__sieve1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(filter2(X1, X2)) -> a__filter2(mark1(X1), mark1(X2))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(true) -> true
mark1(false) -> false
mark1(divides2(X1, X2)) -> divides2(mark1(X1), mark1(X2))
a__primes -> primes
a__sieve1(X) -> sieve1(X)
a__from1(X) -> from1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)
a__filter2(X1, X2) -> filter2(X1, X2)

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:

A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> A__IF3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
A__TAIL1(cons2(X, Y)) -> MARK1(Y)
MARK1(divides2(X1, X2)) -> MARK1(X1)
A__SIEVE1(cons2(X, Y)) -> MARK1(X)
A__IF3(false, X, Y) -> MARK1(Y)
A__PRIMES -> A__FROM1(s1(s1(0)))
MARK1(divides2(X1, X2)) -> MARK1(X2)
A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> MARK1(Y)
MARK1(head1(X)) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(filter2(X1, X2)) -> A__FILTER2(mark1(X1), mark1(X2))
MARK1(filter2(X1, X2)) -> MARK1(X1)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__PRIMES -> A__SIEVE1(a__from1(s1(s1(0))))
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(sieve1(X)) -> MARK1(X)
MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(filter2(X1, X2)) -> MARK1(X2)
MARK1(head1(X)) -> A__HEAD1(mark1(X))
MARK1(s1(X)) -> MARK1(X)
MARK1(primes) -> A__PRIMES
MARK1(sieve1(X)) -> A__SIEVE1(mark1(X))
MARK1(tail1(X)) -> MARK1(X)
A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> MARK1(X)
A__HEAD1(cons2(X, Y)) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__primes -> a__sieve1(a__from1(s1(s1(0))))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__head1(cons2(X, Y)) -> mark1(X)
a__tail1(cons2(X, Y)) -> mark1(Y)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
a__filter2(s1(s1(X)), cons2(Y, Z)) -> a__if3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
a__sieve1(cons2(X, Y)) -> cons2(mark1(X), filter2(X, sieve1(Y)))
mark1(primes) -> a__primes
mark1(sieve1(X)) -> a__sieve1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(filter2(X1, X2)) -> a__filter2(mark1(X1), mark1(X2))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(true) -> true
mark1(false) -> false
mark1(divides2(X1, X2)) -> divides2(mark1(X1), mark1(X2))
a__primes -> primes
a__sieve1(X) -> sieve1(X)
a__from1(X) -> from1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)
a__filter2(X1, X2) -> filter2(X1, X2)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

A__TAIL1(cons2(X, Y)) -> MARK1(Y)
A__SIEVE1(cons2(X, Y)) -> MARK1(X)
MARK1(divides2(X1, X2)) -> MARK1(X1)
A__IF3(false, X, Y) -> MARK1(Y)
A__PRIMES -> A__FROM1(s1(s1(0)))
MARK1(divides2(X1, X2)) -> MARK1(X2)
A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> MARK1(Y)
MARK1(head1(X)) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(filter2(X1, X2)) -> A__FILTER2(mark1(X1), mark1(X2))
MARK1(filter2(X1, X2)) -> MARK1(X1)
MARK1(from1(X)) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__PRIMES -> A__SIEVE1(a__from1(s1(s1(0))))
MARK1(sieve1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(filter2(X1, X2)) -> MARK1(X2)
MARK1(head1(X)) -> A__HEAD1(mark1(X))
MARK1(primes) -> A__PRIMES
MARK1(s1(X)) -> MARK1(X)
MARK1(sieve1(X)) -> A__SIEVE1(mark1(X))
MARK1(tail1(X)) -> MARK1(X)
A__FILTER2(s1(s1(X)), cons2(Y, Z)) -> MARK1(X)
A__HEAD1(cons2(X, Y)) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__primes -> a__sieve1(a__from1(s1(s1(0))))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__head1(cons2(X, Y)) -> mark1(X)
a__tail1(cons2(X, Y)) -> mark1(Y)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
a__filter2(s1(s1(X)), cons2(Y, Z)) -> a__if3(divides2(s1(s1(mark1(X))), mark1(Y)), filter2(s1(s1(X)), Z), cons2(Y, filter2(X, sieve1(Y))))
a__sieve1(cons2(X, Y)) -> cons2(mark1(X), filter2(X, sieve1(Y)))
mark1(primes) -> a__primes
mark1(sieve1(X)) -> a__sieve1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(filter2(X1, X2)) -> a__filter2(mark1(X1), mark1(X2))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(true) -> true
mark1(false) -> false
mark1(divides2(X1, X2)) -> divides2(mark1(X1), mark1(X2))
a__primes -> primes
a__sieve1(X) -> sieve1(X)
a__from1(X) -> from1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)
a__filter2(X1, X2) -> filter2(X1, X2)

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