Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM99

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] 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.