Termination w.r.t. Q proof of /tmp/tmpkJ6qAc/ExIntrod_GM99

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SIEVE(cons(X, Y)) → ACTIVATE(Y)
FROM(X) → CONS(X, n__from(s(X)))
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
SIEVE(cons(X, Y)) → CONS(X, n__filter(X, sieve(activate(Y))))
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
IF(true, X, Y) → ACTIVATE(X)
PRIMES → FROM(s(s(0)))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
IF(false, X, Y) → ACTIVATE(Y)
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
PRIMES → SIEVE(from(s(s(0))))
TAIL(cons(X, Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(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:

SIEVE(cons(X, Y)) → ACTIVATE(Y)
FROM(X) → CONS(X, n__from(s(X)))
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
SIEVE(cons(X, Y)) → CONS(X, n__filter(X, sieve(activate(Y))))
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
IF(true, X, Y) → ACTIVATE(X)
PRIMES → FROM(s(s(0)))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
IF(false, X, Y) → ACTIVATE(Y)
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
PRIMES → SIEVE(from(s(s(0))))
TAIL(cons(X, Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

SIEVE(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SIEVE(cons(X, Y)) → ACTIVATE(Y)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( cons(x₁, x₂) ) =

/	1	\
\	0	/

/	1	0	\
\	1	0	/

x₁

/	1	0	\
\	1	0	/

x₂

M( from(x₁) ) =

/	1	\
\	1	/

/	1	0	\
\	1	0	/

x₁

M( if(x₁, ..., x₃) ) =

/	0	\
\	0	/

/	0	0	\
\	1	0	/

x₁

/	0	0	\
\	0	0	/

x₂

/	0	0	\
\	0	1	/

x₃

M( n__from(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( filter(x₁, x₂) ) =

/	1	\
\	1	/

/	0	0	\
\	0	0	/

x₁

/	0	1	\
\	1	1	/

x₂

M( n__cons(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	1	0	/

x₁

/	1	0	\
\	0	0	/

x₂

M( s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( activate(x₁) ) =

/	1	\
\	1	/

/	1	0	\
\	1	1	/

x₁

M( divides(x₁, x₂) ) =

/	1	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( sieve(x₁) ) =

/	1	\
\	0	/

/	1	0	\
\	1	0	/

x₁

M( n__filter(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	1	\
\	1	0	/

x₂

Tuple symbols:

M( SIEVE(x₁) ) =

[

]

x₁

M( ACTIVATE(x₁) ) =

[

]

x₁

M( FILTER(x₁, x₂) ) =

[

]

x₁

[

]

x₂

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
cons(X1, X2) → n__cons(X1, X2)
filter(X1, X2) → n__filter(X1, X2)
from(X) → n__from(X)
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
activate(X) → X
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__from(X)) → from(X)
from(X) → cons(X, n__from(s(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                  ↳ QDP

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

SIEVE(cons(X, Y)) → SIEVE(activate(Y))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule SIEVE(cons(X, Y)) → SIEVE(activate(Y)) at position [0] we obtained the following new rules:

SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0)) at position [0] we obtained the following new rules:

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__from(x0))) → SIEVE(n__from(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__from(x0))) → SIEVE(n__from(x0))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1)) at position [0] we obtained the following new rules:

SIEVE(cons(y0, n__filter(s(s(x0)), cons(x1, x2)))) → SIEVE(if(divides(s(s(x0)), x1), n__filter(s(s(x0)), activate(x2)), n__cons(x1, n__filter(x0, sieve(x1)))))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(n__filter(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__filter(s(s(x0)), cons(x1, x2)))) → SIEVE(if(divides(s(s(x0)), x1), n__filter(s(s(x0)), activate(x2)), n__cons(x1, n__filter(x0, sieve(x1)))))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(n__filter(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SIEVE(cons(y0, x0)) → SIEVE(x0)

The remaining pairs can at least be oriented weakly.

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))

Used ordering: Polynomial interpretation [25]:

POL(SIEVE(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₂
POL(n__cons(x₁, x₂)) = x₂
POL(n__from(x₁)) = x₁
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

cons(X1, X2) → n__cons(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))

The remaining pairs can at least be oriented weakly.

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(SIEVE(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₂
POL(n__cons(x₁, x₂)) = 1 + x₂
POL(n__from(x₁)) = 0
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

cons(X1, X2) → n__cons(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ NonTerminationProof

                  ↳ QDP

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

s = SIEVE(cons(y0, n__from(x0))) evaluates to t =SIEVE(cons(x0, n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [y0 / x0, x0 / s(x0)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from SIEVE(cons(y0, n__from(x0))) to SIEVE(cons(x0, n__from(s(x0)))).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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