Termination w.r.t. Q proof of TRS/TRCSREx7_BLR02

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

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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:

ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
2ND₁(cons₂(X, XS)) -> HEAD₁(activate₁(XS))
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(activate₁(X1), activate₁(X2))
2ND₁(cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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:

ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
2ND₁(cons₂(X, XS)) -> HEAD₁(activate₁(XS))
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(activate₁(X1), activate₁(X2))
2ND₁(cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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 2 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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__from₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least be oriented weakly.

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = x₁ + 1

POL( ACTIVATE₁(x₁) ) = x₁

POL( n__from₁(x₁) ) = x₁ + 1

POL( TAKE₂(x₁, x₂) ) = x₂

POL( n__s₁(x₁) ) = x₁

POL( s₁(x₁) ) = x₁

POL( 0 ) = 1

POL( nil ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = x₂

POL( take₂(x₁, x₂) ) = x₁ + x₂

POL( activate₁(x₁) ) = x₁

POL( n__take₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented:

take₂(X1, X2) -> n__take₂(X1, X2)
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X
take₂(0, XS) -> nil
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(X2))
s₁(X) -> n__s₁(X)
activate₁(n__from₁(X)) -> from₁(activate₁(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X1)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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__take₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__take₂(X1, X2)) -> ACTIVATE₁(X1)

The remaining pairs can at least be oriented weakly.

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = max{0, -1}

POL( ACTIVATE₁(x₁) ) = x₁ + 1

POL( n__from₁(x₁) ) = max{0, -1}

POL( TAKE₂(x₁, x₂) ) = x₂ + 1

POL( n__s₁(x₁) ) = x₁ + 1

POL( s₁(x₁) ) = x₁ + 1

POL( 0 ) = 1

POL( nil ) = 1

POL( cons₂(x₁, x₂) ) = x₂

POL( take₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( activate₁(x₁) ) = x₁

POL( n__take₂(x₁, x₂) ) = x₁ + x₂ + 1

The following usable rules [14] were oriented:

take₂(X1, X2) -> n__take₂(X1, X2)
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X
take₂(0, XS) -> nil
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(X2))
s₁(X) -> n__s₁(X)
activate₁(n__from₁(X)) -> from₁(activate₁(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

          ↳ QDP

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

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(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.

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = max{0, x₁ - 1}

POL( SEL₂(x₁, x₂) ) = x₁

POL( n__from₁(x₁) ) = x₁ + 1

POL( n__s₁(x₁) ) = 1

POL( s₁(x₁) ) = x₁ + 1

POL( 0 ) = max{0, -1}

POL( nil ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( take₂(x₁, x₂) ) = max{0, x₂ - 1}

POL( activate₁(x₁) ) = 0

POL( n__take₂(x₁, x₂) ) = max{0, x₁ - 1}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__take₂(X1, X2)) -> take₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.