Termination w.r.t. Q proof of TRS/TRCSREx4_7_37_Bor03

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)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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:

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(minus₂(X, Y), s₁(Y))
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ZWQUOT₂(activate₁(X1), activate₁(X2))
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> QUOT₂(X, Y)
ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X1)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X2)
QUOT₂(s₁(X), s₁(Y)) -> S₁(quot₂(minus₂(X, Y), s₁(Y)))
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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:

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(minus₂(X, Y), s₁(Y))
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ZWQUOT₂(activate₁(X1), activate₁(X2))
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> QUOT₂(X, Y)
ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X1)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X2)
QUOT₂(s₁(X), s₁(Y)) -> S₁(quot₂(minus₂(X, Y), s₁(Y)))
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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 4 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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.

MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

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

POL( MINUS₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(minus₂(X, Y), s₁(Y))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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.

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(minus₂(X, Y), s₁(Y))

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

POL( n__s₁(x₁) ) = 1

POL( 0 ) = max{0, -3}

POL( s₁(x₁) ) = 3x₁ + 3

POL( minus₂(x₁, x₂) ) = max{0, -2}

POL( QUOT₂(x₁, x₂) ) = 3x₁ + 2

The following usable rules [14] were oriented:

minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X1)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ZWQUOT₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X2)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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__zWquot₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ZWQUOT₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ACTIVATE₁(X2)

The remaining pairs can at least be oriented weakly.

ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

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

POL( from₁(x₁) ) = 2x₁

POL( minus₂(x₁, x₂) ) = 2x₁ + 3

POL( ZWQUOT₂(x₁, x₂) ) = 2x₁ + 3x₂

POL( n__from₁(x₁) ) = 2x₁

POL( quot₂(x₁, x₂) ) = 0

POL( n__s₁(x₁) ) = x₁

POL( 0 ) = max{0, -3}

POL( zWquot₂(x₁, x₂) ) = x₁ + 3x₂ + 2

POL( nil ) = max{0, -3}

POL( cons₂(x₁, x₂) ) = x₂

POL( activate₁(x₁) ) = x₁

POL( ACTIVATE₁(x₁) ) = 2x₁

POL( n__zWquot₂(x₁, x₂) ) = x₁ + 3x₂ + 2

POL( s₁(x₁) ) = x₁

The following usable rules [14] were oriented:

s₁(X) -> n__s₁(X)
zWquot₂(XS, nil) -> nil
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(X) -> X
zWquot₂(nil, XS) -> nil
activate₁(n__from₁(X)) -> from₁(activate₁(X))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(activate₁(X1), activate₁(X2))
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
from₁(X) -> n__from₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

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

POL( n__s₁(x₁) ) = 2x₁ + 2

POL( ACTIVATE₁(x₁) ) = 2x₁

POL( n__from₁(x₁) ) = 3x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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) ) = 1

POL( minus₂(x₁, x₂) ) = max{0, -2}

POL( n__from₁(x₁) ) = 0

POL( quot₂(x₁, x₂) ) = 0

POL( n__s₁(x₁) ) = 2

POL( 0 ) = 1

POL( zWquot₂(x₁, x₂) ) = max{0, 2x₂ - 3}

POL( nil ) = 2

POL( cons₂(x₁, x₂) ) = x₁ + 2

POL( activate₁(x₁) ) = max{0, -3}

POL( SEL₂(x₁, x₂) ) = max{0, 2x₁ - 3}

POL( s₁(x₁) ) = 3x₁ + 2

POL( n__zWquot₂(x₁, x₂) ) = 3x₁ + 3x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
minus₂(X, 0) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(minus₂(X, Y), s₁(Y)))
zWquot₂(XS, nil) -> nil
zWquot₂(nil, XS) -> nil
zWquot₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(quot₂(X, Y), n__zWquot₂(activate₁(XS), activate₁(YS)))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(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.