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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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))
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
QUOT₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ZWQUOT₂(X1, X2)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> QUOT₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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))
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(XS)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> ACTIVATE₁(YS)
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
QUOT₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
ACTIVATE₁(n__zWquot₂(X1, X2)) -> ZWQUOT₂(X1, X2)
ZWQUOT₂(cons₂(X, XS), cons₂(Y, YS)) -> QUOT₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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 4 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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 interpretation [21]:

POL(MINUS₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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 interpretation [21]:

POL(0) = 0
POL(QUOT₂(x₁, x₂)) = 2·x₁
POL(minus₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented:

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

↳ 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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:

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = 2 + 2·x₁
POL(ZWQUOT₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = x₂
POL(n__zWquot₂(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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 interpretation [21]:

POL(0) = 0
POL(SEL₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(activate₁(x₁)) = 2 + x₁
POL(cons₂(x₁, x₂)) = 1 + 2·x₂
POL(from₁(x₁)) = 2
POL(minus₂(x₁, x₂)) = 0
POL(n__from₁(x₁)) = 0
POL(n__zWquot₂(x₁, x₂)) = 0
POL(nil) = 0
POL(quot₂(x₁, x₂)) = 1 + x₁ + 3·x₂
POL(s₁(x₁)) = 3 + x₁
POL(zWquot₂(x₁, x₂)) = 2

The following usable rules [14] were oriented:

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

↳ 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(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)
zWquot₂(X1, X2) -> n__zWquot₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__zWquot₂(X1, X2)) -> zWquot₂(X1, 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.