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

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:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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:

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(activate(X1), activate(X2))
MINUS(s(X), s(Y)) → MINUS(X, Y)
QUOT(s(X), s(Y)) → MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) → QUOT(X, Y)
ACTIVATE(n__from(X)) → FROM(activate(X))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X1)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X2)
QUOT(s(X), s(Y)) → S(quot(minus(X, Y), s(Y)))
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(activate(X1), activate(X2))
MINUS(s(X), s(Y)) → MINUS(X, Y)
QUOT(s(X), s(Y)) → MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) → QUOT(X, Y)
ACTIVATE(n__from(X)) → FROM(activate(X))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X1)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X2)
QUOT(s(X), s(Y)) → S(quot(minus(X, Y), s(Y)))
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(activate(X1), activate(X2))
MINUS(s(X), s(Y)) → MINUS(X, Y)
QUOT(s(X), s(Y)) → MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) → QUOT(X, Y)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__s(X)) → S(activate(X))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X1)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
QUOT(s(X), s(Y)) → S(quot(minus(X, Y), s(Y)))
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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.

MINUS(s(X), s(Y)) → MINUS(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = MINUS(x2)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > MINUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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.

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2) = x1
s(x1) = s(x1)
minus(x1, x2) = minus(x1)
0 = 0
n__s(x1) = n__s(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > minus₁ > 0

The following usable rules [14] were oriented:

minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X1)
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(activate(X1), activate(X2))
ACTIVATE(n__zWquot(X1, X2)) → ACTIVATE(X2)
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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.

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2) = SEL(x1)
s(x1) = s(x1)
cons(x1, x2) = cons(x1)
activate(x1) = activate
minus(x1, x2) = minus(x2)
0 = 0
n__s(x1) = n__s
quot(x1, x2) = quot
zWquot(x1, x2) = zWquot(x1, x2)
nil = nil
n__zWquot(x1, x2) = n__zWquot(x1, x2)
n__from(x1) = n__from
from(x1) = x1

Lexicographic Path Order [19].
Precedence:

SEL₁ > activate > zWquot₂ > nil > n_{_from}
s₁ > activate > zWquot₂ > nil > n_{_from}
s₁ > 0 > n_{_from}
s₁ > quot > n_{_from}
cons₁ > n_{_from}
minus₁ > 0 > n_{_from}
n_{_s} > activate > zWquot₂ > nil > n_{_from}
n_{_zWquot₂} > zWquot₂ > nil > n_{_from}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(activate(X1), activate(X2))
activate(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.