Termination w.r.t. Q proof of TRS/TRCSRExSec11_1_Luc02a

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:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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:

ACTIVATE(n__terms(X)) → TERMS(activate(X))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ADD(s(X), Y) → ADD(X, Y)
HALF(s(s(X))) → S(half(X))
SQR(s(X)) → SQR(X)
TERMS(N) → SQR(N)
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
SQR(s(X)) → ADD(sqr(X), dbl(X))
ACTIVATE(n__s(X)) → S(activate(X))
DBL(s(X)) → S(dbl(X))
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
SQR(s(X)) → DBL(X)
ACTIVATE(n__terms(X)) → ACTIVATE(X)
SQR(s(X)) → S(add(sqr(X), dbl(X)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
DBL(s(X)) → DBL(X)
DBL(s(X)) → S(s(dbl(X)))
ACTIVATE(n__s(X)) → ACTIVATE(X)
HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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:

ACTIVATE(n__terms(X)) → TERMS(activate(X))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ADD(s(X), Y) → ADD(X, Y)
HALF(s(s(X))) → S(half(X))
SQR(s(X)) → SQR(X)
TERMS(N) → SQR(N)
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
SQR(s(X)) → ADD(sqr(X), dbl(X))
ACTIVATE(n__s(X)) → S(activate(X))
DBL(s(X)) → S(dbl(X))
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
SQR(s(X)) → DBL(X)
ACTIVATE(n__terms(X)) → ACTIVATE(X)
SQR(s(X)) → S(add(sqr(X), dbl(X)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
DBL(s(X)) → DBL(X)
DBL(s(X)) → S(s(dbl(X)))
ACTIVATE(n__s(X)) → ACTIVATE(X)
HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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:

ACTIVATE(n__terms(X)) → TERMS(activate(X))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ADD(s(X), Y) → ADD(X, Y)
HALF(s(s(X))) → S(half(X))
SQR(s(X)) → SQR(X)
TERMS(N) → SQR(N)
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
SQR(s(X)) → ADD(sqr(X), dbl(X))
DBL(s(X)) → S(dbl(X))
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
SQR(s(X)) → DBL(X)
ACTIVATE(n__terms(X)) → ACTIVATE(X)
SQR(s(X)) → S(add(sqr(X), dbl(X)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
DBL(s(X)) → DBL(X)
DBL(s(X)) → S(s(dbl(X)))
HALF(s(s(X))) → HALF(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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 5 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

HALF(s(s(X))) → HALF(X)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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

              ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

ADD(s(X), Y) → ADD(X, Y)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

DBL(s(X)) → DBL(X)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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

                ↳ QDPOrderProof

              ↳ QDP

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

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

SQR(s(X)) → SQR(X)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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

              ↳ QDP

                ↳ QDPOrderProof

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

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__terms(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__terms(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__first(x1, x2) = n__first(x1, x2)
FIRST(x1, x2) = x2
activate(x1) = x1
n__terms(x1) = x1
cons(x1, x2) = x2
n__s(x1) = x1
first(x1, x2) = first(x1, x2)
s(x1) = x1
terms(x1) = x1
0 = 0
nil = nil

Lexicographic Path Order [19].
Precedence:

[n_{_first₂}, first_2] > nil

The following usable rules [14] were oriented:

activate(X) → X
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
first(X1, X2) → n__first(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
terms(X) → n__terms(X)
activate(n__s(X)) → s(activate(X))
s(X) → n__s(X)
first(0, X) → nil
activate(n__terms(X)) → terms(activate(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

ACTIVATE(n__terms(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

ACTIVATE(n__terms(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

ACTIVATE(n__s(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__terms(X)) → ACTIVATE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__terms(x1) = x1
n__s(x1) = n__s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

ACTIVATE(n__terms(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.

ACTIVATE(n__terms(X)) → ACTIVATE(X)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(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.