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:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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__terms₁(X)) -> TERMS₁(activate₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
HALF₁(s₁(s₁(X))) -> S₁(half₁(X))
SQR₁(s₁(X)) -> SQR₁(X)
TERMS₁(N) -> SQR₁(N)
ADD₂(s₁(X), Y) -> S₁(add₂(X, Y))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
SQR₁(s₁(X)) -> ADD₂(sqr₁(X), dbl₁(X))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
DBL₁(s₁(X)) -> S₁(dbl₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
SQR₁(s₁(X)) -> DBL₁(X)
ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> S₁(add₂(sqr₁(X), dbl₁(X)))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
DBL₁(s₁(X)) -> DBL₁(X)
DBL₁(s₁(X)) -> S₁(s₁(dbl₁(X)))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
HALF₁(s₁(s₁(X))) -> HALF₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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__terms₁(X)) -> TERMS₁(activate₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
HALF₁(s₁(s₁(X))) -> S₁(half₁(X))
SQR₁(s₁(X)) -> SQR₁(X)
TERMS₁(N) -> SQR₁(N)
ADD₂(s₁(X), Y) -> S₁(add₂(X, Y))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
SQR₁(s₁(X)) -> ADD₂(sqr₁(X), dbl₁(X))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
DBL₁(s₁(X)) -> S₁(dbl₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
SQR₁(s₁(X)) -> DBL₁(X)
ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> S₁(add₂(sqr₁(X), dbl₁(X)))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
DBL₁(s₁(X)) -> DBL₁(X)
DBL₁(s₁(X)) -> S₁(s₁(dbl₁(X)))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
HALF₁(s₁(s₁(X))) -> HALF₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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 5 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HALF₁(s₁(s₁(X))) -> HALF₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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.

HALF₁(s₁(s₁(X))) -> HALF₁(X)

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

POL(HALF₁(x₁)) = x₁
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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

          ↳ QDP

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

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

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

POL(ADD₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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

          ↳ QDP

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

DBL₁(s₁(X)) -> DBL₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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.

DBL₁(s₁(X)) -> DBL₁(X)

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

POL(DBL₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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

          ↳ QDP

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

SQR₁(s₁(X)) -> SQR₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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.

SQR₁(s₁(X)) -> SQR₁(X)

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

POL(SQR₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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

          ↳ QDP

            ↳ QDPOrderProof

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

ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

Used ordering: Polynomial interpretation [21]:

POL(0) = 1
POL(ACTIVATE₁(x₁)) = 2·x₁
POL(FIRST₂(x₁, x₂)) = 2·x₂
POL(activate₁(x₁)) = x₁
POL(add₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = x₂
POL(dbl₁(x₁)) = 1
POL(first₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(n__first₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(n__s₁(x₁)) = x₁
POL(n__terms₁(x₁)) = 2·x₁
POL(nil) = 2
POL(recip₁(x₁)) = 2 + 2·x₁
POL(s₁(x₁)) = x₁
POL(sqr₁(x₁)) = x₁
POL(terms₁(x₁)) = 2·x₁

The following usable rules [14] were oriented:

first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
s₁(X) -> n__s₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
terms₁(X) -> n__terms₁(X)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
first₂(0, X) -> nil
activate₁(X) -> X
terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
first₂(X1, X2) -> n__first₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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__terms₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

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

POL(ACTIVATE₁(x₁)) = 2·x₁
POL(n__s₁(x₁)) = 2 + 2·x₁
POL(n__terms₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(add₂(sqr₁(X), dbl₁(X)))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X
terms₁(X) -> n__terms₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(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.