Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx26_Luc03b

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:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, 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(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
SQR₁(s₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
TERMS₁(N) -> SQR₁(N)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> DBL₁(activate₁(X))
SQR₁(s₁(X)) -> SQR₁(activate₁(X))
ACTIVATE₁(n__terms₁(X)) -> TERMS₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
SQR₁(s₁(X)) -> S₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
TERMS₁(N) -> S₁(N)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), Y))
DBL₁(s₁(X)) -> S₁(n__s₁(n__dbl₁(activate₁(X))))

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(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:

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
SQR₁(s₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
TERMS₁(N) -> SQR₁(N)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> DBL₁(activate₁(X))
SQR₁(s₁(X)) -> SQR₁(activate₁(X))
ACTIVATE₁(n__terms₁(X)) -> TERMS₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
SQR₁(s₁(X)) -> S₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
TERMS₁(N) -> S₁(N)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), Y))
DBL₁(s₁(X)) -> S₁(n__s₁(n__dbl₁(activate₁(X))))

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
SQR₁(s₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
TERMS₁(N) -> SQR₁(N)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> DBL₁(activate₁(X))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SQR₁(s₁(X)) -> SQR₁(activate₁(X))
ACTIVATE₁(n__terms₁(X)) -> TERMS₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

Used argument filtering: ACTIVATE₁(x1) = x1
n__dbl₁(x1) = x1
DBL₁(x1) = x1
SQR₁(x1) = x1
s₁(x1) = x1
FIRST₂(x1, x2) = FIRST₂(x1, x2)
cons₂(x1, x2) = x2
ADD₂(x1, x2) = x1
n__add₂(x1, x2) = n__add₂(x1, x2)
TERMS₁(x1) = x1
n__first₂(x1, x2) = n__first₂(x1, x2)
activate₁(x1) = x1
n__terms₁(x1) = x1
terms₁(x1) = x1
add₂(x1, x2) = add₂(x1, x2)
n__s₁(x1) = x1
dbl₁(x1) = x1
first₂(x1, x2) = first₂(x1, x2)
0 = 0
nil = nil
Used ordering: Quasi Precedence: [FIRST_2, n__first_2, first_2] > nil [n__add_2, add_2] > nil 0 > nil

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

SQR₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
TERMS₁(N) -> SQR₁(N)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
SQR₁(s₁(X)) -> DBL₁(activate₁(X))
SQR₁(s₁(X)) -> SQR₁(activate₁(X))
ACTIVATE₁(n__terms₁(X)) -> TERMS₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
SQR₁(s₁(X)) -> ACTIVATE₁(X)
TERMS₁(N) -> SQR₁(N)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> DBL₁(activate₁(X))
SQR₁(s₁(X)) -> SQR₁(activate₁(X))
ACTIVATE₁(n__terms₁(X)) -> TERMS₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__terms₁(X)) -> TERMS₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__dbl₁(x1) = x1
DBL₁(x1) = x1
SQR₁(x1) = x1
s₁(x1) = x1
TERMS₁(x1) = x1
activate₁(x1) = x1
n__terms₁(x1) = n__terms₁(x1)
terms₁(x1) = terms₁(x1)
n__add₂(x1, x2) = x2
add₂(x1, x2) = x2
n__s₁(x1) = x1
dbl₁(x1) = x1
n__first₂(x1, x2) = n__first
first₂(x1, x2) = first
nil = nil
cons₂(x1, x2) = cons
0 = 0
Used ordering: Quasi Precedence: [n__terms_1, terms_1, n__first, first, nil, cons]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

SQR₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
TERMS₁(N) -> SQR₁(N)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
SQR₁(s₁(X)) -> DBL₁(activate₁(X))
SQR₁(s₁(X)) -> SQR₁(activate₁(X))

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                          ↳ QDP

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DBL₁(s₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__dbl₁(x1) = x1
DBL₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                          ↳ QDP

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

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

SQR₁(s₁(X)) -> SQR₁(activate₁(X))

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), n__terms₁(s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(sqr₁(activate₁(X)), dbl₁(activate₁(X))))
dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(activate₁(X), activate₁(Z)))
terms₁(X) -> n__terms₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(X) -> X

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