Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx2_Luc02a

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(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)))
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)))
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.

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)
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)

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)))
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)
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)

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)))
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 contains 4 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

Used argument filtering: ADD₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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)))
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

            ↳ QDPAfsSolverProof

          ↳ 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)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DBL₁(s₁(X)) -> DBL₁(X)

Used argument filtering: DBL₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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)))
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

            ↳ QDPAfsSolverProof

          ↳ 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)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SQR₁(s₁(X)) -> SQR₁(X)

Used argument filtering: SQR₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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)))
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

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)))
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.