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(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(n__sqr₁(activate₁(X)), n__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)
s₁(X) -> n__s₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
sqr₁(X) -> n__sqr₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(n__sqr₁(X)) -> sqr₁(activate₁(X))
activate₁(n__dbl₁(X)) -> dbl₁(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(n__add₂(n__sqr₁(activate₁(X)), n__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)
s₁(X) -> n__s₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
sqr₁(X) -> n__sqr₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(n__sqr₁(X)) -> sqr₁(activate₁(X))
activate₁(n__dbl₁(X)) -> dbl₁(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))
SQR₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__sqr₁(X)) -> SQR₁(activate₁(X))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
TERMS₁(N) -> SQR₁(N)
ACTIVATE₁(n__sqr₁(X)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__dbl₁(X)) -> DBL₁(activate₁(X))
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
SQR₁(s₁(X)) -> S₁(n__add₂(n__sqr₁(activate₁(X)), n__dbl₁(activate₁(X))))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__dbl₁(X)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), Y))
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)
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(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(n__sqr₁(activate₁(X)), n__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)
s₁(X) -> n__s₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
sqr₁(X) -> n__sqr₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(n__sqr₁(X)) -> sqr₁(activate₁(X))
activate₁(n__dbl₁(X)) -> dbl₁(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))
SQR₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__sqr₁(X)) -> SQR₁(activate₁(X))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
TERMS₁(N) -> SQR₁(N)
ACTIVATE₁(n__sqr₁(X)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__dbl₁(X)) -> DBL₁(activate₁(X))
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
SQR₁(s₁(X)) -> S₁(n__add₂(n__sqr₁(activate₁(X)), n__dbl₁(activate₁(X))))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__dbl₁(X)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), Y))
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)
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(n__s₁(N)))
sqr₁(0) -> 0
sqr₁(s₁(X)) -> s₁(n__add₂(n__sqr₁(activate₁(X)), n__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)
s₁(X) -> n__s₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
sqr₁(X) -> n__sqr₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(n__sqr₁(X)) -> sqr₁(activate₁(X))
activate₁(n__dbl₁(X)) -> dbl₁(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 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

ACTIVATE₁(n__terms₁(X)) -> TERMS₁(activate₁(X))
SQR₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__sqr₁(X)) -> SQR₁(activate₁(X))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
TERMS₁(N) -> SQR₁(N)
ACTIVATE₁(n__sqr₁(X)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__dbl₁(X)) -> DBL₁(activate₁(X))
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__dbl₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__terms₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)

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(n__add₂(n__sqr₁(activate₁(X)), n__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)
s₁(X) -> n__s₁(X)
add₂(X1, X2) -> n__add₂(X1, X2)
sqr₁(X) -> n__sqr₁(X)
dbl₁(X) -> n__dbl₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__terms₁(X)) -> terms₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(n__sqr₁(X)) -> sqr₁(activate₁(X))
activate₁(n__dbl₁(X)) -> dbl₁(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.