Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx26_Luc03b

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(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.

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__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 [13,14,18] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The remaining pairs can at least by weakly be oriented.

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)

Used ordering: Combined order from the following AFS and order.
ACTIVATE₁(x1) = ACTIVATE₁(x1)
n__dbl₁(x1) = n__dbl₁(x1)
DBL₁(x1) = DBL₁(x1)
SQR₁(x1) = SQR₁(x1)
s₁(x1) = x1
FIRST₂(x1, x2) = FIRST₂(x1, x2)
cons₂(x1, x2) = x2
ADD₂(x1, x2) = ADD₂(x1, x2)
n__add₂(x1, x2) = n__add₂(x1, x2)
TERMS₁(x1) = TERMS₁(x1)
n__first₂(x1, x2) = n__first₂(x1, x2)
activate₁(x1) = x1
n__terms₁(x1) = x1
add₂(x1, x2) = add₂(x1, x2)
sqr₁(x1) = sqr
0 = 0
terms₁(x1) = x1
recip₁(x1) = recip₁(x1)
dbl₁(x1) = dbl₁(x1)
n__s₁(x1) = x1
first₂(x1, x2) = first₂(x1, x2)
nil = nil

Lexicographic Path Order [19].
Precedence:

[FIRST₂, n_{_first₂}, 0, first₂, nil] > [ACTIVATE₁, DBL₁, SQR₁, TERMS_1]
[ADD₂, n_{_add₂}, add_2] > [ACTIVATE₁, DBL₁, SQR₁, TERMS_1]
sqr > [n_{_dbl₁}, dbl_1] > [ACTIVATE₁, DBL₁, SQR₁, TERMS_1]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

TERMS₁(N) -> SQR₁(N)

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
SQR₁(x1) = x1
s₁(x1) = x1
ACTIVATE₁(x1) = x1
TERMS₁(x1) = TERMS₁(x1)
DBL₁(x1) = x1
activate₁(x1) = x1
n__terms₁(x1) = n__terms₁(x1)
n__add₂(x1, x2) = x2
add₂(x1, x2) = x2
terms₁(x1) = terms₁(x1)
cons₂(x1, x2) = cons
recip₁(x1) = recip₁(x1)
sqr₁(x1) = sqr
0 = 0
dbl₁(x1) = x1
n__s₁(x1) = x1
n__dbl₁(x1) = x1
first₂(x1, x2) = x1
n__first₂(x1, x2) = x1
nil = nil

Lexicographic Path Order [19].
Precedence:

[TERMS₁, n_{_terms₁}, terms₁, recip_1] > sqr > [0, nil] > cons

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

SQR₁(s₁(X)) -> ACTIVATE₁(X)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

SQR₁(s₁(X)) -> SQR₁(activate₁(X))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SQR₁(x1) = SQR₁(x1)
s₁(x1) = s₁(x1)
activate₁(x1) = x1
n__add₂(x1, x2) = n__add₂(x1, x2)
add₂(x1, x2) = add₂(x1, x2)
terms₁(x1) = x1
cons₂(x1, x2) = cons
recip₁(x1) = recip₁(x1)
sqr₁(x1) = sqr
n__terms₁(x1) = x1
0 = 0
dbl₁(x1) = dbl₁(x1)
n__s₁(x1) = n__s₁(x1)
n__dbl₁(x1) = n__dbl₁(x1)
first₂(x1, x2) = x2
n__first₂(x1, x2) = x2
nil = nil

Lexicographic Path Order [19].
Precedence:

sqr > [n_{_add₂}, add_2]
[0, dbl₁, n_{_{dbl_1]}}

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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(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 TRS P is empty. Hence, there is no (P,Q,R) chain.