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)))
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: Quasi 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: Quasi 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: Quasi 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
            ↳ QDPAfsSolverProof

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

ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
Used argument filtering: ACTIVATE1(x1)  =  x1
n__first2(x1, x2)  =  n__first2(x1, x2)
FIRST2(x1, x2)  =  x2
activate1(x1)  =  x1
n__terms1(x1)  =  x1
cons2(x1, x2)  =  x2
n__s1(x1)  =  x1
terms1(x1)  =  x1
s1(x1)  =  x1
first2(x1, x2)  =  first2(x1, x2)
0  =  0
nil  =  nil
Used ordering: Quasi Precedence: [n__first_2, first_2, nil]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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)))
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 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ DependencyGraphProof
QDP
                    ↳ QDPAfsSolverProof

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)))
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.

ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__terms1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof

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

ACTIVATE1(n__terms1(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.
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)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__terms1(x1)  =  n__terms1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
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)))
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.