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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.

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__terms1(X)) -> TERMS1(activate1(X))
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ADD2(s1(X), Y) -> ADD2(X, Y)
HALF1(s1(s1(X))) -> S1(half1(X))
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)
HALF1(s1(s1(X))) -> HALF1(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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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)
HALF1(s1(s1(X))) -> S1(half1(X))
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)
HALF1(s1(s1(X))) -> HALF1(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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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 [13,14,18] contains 5 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

HALF1(s1(s1(X))) -> HALF1(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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


HALF1(s1(s1(X))) -> HALF1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
HALF1(x1)  =  HALF1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[HALF1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ 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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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
            ↳ QDPOrderProof
          ↳ 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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADD2(x1, x2)  =  ADD1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[ADD1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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
            ↳ QDPOrderProof
          ↳ 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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DBL1(s1(X)) -> DBL1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBL1(x1)  =  DBL1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > DBL1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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
            ↳ QDPOrderProof
          ↳ 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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


SQR1(s1(X)) -> SQR1(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)

Lexicographic Path Order [19].
Precedence:
s1 > SQR1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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
QDP
            ↳ QDPOrderProof

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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
The remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__terms1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  x1
n__first2(x1, x2)  =  n__first2(x1, x2)
FIRST2(x1, x2)  =  x2
activate1(x1)  =  x1
n__terms1(x1)  =  x1
s1(x1)  =  x1
cons2(x1, x2)  =  x2
n__s1(x1)  =  x1
terms1(x1)  =  x1
add2(x1, x2)  =  add1(x2)
sqr1(x1)  =  sqr
0  =  0
first2(x1, x2)  =  first2(x1, x2)
dbl1(x1)  =  x1
nil  =  nil
recip1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[nfirst2, first2] > nil
sqr > add1


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
QDP
                    ↳ QDPOrderProof

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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__terms1(X)) -> ACTIVATE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__terms1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof

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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVATE1(n__terms1(X)) -> ACTIVATE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__terms1(x1)  =  n__terms1(x1)

Lexicographic Path Order [19].
Precedence:
nterms1 > ACTIVATE1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ 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(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)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
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.