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(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 oriented strictly and are deleted.


TERMS1(N) -> SQR1(N)
The remaining pairs can at least be oriented weakly.

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)
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)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(ACTIVATE1(x1)) = x1   
POL(ADD2(x1, x2)) = x1   
POL(DBL1(x1)) = x1   
POL(FIRST2(x1, x2)) = x1 + x2   
POL(SQR1(x1)) = x1   
POL(TERMS1(x1)) = 1 + x1   
POL(activate1(x1)) = x1   
POL(add2(x1, x2)) = x1 + x2   
POL(cons2(x1, x2)) = x2   
POL(dbl1(x1)) = x1   
POL(first2(x1, x2)) = x1 + x2   
POL(n__add2(x1, x2)) = x1 + x2   
POL(n__dbl1(x1)) = x1   
POL(n__first2(x1, x2)) = x1 + x2   
POL(n__s1(x1)) = x1   
POL(n__terms1(x1)) = 1 + x1   
POL(nil) = 0   
POL(recip1(x1)) = 0   
POL(s1(x1)) = x1   
POL(sqr1(x1)) = 0   
POL(terms1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

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



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

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

SQR1(s1(X)) -> ACTIVATE1(X)
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)
DBL1(s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
SQR1(s1(X)) -> DBL1(activate1(X))
SQR1(s1(X)) -> SQR1(activate1(X))
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
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 2 SCCs with 3 less nodes.

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

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

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)
DBL1(s1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

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 oriented strictly 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)
DBL1(s1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
The remaining pairs can at least be oriented weakly.

ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = x1   
POL(ADD2(x1, x2)) = x1 + x2   
POL(DBL1(x1)) = x1   
POL(FIRST2(x1, x2)) = x1 + x2   
POL(cons2(x1, x2)) = 1 + x2   
POL(n__add2(x1, x2)) = 1 + x1 + x2   
POL(n__dbl1(x1)) = 1 + x1   
POL(n__first2(x1, x2)) = x1 + x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)

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 0 SCCs with 1 less node.

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

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.