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.


ACTIVATE1(n__dbl1(X)) -> DBL1(X)
The remaining pairs can at least be oriented weakly.

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)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVATE1(x1) ) = x1


POL( n__dbl1(x1) ) = x1 + 1


POL( DBL1(x1) ) = x1


POL( SQR1(x1) ) = x1


POL( s1(x1) ) = x1


POL( FIRST2(x1, x2) ) = x1 + x2


POL( ADD2(x1, x2) ) = x1


POL( n__add2(x1, x2) ) = x1 + x2


POL( TERMS1(x1) ) = x1


POL( n__first2(x1, x2) ) = x1 + x2


POL( activate1(x1) ) = x1


POL( cons2(x1, x2) ) = x2


POL( n__terms1(x1) ) = x1


POL( add2(x1, x2) ) = x1 + x2


POL( n__s1(x1) ) = x1


POL( first2(x1, x2) ) = x1 + x2


POL( dbl1(x1) ) = x1 + 1


POL( terms1(x1) ) = x1


POL( 0 ) = 0


POL( nil ) = 0



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
              ↳ QDPOrderProof

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

SQR1(s1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
TERMS1(N) -> SQR1(N)
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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SQR1(s1(X)) -> ACTIVATE1(X)
SQR1(s1(X)) -> DBL1(activate1(X))
The remaining pairs can at least be oriented weakly.

FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
TERMS1(N) -> SQR1(N)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
DBL1(s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
SQR1(s1(X)) -> SQR1(activate1(X))
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__terms1(X)) -> TERMS1(X)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( SQR1(x1) ) = x1


POL( s1(x1) ) = x1 + 1


POL( ACTIVATE1(x1) ) = x1


POL( FIRST2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( ADD2(x1, x2) ) = max{0, x1 - 1}


POL( TERMS1(x1) ) = x1


POL( n__add2(x1, x2) ) = x1 + x2


POL( DBL1(x1) ) = max{0, x1 - 1}


POL( n__first2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( activate1(x1) ) = x1 + 1


POL( cons2(x1, x2) ) = x2


POL( n__terms1(x1) ) = x1


POL( add2(x1, x2) ) = x1 + x2 + 1


POL( n__s1(x1) ) = x1


POL( first2(x1, x2) ) = x1 + x2


POL( n__dbl1(x1) ) = 0


POL( dbl1(x1) ) = 1


POL( terms1(x1) ) = x1 + 1


POL( 0 ) = 0


POL( nil ) = 0



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
              ↳ QDPOrderProof
QDP
                  ↳ DependencyGraphProof

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

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)
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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

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

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

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

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 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.


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

ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIRST2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( s1(x1) ) = x1 + 1


POL( cons2(x1, x2) ) = x2 + 1


POL( ACTIVATE1(x1) ) = max{0, x1 - 1}


POL( ADD2(x1, x2) ) = max{0, x1 - 1}


POL( n__add2(x1, x2) ) = x1


POL( n__first2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented: none



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

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

ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
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 1 SCC with 1 less node.

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

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

ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ADD2(s1(X), Y) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ADD2(x1, x2) ) = x1


POL( s1(x1) ) = x1 + 1


POL( ACTIVATE1(x1) ) = max{0, x1 - 1}


POL( n__add2(x1, x2) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

ACTIVATE1(n__add2(X1, X2)) -> ADD2(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.