Termination w.r.t. Q proof of TRS/TRCSREx26_Luc03b

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__terms₁(X)) -> TERMS₁(X)

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)
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))

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)) = 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)
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)

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__add₂(X1, X2)) -> ADD₂(X1, X2)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(ADD₂(x₁, x₂)) = x₁
POL(DBL₁(x₁)) = x₁
POL(FIRST₂(x₁, x₂)) = x₁ + x₂
POL(cons₂(x₁, x₂)) = x₂
POL(n__add₂(x₁, x₂)) = 1 + x₁
POL(n__dbl₁(x₁)) = x₁
POL(n__first₂(x₁, x₂)) = x₁ + x₂
POL(s₁(x₁)) = 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__dbl₁(X)) -> DBL₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ 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)
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__first₂(X1, X2)) -> FIRST₂(X1, X2)

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                  ↳ 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)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                  ↳ QDP

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
DBL₁(s₁(X)) -> 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 oriented strictly and are deleted.

DBL₁(s₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(DBL₁(x₁)) = x₁
POL(n__dbl₁(x₁)) = x₁
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

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                  ↳ QDP

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(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 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.