Termination w.r.t. Q proof of TRS/nontermin/TRCSRExSec11_1

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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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:

SQR₁(s₁(X)) -> ADD₂(sqr₁(X), dbl₁(X))
TERMS₁(N) -> TERMS₁(s₁(N))
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
SQR₁(s₁(X)) -> SQR₁(X)
SQR₁(s₁(X)) -> DBL₁(X)
TERMS₁(N) -> SQR₁(N)
DBL₁(s₁(X)) -> DBL₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)
HALF₁(s₁(s₁(X))) -> HALF₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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:

SQR₁(s₁(X)) -> ADD₂(sqr₁(X), dbl₁(X))
TERMS₁(N) -> TERMS₁(s₁(N))
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
SQR₁(s₁(X)) -> SQR₁(X)
SQR₁(s₁(X)) -> DBL₁(X)
TERMS₁(N) -> SQR₁(N)
DBL₁(s₁(X)) -> DBL₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)
HALF₁(s₁(s₁(X))) -> HALF₁(X)

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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 6 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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.

HALF₁(s₁(s₁(X))) -> HALF₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HALF₁(x₁)) = 3·x₁
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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

          ↳ QDP

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

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

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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)) -> FIRST₂(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FIRST₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(cons₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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) -> ADD₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADD₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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)) -> DBL₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DBL₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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)) -> SQR₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SQR₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(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

          ↳ QDP

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

TERMS₁(N) -> TERMS₁(s₁(N))

The TRS R consists of the following rules:

terms₁(N) -> cons₂(recip₁(sqr₁(N)), terms₁(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, first₂(X, Z))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(X))) -> s₁(half₁(X))
half₁(dbl₁(X)) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.