Termination w.r.t. Q proof of TRS/TRCSREx2_Luc02a

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) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

terms₁(N) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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))
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)

The TRS R consists of the following rules:

terms₁(N) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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))
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)

The TRS R consists of the following rules:

terms₁(N) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

terms₁(N) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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

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) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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)) = 2·x₁²
POL(s₁(x₁)) = 2 + 2·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

terms₁(N) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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

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) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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)) = 2·x₁²
POL(s₁(x₁)) = 2 + 2·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

terms₁(N) -> cons₁(recip₁(sqr₁(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), cons₁(Y)) -> cons₁(Y)

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.