Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx2_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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

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

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADD₂(x1, x2) = ADD₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[ADD₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

DBL₁(s₁(X)) -> DBL₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBL₁(x1) = DBL₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > DBL₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

SQR₁(s₁(X)) -> SQR₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SQR₁(x1) = SQR₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > SQR₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

terms₁(x0)
sqr₁(0)
sqr₁(s₁(x0))
dbl₁(0)
dbl₁(s₁(x0))
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.