Termination w.r.t. Q proof of TRS/SK902.19.trs

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:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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:

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)
SQR₁(s₁(x)) -> DOUBLE₁(x)
SQR₁(s₁(x)) -> +¹₂(sqr₁(x), double₁(x))
+¹₂(x, s₁(y)) -> +¹₂(x, y)
SQR₁(s₁(x)) -> SQR₁(x)
SQR₁(s₁(x)) -> +¹₂(sqr₁(x), s₁(double₁(x)))

The TRS R consists of the following rules:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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:

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)
SQR₁(s₁(x)) -> DOUBLE₁(x)
SQR₁(s₁(x)) -> +¹₂(sqr₁(x), double₁(x))
+¹₂(x, s₁(y)) -> +¹₂(x, y)
SQR₁(s₁(x)) -> SQR₁(x)
SQR₁(s₁(x)) -> +¹₂(sqr₁(x), s₁(double₁(x)))

The TRS R consists of the following rules:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(x)))

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:

+¹₂(x, s₁(y)) -> +¹₂(x, y)

The TRS R consists of the following rules:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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.

+¹₂(x, s₁(y)) -> +¹₂(x, y)

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

POL(+¹₂(x₁, x₂)) = 2·x₁·x₂ + 2·x₂
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:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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

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

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)

The TRS R consists of the following rules:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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.

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)

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

POL(DOUBLE₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 2 + x₁

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:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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

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

SQR₁(s₁(x)) -> SQR₁(x)

The TRS R consists of the following rules:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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)) = 2·x₁
POL(s₁(x₁)) = 2 + x₁

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:

sqr₁(0) -> 0
sqr₁(s₁(x)) -> +₂(sqr₁(x), s₁(double₁(x)))
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
sqr₁(s₁(x)) -> s₁(+₂(sqr₁(x), double₁(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.