Termination w.r.t. Q proof of TRS/SK902.14.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:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(x)) -> x

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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:

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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:

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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 3 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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.

-¹₂(s₁(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 + 2·x₂
POL(s₁(x₁)) = 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:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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

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:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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)) = 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:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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

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

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)

The TRS R consists of the following rules:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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.

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)) = 1 + 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:

double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
if₃(0, y, z) -> y
if₃(s₁(x), y, z) -> z
half₁(double₁(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.