Termination w.r.t. Q proof of TRS/AG01#3.18.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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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:

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(s₁(x), y) -> MINUS₂(x, y)
PLUS₂(s₁(x), y) -> DOUBLE₁(y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
PLUS₂(s₁(x), y) -> PLUS₂(minus₂(x, y), double₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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:

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(s₁(x), y) -> MINUS₂(x, y)
PLUS₂(s₁(x), y) -> DOUBLE₁(y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
PLUS₂(s₁(x), y) -> PLUS₂(minus₂(x, y), double₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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.

DOUBLE₁(s₁(x)) -> DOUBLE₁(x)

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

POL(DOUBLE₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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:

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

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

POL(MINUS₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))
PLUS₂(s₁(x), y) -> PLUS₂(minus₂(x, y), double₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))
PLUS₂(s₁(x), y) -> PLUS₂(minus₂(x, y), double₁(y))

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

POL(0) = 0
POL(PLUS₂(x₁, x₂)) = x₁
POL(double₁(x₁)) = 0
POL(minus₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)

↳ 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
double₁(0) -> 0
double₁(s₁(x)) -> s₁(s₁(double₁(x)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(s₁(x), y) -> s₁(plus₂(minus₂(x, y), double₁(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.