Termination w.r.t. Q proof of TRS/D3307.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:

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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:

B₁(w₁(x)) -> W₁(b₁(x))
B₁(w₁(x)) -> B₁(x)
W₁(r₁(x)) -> W₁(x)
B₁(r₁(x)) -> B₁(x)

The TRS R consists of the following rules:

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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:

B₁(w₁(x)) -> W₁(b₁(x))
B₁(w₁(x)) -> B₁(x)
W₁(r₁(x)) -> W₁(x)
B₁(r₁(x)) -> B₁(x)

The TRS R consists of the following rules:

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

W₁(r₁(x)) -> W₁(x)

The TRS R consists of the following rules:

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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.

W₁(r₁(x)) -> W₁(x)

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

POL(W₁(x₁)) = 2·x₁
POL(r₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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

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

B₁(w₁(x)) -> B₁(x)
B₁(r₁(x)) -> B₁(x)

The TRS R consists of the following rules:

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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.

B₁(w₁(x)) -> B₁(x)

The remaining pairs can at least be oriented weakly.

B₁(r₁(x)) -> B₁(x)

Used ordering: Polynomial interpretation [21]:

POL(B₁(x₁)) = x₁
POL(r₁(x₁)) = 3·x₁
POL(w₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

B₁(r₁(x)) -> B₁(x)

The TRS R consists of the following rules:

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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.

B₁(r₁(x)) -> B₁(x)

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

POL(B₁(x₁)) = 2·x₁
POL(r₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

w₁(r₁(x)) -> r₁(w₁(x))
b₁(r₁(x)) -> r₁(b₁(x))
b₁(w₁(x)) -> w₁(b₁(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.