Termination w.r.t. Q proof of TRS/SK902.49.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:

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> b₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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:

V₁(a₁(c₁(x))) -> U₁(b₁(d₁(x)))
W₁(a₁(a₁(x))) -> U₁(w₁(x))
V₁(a₁(a₁(x))) -> V₁(x)
V₁(a₁(a₁(x))) -> U₁(v₁(x))
W₁(a₁(c₁(x))) -> U₁(b₁(d₁(x)))
W₁(a₁(a₁(x))) -> W₁(x)

The TRS R consists of the following rules:

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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:

V₁(a₁(c₁(x))) -> U₁(b₁(d₁(x)))
W₁(a₁(a₁(x))) -> U₁(w₁(x))
V₁(a₁(a₁(x))) -> V₁(x)
V₁(a₁(a₁(x))) -> U₁(v₁(x))
W₁(a₁(c₁(x))) -> U₁(b₁(d₁(x)))
W₁(a₁(a₁(x))) -> W₁(x)

The TRS R consists of the following rules:

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

W₁(a₁(a₁(x))) -> W₁(x)

The TRS R consists of the following rules:

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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(a₁(a₁(x))) -> W₁(x)

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

POL(W₁(x₁)) = x₁
POL(a₁(x₁)) = 2 + 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:

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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:

V₁(a₁(a₁(x))) -> V₁(x)

The TRS R consists of the following rules:

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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.

V₁(a₁(a₁(x))) -> V₁(x)

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

POL(V₁(x₁)) = x₁
POL(a₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

a₁(c₁(d₁(x))) -> c₁(x)
u₁(b₁(d₁(d₁(x)))) -> b₁(x)
v₁(a₁(a₁(x))) -> u₁(v₁(x))
v₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
v₁(c₁(x)) -> b₁(x)
w₁(a₁(a₁(x))) -> u₁(w₁(x))
w₁(a₁(c₁(x))) -> u₁(b₁(d₁(x)))
w₁(c₁(x)) -> 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.