Termination w.r.t. Q proof of TRS/Zantemaz07.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:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, 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:

F(a, f(c, x)) → F(c, f(a, x))
F(a, x) → F(b, f(c, x))
F(d, f(c, x)) → F(d, f(a, x))
F(a, x) → F(c, x)
F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(b, f(a, x))
F(a, f(c, x)) → F(a, x)
F(d, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

F(a, f(c, x)) → F(c, f(a, x))
F(a, x) → F(b, f(c, x))
F(d, f(c, x)) → F(d, f(a, x))
F(a, x) → F(c, x)
F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(b, f(a, x))
F(a, f(c, x)) → F(a, x)
F(d, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F(a, x) → F(b, f(c, x))
F(a, f(c, x)) → F(c, f(a, x))
F(d, f(c, x)) → F(d, f(a, x))
F(a, x) → F(c, x)
F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(b, f(a, x))
F(a, f(c, x)) → F(a, x)
F(d, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, 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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

F(a, f(b, x)) → F(a, x)
F(a, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

A(B(x)) → A(x)
A(C(x)) → A(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

The following pairs can be oriented strictly and are deleted.

F(a, f(b, x)) → F(a, x)

The remaining pairs can at least be oriented weakly.

F(a, f(c, x)) → F(a, x)

Used ordering: Combined order from the following AFS and order.
A(x1) = A(x1)
B(x1) = B(x1)
C(x1) = x1

Lexicographic path order with status [19].
Precedence:

B₁ > A₁

Status:

A₁: [1]
B₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

F(a, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

A(C(x)) → A(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

The following pairs can be oriented strictly and are deleted.

F(a, f(c, x)) → F(a, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1) = A(x1)
C(x1) = C(x1)

Lexicographic path order with status [19].
Precedence:

C₁ > A₁

Status:

C₁: multiset
A₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

F(d, f(c, x)) → F(d, f(a, x))

The TRS R consists of the following rules:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

D(C(x)) → D(A(x))

The TRS R consists of the following rules:

A(x) → B(C(x))
A(B(x)) → B(A(x))
A(C(x)) → C(A(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

The following pairs can be oriented strictly and are deleted.

F(d, f(c, x)) → F(d, f(a, x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D(x1) = D(x1)
C(x1) = C(x1)
A(x1) = x1
B(x1) = B

Lexicographic path order with status [19].
Precedence:

D₁ > B
C₁ > B

Status:

C₁: [1]
D₁: [1]
B: []

The following usable rules [14] were oriented:

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(a, f(c, x)) → f(c, f(a, x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

f(a, x) → f(b, f(c, x))
f(a, f(b, x)) → f(b, f(a, x))
f(d, f(c, x)) → f(d, f(a, x))
f(a, f(c, x)) → f(c, f(a, 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.