Termination w.r.t. Q proof of /tmp/tmpelgZSH/01.srs

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:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

D(x1) → A(x1)
R(a(x1)) → D(r(x1))
D(a(x1)) → D(x1)
R(a(x1)) → R(x1)
D(a(x1)) → A(d(x1))
R(x1) → D(x1)
A(b(x1)) → R(x1)
D(a(x1)) → A(a(d(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

D(x1) → A(x1)
R(a(x1)) → D(r(x1))
D(a(x1)) → D(x1)
R(a(x1)) → R(x1)
D(a(x1)) → A(d(x1))
R(x1) → D(x1)
A(b(x1)) → R(x1)
D(a(x1)) → A(a(d(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

R(a(x1)) → D(r(x1))
R(x1) → D(x1)
A(b(x1)) → R(x1)

The remaining pairs can at least be oriented weakly.

D(x1) → A(x1)
D(a(x1)) → D(x1)
R(a(x1)) → R(x1)
D(a(x1)) → A(d(x1))
D(a(x1)) → A(a(d(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(D(x₁)) = (4)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = (4)x_1
POL(r(x₁)) = x_1
POL(d(x₁)) = x_1
POL(b(x₁)) = 4 + (2)x_1
POL(R(x₁)) = 9/4 + (4)x_1

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] were oriented:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

D(x1) → A(x1)
R(a(x1)) → R(x1)
D(a(x1)) → D(x1)
D(a(x1)) → A(d(x1))
D(a(x1)) → A(a(d(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

R(a(x1)) → R(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

R(a(x1)) → R(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(a(x₁)) = 1 + (4)x_1
POL(R(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

D(a(x1)) → D(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

D(a(x1)) → D(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(D(x₁)) = (4)x_1
POL(a(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)

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.