Termination w.r.t. Q proof of /tmp/tmpjeAvhl/beans3.srs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(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:

A(d(x1)) → A(x1)
D(x1) → A(x1)
L¹(a(a(x1))) → A(b(c(x1)))
C(a(x1)) → C(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(c(a(x1))) → A(b(c(x1)))
C(R(x1)) → C(b(R(x1)))
C(b(x1)) → D(x1)
L¹(a(a(x1))) → C(x1)
C(R(x1)) → B(R(x1))
L¹(a(a(x1))) → L¹(a(b(c(x1))))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
L¹(a(a(x1))) → B(c(x1))
B(a(a(x1))) → A(b(c(x1)))
B(c(a(x1))) → B(c(x1))
B(c(a(x1))) → C(x1)
C(a(x1)) → A(c(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(x1)))

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:

A(d(x1)) → A(x1)
D(x1) → A(x1)
L¹(a(a(x1))) → A(b(c(x1)))
C(a(x1)) → C(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(c(a(x1))) → A(b(c(x1)))
C(R(x1)) → C(b(R(x1)))
C(b(x1)) → D(x1)
L¹(a(a(x1))) → C(x1)
C(R(x1)) → B(R(x1))
L¹(a(a(x1))) → L¹(a(b(c(x1))))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
L¹(a(a(x1))) → B(c(x1))
B(a(a(x1))) → A(b(c(x1)))
B(c(a(x1))) → B(c(x1))
B(c(a(x1))) → C(x1)
C(a(x1)) → A(c(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

A(d(x1)) → A(x1)
D(x1) → A(x1)
C(a(x1)) → C(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(c(a(x1))) → A(b(c(x1)))
C(R(x1)) → C(b(R(x1)))
C(b(x1)) → D(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(a(a(x1))) → A(b(c(x1)))
B(c(a(x1))) → B(c(x1))
B(c(a(x1))) → C(x1)
C(a(x1)) → A(c(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(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.

A(d(x1)) → A(x1)
C(a(x1)) → C(x1)
C(b(x1)) → D(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(c(a(x1))) → B(c(x1))
B(c(a(x1))) → C(x1)
C(a(x1)) → A(c(x1))

The remaining pairs can at least be oriented weakly.

D(x1) → A(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(c(a(x1))) → A(b(c(x1)))
C(R(x1)) → C(b(R(x1)))
B(a(a(x1))) → A(b(c(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 9/4 + (9/4)x_1
POL(c(x₁)) = 1/2 + (3/2)x_1
POL(D(x₁)) = 2 + (9/4)x_1
POL(B(x₁)) = 5/4 + (3/2)x_1
POL(a(x₁)) = 1/2 + (3/2)x_1
POL(A(x₁)) = 2 + (9/4)x_1
POL(d(x₁)) = 1/2 + (3/2)x_1
POL(b(x₁)) = x_1
POL(R(x₁)) = 0

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

c(a(x1)) → a(c(x1))
c(b(x1)) → d(x1)
d(x1) → b(a(x1))
b(a(a(x1))) → a(b(c(x1)))
b(c(a(x1))) → a(b(c(x1)))
a(d(x1)) → d(a(x1))
c(R(x1)) → c(b(R(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

D(x1) → A(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(a(a(x1))) → A(b(c(x1)))
B(c(a(x1))) → A(b(c(x1)))
C(R(x1)) → C(b(R(x1)))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

D(x1) → A(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(c(a(x1))) → A(b(c(x1)))
B(a(a(x1))) → A(b(c(x1)))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(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.

B(c(a(x1))) → A(b(c(x1)))

The remaining pairs can at least be oriented weakly.

D(x1) → A(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(a(a(x1))) → A(b(c(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 4
POL(D(x₁)) = 0
POL(B(x₁)) = (1/2)x_1
POL(a(x₁)) = 0
POL(A(x₁)) = 0
POL(d(x₁)) = 0
POL(b(x₁)) = 0
POL(R(x₁)) = 2 + (1/4)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

c(a(x1)) → a(c(x1))
c(b(x1)) → d(x1)
d(x1) → b(a(x1))
b(a(a(x1))) → a(b(c(x1)))
b(c(a(x1))) → a(b(c(x1)))
a(d(x1)) → d(a(x1))
c(R(x1)) → c(b(R(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

          ↳ QDP

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

D(x1) → A(x1)
D(x1) → B(a(x1))
A(d(x1)) → D(a(x1))
B(a(a(x1))) → A(b(c(x1)))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(x1)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(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.

L¹(a(a(x1))) → L¹(a(b(c(x1))))

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

POL(c(x₁)) = 1/2 + (4)x_1
POL(L¹(x₁)) = (1/2)x_1
POL(a(x₁)) = 1 + (4)x_1
POL(b(x₁)) = (1/2)x_1
POL(d(x₁)) = 1/2 + (2)x_1
POL(R(x₁)) = (3/4)x_1

The value of delta used in the strict ordering is 3/2.
The following usable rules [17] were oriented:

c(a(x1)) → a(c(x1))
c(b(x1)) → d(x1)
d(x1) → b(a(x1))
b(a(a(x1))) → a(b(c(x1)))
b(c(a(x1))) → a(b(c(x1)))
a(d(x1)) → d(a(x1))
c(R(x1)) → c(b(R(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → d(x1)
a(d(x1)) → d(a(x1))
d(x1) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → c(b(R(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.