Termination w.r.t. Q proof of /tmp/tmp53iCXN/secr9.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 11 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

C(c(x1)) → C(b(c(b(c(x1)))))
C(c(c(x1))) → B(x1)
C(c(c(x1))) → B(b(x1))
B(b(b(x1))) → C(b(x1))
C(c(c(x1))) → C(b(b(x1)))
C(c(x1)) → C(b(c(x1)))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

D(d(x1)) → D(b(d(x1)))
D(d(x1)) → D(b(d(b(d(x1)))))

The TRS R consists of the following rules:

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

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

POL(c(x₁)) = 0
POL(D(x₁)) = (5/4)x_1
POL(d(x₁)) = 15/4 + (1/4)x_1
POL(b(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 3525/1024.
The following usable rules [17] were oriented:

d(d(x1)) → d(b(d(b(d(x1)))))
b(b(b(x1))) → c(b(x1))
c(c(x1)) → c(b(c(b(c(x1)))))
c(c(c(x1))) → c(b(b(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

A(a(x1)) → A(d(a(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 7/4 + (4)x_1
POL(a(x₁)) = (9/4)x_1
POL(A(x₁)) = x_1
POL(d(x₁)) = 0
POL(b(x₁)) = 4 + (3/2)x_1

The value of delta used in the strict ordering is 7/4.
The following usable rules [17] were oriented:

d(d(x1)) → d(b(d(b(d(x1)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

A(a(x1)) → A(d(a(x1)))

The TRS R consists of the following rules:

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

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

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

The value of delta used in the strict ordering is 7.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

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