Termination w.r.t. Q proof of /tmp/tmp19G0Nh/dup12.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(b(d(d(b(b(x1)))))) → c(c(d(d(b(b(x1))))))
b(b(a(a(c(c(x1)))))) → b(b(c(c(x1))))
a(a(d(d(x1)))) → d(d(c(c(x1))))
b(b(b(b(b(b(x1)))))) → a(a(b(b(c(c(x1))))))
d(d(c(c(x1)))) → b(b(d(d(x1))))
d(d(c(c(x1)))) → d(d(b(b(d(d(x1))))))
d(d(a(a(c(c(x1)))))) → b(b(b(b(x1))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

b(b(d(d(b(b(x1)))))) → c(c(d(d(b(b(x1))))))
b(b(a(a(c(c(x1)))))) → b(b(c(c(x1))))
a(a(d(d(x1)))) → d(d(c(c(x1))))
b(b(b(b(b(b(x1)))))) → a(a(b(b(c(c(x1))))))
d(d(c(c(x1)))) → b(b(d(d(x1))))
d(d(c(c(x1)))) → d(d(b(b(d(d(x1))))))
d(d(a(a(c(c(x1)))))) → b(b(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 1 SCC with 14 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 1/2 + (2)x_1
POL(D(x₁)) = (2)x_1
POL(a(x₁)) = 1/2 + (2)x_1
POL(d(x₁)) = x_1
POL(b(x₁)) = 1/2 + (2)x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

D(d(c(c(x1)))) → D(d(b(b(d(d(x1))))))

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 0
POL(D(x₁)) = x_1
POL(a(x₁)) = 0
POL(d(x₁)) = 1/4
POL(b(x₁)) = 0

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

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

D(d(c(c(x1)))) → D(d(b(b(d(d(x1))))))

The TRS R consists of the following rules:

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

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