Termination w.r.t. Q proof of /tmp/tmpBcM-Wu/z003.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(c(a(x1))) → a(b(a(b(c(x1)))))
b(x1) → c(c(x1))
c(d(x1)) → a(b(c(a(x1))))
a(a(x1)) → a(c(b(a(x1))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

b(c(a(x1))) → a(b(a(b(c(x1)))))
b(x1) → c(c(x1))
c(d(x1)) → a(b(c(a(x1))))
a(a(x1)) → a(c(b(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 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 1 + (2)x_1
POL(c(x₁)) = x_1
POL(B(x₁)) = 4 + (4)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = 4 + (4)x_1
POL(d(x₁)) = 2 + (2)x_1
POL(b(x₁)) = x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

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

A(a(x1)) → A(c(b(a(x1))))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

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

B(c(a(x1))) → B(c(x1))

The TRS R consists of the following rules:

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

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