Termination w.r.t. Q proof of /tmp/tmpxLrqa

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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

D(a(x1)) → B(d(x1))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (1/4)x_1
POL(c(x₁)) = 1/4 + x_1
POL(D(x₁)) = (3/4)x_1
POL(B(x₁)) = 3/4 + (1/4)x_1
POL(a(x₁)) = 1 + x_1
POL(d(x₁)) = (3)x_1
POL(b(x₁)) = 3 + x_1

The value of delta used in the strict ordering is 3/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

D(a(x1)) → B(d(x1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.