Termination w.r.t. Q proof of /tmp/tmpEuf70o/aprove5.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(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(x1))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(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(a(x1)) → D(d(c(x1)))
A(a(x1)) → A(d(d(d(x1))))
A(b(x1)) → B(c(a(x1)))
B(d(x1)) → D(d(x1))
A(d(d(c(x1)))) → D(x1)
A(c(x1)) → B(x1)
A(b(x1)) → A(x1)
B(c(x1)) → B(b(x1))
E(x1) → A(x1)
D(a(x1)) → D(c(x1))
E(e(f(f(x1)))) → E(x1)
A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(a(x1)) → D(d(x1))
E(e(f(f(x1)))) → E(e(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(a(x1)) → D(d(d(x1)))
A(a(x1)) → D(x1)
A(d(d(c(x1)))) → A(a(d(x1)))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(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(a(x1)) → D(d(c(x1)))
A(a(x1)) → A(d(d(d(x1))))
A(b(x1)) → B(c(a(x1)))
B(d(x1)) → D(d(x1))
A(d(d(c(x1)))) → D(x1)
A(c(x1)) → B(x1)
A(b(x1)) → A(x1)
B(c(x1)) → B(b(x1))
E(x1) → A(x1)
D(a(x1)) → D(c(x1))
E(e(f(f(x1)))) → E(x1)
A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(a(x1)) → D(d(x1))
E(e(f(f(x1)))) → E(e(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(a(x1)) → D(d(d(x1)))
A(a(x1)) → D(x1)
A(d(d(c(x1)))) → A(a(d(x1)))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(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 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(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(x1)) → B(b(x1))
B(c(x1)) → B(x1)

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

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

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

E(e(f(f(x1)))) → E(x1)
E(e(f(f(x1)))) → E(e(x1))

The TRS R consists of the following rules:

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

E(e(f(f(x1)))) → E(x1)

The remaining pairs can at least be oriented weakly.

E(e(f(f(x1)))) → E(e(x1))

Used ordering: Polynomial interpretation [25,35]:

POL(E(x₁)) = (4)x_1
POL(f(x₁)) = x_1
POL(c(x₁)) = (1/4)x_1
POL(a(x₁)) = 0
POL(e(x₁)) = 1/4 + (4)x_1
POL(b(x₁)) = (1/4)x_1
POL(d(x₁)) = 0

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

E(e(f(f(x1)))) → E(e(x1))

The TRS R consists of the following rules:

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

E(e(f(f(x1)))) → E(e(x1))

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

POL(E(x₁)) = (4)x_1
POL(f(x₁)) = 1 + x_1
POL(c(x₁)) = 0
POL(a(x₁)) = 0
POL(e(x₁)) = (4)x_1
POL(b(x₁)) = 0
POL(d(x₁)) = 0

The value of delta used in the strict ordering is 32.
The following usable rules [17] were oriented:

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