Termination w.r.t. Q proof of /tmp/tmpKNfQfr/z121.srs

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:

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = 5 + x₁
POL(C(x₁)) = 2 + x₁
POL(D(x₁)) = x₁
POL(F(x₁)) = 9 + x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 9 + x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 4 + x₁
POL(f(x₁)) = 14 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

f(f(x1)) → b(b(b(x1)))
a(f(x1)) → f(a(a(x1)))
b(b(x1)) → c(c(a(c(x1))))
d(b(x1)) → d(a(b(x1)))
c(c(x1)) → d(d(d(x1)))
b(d(x1)) → d(b(x1))
c(d(d(x1))) → f(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 5 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

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

D(b(x1)) → D(a(b(x1)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule D(b(x1)) → D(a(b(x1))) at position [0] we obtained the following new rules:

D(b(b(x0))) → D(a(c(c(a(c(x0))))))
D(b(d(x0))) → D(a(d(b(x0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

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

D(b(b(x0))) → D(a(c(c(a(c(x0))))))
D(b(d(x0))) → D(a(d(b(x0))))

The TRS R consists of the following rules:

f(f(x1)) → b(b(b(x1)))
a(f(x1)) → f(a(a(x1)))
b(b(x1)) → c(c(a(c(x1))))
d(b(x1)) → d(a(b(x1)))
c(c(x1)) → d(d(d(x1)))
b(d(x1)) → d(b(x1))
c(d(d(x1))) → f(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(b(d(x0))) → D(a(d(b(x0))))

The remaining pairs can at least be oriented weakly.

D(b(b(x0))) → D(a(c(c(a(c(x0))))))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( c(x₁) ) = 1

POL( f(x₁) ) = 1

POL( d(x₁) ) = max{0, -1}

POL( b(x₁) ) = 1

POL( D(x₁) ) = x₁

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

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

D(b(b(x0))) → D(a(c(c(a(c(x0))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

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

f(f(x1)) → b(b(b(x1)))
a(f(x1)) → f(a(a(x1)))
b(b(x1)) → c(c(a(c(x1))))
d(b(x1)) → d(a(b(x1)))
c(c(x1)) → d(d(d(x1)))
b(d(x1)) → d(b(x1))
c(d(d(x1))) → f(x1)
D(b(b(x0))) → D(a(c(c(a(c(x0))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

f(f(x1)) → b(b(b(x1)))
a(f(x1)) → f(a(a(x1)))
b(b(x1)) → c(c(a(c(x1))))
d(b(x1)) → d(a(b(x1)))
c(c(x1)) → d(d(d(x1)))
b(d(x1)) → d(b(x1))
c(d(d(x1))) → f(x1)
D(b(b(x0))) → D(a(c(c(a(c(x0))))))

The set Q is empty.
We have obtained the following QTRS:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

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

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

The set Q is empty.
We have obtained the following QTRS:

f(f(x)) → b(b(b(x)))
a(f(x)) → f(a(a(x)))
b(b(x)) → c(c(a(c(x))))
d(b(x)) → d(a(b(x)))
c(c(x)) → d(d(d(x)))
b(d(x)) → d(b(x))
c(d(d(x))) → f(x)
D(b(b(x))) → D(a(c(c(a(c(x))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

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

f(f(x)) → b(b(b(x)))
a(f(x)) → f(a(a(x)))
b(b(x)) → c(c(a(c(x))))
d(b(x)) → d(a(b(x)))
c(c(x)) → d(d(d(x)))
b(d(x)) → d(b(x))
c(d(d(x))) → f(x)
D(b(b(x))) → D(a(c(c(a(c(x))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

The set Q is empty.
We have obtained the following QTRS:

f(f(x)) → b(b(b(x)))
a(f(x)) → f(a(a(x)))
b(b(x)) → c(c(a(c(x))))
d(b(x)) → d(a(b(x)))
c(c(x)) → d(d(d(x)))
b(d(x)) → d(b(x))
c(d(d(x))) → f(x)
D(b(b(x))) → D(a(c(c(a(c(x))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                                ↳ QTRS

                              ↳ DependencyPairsProof

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

f(f(x)) → b(b(b(x)))
a(f(x)) → f(a(a(x)))
b(b(x)) → c(c(a(c(x))))
d(b(x)) → d(a(b(x)))
c(c(x)) → d(d(d(x)))
b(d(x)) → d(b(x))
c(d(d(x))) → f(x)
D(b(b(x))) → D(a(c(c(a(c(x))))))

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(b(D(x))) → C(a(D(x)))
B(b(x)) → C(a(c(c(x))))
C(c(x)) → D¹(d(x))
D¹(b(x)) → B(d(x))
B(d(x)) → B(a(d(x)))
D¹(b(x)) → D¹(x)
B(b(D(x))) → C(c(a(D(x))))
D¹(d(c(x))) → F(x)
F(a(x)) → F(x)
B(b(x)) → C(c(x))
C(c(x)) → D¹(x)
C(c(x)) → D¹(d(d(x)))
F(f(x)) → B(b(x))
F(f(x)) → B(x)
B(b(x)) → C(x)
B(b(D(x))) → C(a(c(c(a(D(x))))))
F(f(x)) → B(b(b(x)))

The TRS R consists of the following rules:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

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

B(b(D(x))) → C(a(D(x)))
B(b(x)) → C(a(c(c(x))))
C(c(x)) → D¹(d(x))
D¹(b(x)) → B(d(x))
B(d(x)) → B(a(d(x)))
D¹(b(x)) → D¹(x)
B(b(D(x))) → C(c(a(D(x))))
D¹(d(c(x))) → F(x)
F(a(x)) → F(x)
B(b(x)) → C(c(x))
C(c(x)) → D¹(x)
C(c(x)) → D¹(d(d(x)))
F(f(x)) → B(b(x))
F(f(x)) → B(x)
B(b(x)) → C(x)
B(b(D(x))) → C(a(c(c(a(D(x))))))
F(f(x)) → B(b(b(x)))

The TRS R consists of the following rules:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

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 4 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

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

D¹(b(x)) → B(d(x))
C(c(x)) → D¹(d(x))
D¹(b(x)) → D¹(x)
B(b(D(x))) → C(c(a(D(x))))
D¹(d(c(x))) → F(x)
F(a(x)) → F(x)
B(b(x)) → C(c(x))
C(c(x)) → D¹(x)
C(c(x)) → D¹(d(d(x)))
B(b(x)) → C(x)
F(f(x)) → B(b(x))
F(f(x)) → B(x)
F(f(x)) → B(b(b(x)))

The TRS R consists of the following rules:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

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.

D¹(b(x)) → B(d(x))
C(c(x)) → D¹(d(x))
D¹(b(x)) → D¹(x)
B(b(D(x))) → C(c(a(D(x))))
D¹(d(c(x))) → F(x)
B(b(x)) → C(c(x))
C(c(x)) → D¹(x)
C(c(x)) → D¹(d(d(x)))
B(b(x)) → C(x)
F(f(x)) → B(b(x))
F(f(x)) → B(x)

The remaining pairs can at least be oriented weakly.

F(a(x)) → F(x)
F(f(x)) → B(b(b(x)))

Used ordering: Polynomial interpretation [25]:

POL(B(x₁)) = 6 + 2·x₁
POL(C(x₁)) = 11 + 2·x₁
POL(D(x₁)) = 4 + 2·x₁
POL(D¹(x₁)) = 2·x₁
POL(F(x₁)) = 14 + 2·x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 9 + x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 4 + x₁
POL(f(x₁)) = 14 + x₁

The following usable rules [17] were oriented:

b(b(D(x))) → c(a(c(c(a(D(x))))))
d(d(c(x))) → f(x)
d(b(x)) → b(d(x))
c(c(x)) → d(d(d(x)))
b(d(x)) → b(a(d(x)))
b(b(x)) → c(a(c(c(x))))
f(a(x)) → a(a(f(x)))
f(f(x)) → b(b(b(x)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

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

F(f(x)) → B(b(b(x)))
F(a(x)) → F(x)

The TRS R consists of the following rules:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

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

F(a(x)) → F(x)

The TRS R consists of the following rules:

f(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
b(b(D(x))) → c(a(c(c(a(D(x))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

                                                ↳ QDP

                                                  ↳ UsableRulesReductionPairsProof

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

F(a(x)) → F(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F(a(x)) → F(x)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x₁)) = 2·x₁
POL(a(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

                                                ↳ QDP

                                                  ↳ UsableRulesReductionPairsProof

                                                    ↳ QDP

                                                      ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.