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.