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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(x1) → B(x1) we obtained the following new rules:

A(a(y_2)) → B(a(y_2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
QDP
          ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(a(a(x1))))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(x1)))) → A(x1)
A(a(y_2)) → B(a(y_2))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(a(b(c(x1)))) → A(x1) we obtained the following new rules:

B(a(b(c(a(y_0))))) → A(a(y_0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
QDP
              ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(x1)))) → A(a(a(a(x1))))
B(a(b(c(a(y_0))))) → A(a(y_0))
A(a(y_2)) → B(a(y_2))

The TRS R consists of the following rules:

a(x1) → b(x1)
b(a(b(c(x1)))) → c(c(a(a(a(a(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
QTRS
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(x1) → b(x1)
b(a(b(c(x1)))) → c(c(a(a(a(a(x1))))))
B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(x1)))) → A(a(a(a(x1))))
B(a(b(c(a(y_0))))) → A(a(y_0))
A(a(y_2)) → B(a(y_2))

Q is empty.

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

a(x1) → b(x1)
b(a(b(c(x1)))) → c(c(a(a(a(a(x1))))))
B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(x1)))) → A(a(a(a(x1))))
B(a(b(c(a(y_0))))) → A(a(y_0))
A(a(y_2)) → B(a(y_2))

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                      ↳ DependencyPairsProof
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(x) → b(x)
b(a(b(c(x)))) → c(c(a(a(a(a(x))))))
B(a(b(c(x)))) → A(a(a(x)))
B(a(b(c(x)))) → A(a(x))
B(a(b(c(x)))) → A(a(a(a(x))))
B(a(b(c(a(x))))) → A(a(x))
A(a(x)) → B(a(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:

C(b(a(B(x)))) → A1(a(a(A(x))))
C(b(a(b(x)))) → A1(a(a(a(c(c(x))))))
C(b(a(b(x)))) → A1(a(a(c(c(x)))))
C(b(a(b(x)))) → A1(a(c(c(x))))
A1(A(x)) → A1(B(x))
C(b(a(b(x)))) → C(c(x))
A1(c(b(a(B(x))))) → A1(A(x))
C(b(a(b(x)))) → C(x)
C(b(a(B(x)))) → A1(A(x))
C(b(a(b(x)))) → A1(c(c(x)))
C(b(a(B(x)))) → A1(a(A(x)))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
QDP
                          ↳ DependencyGraphProof
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(a(B(x)))) → A1(a(a(A(x))))
C(b(a(b(x)))) → A1(a(a(a(c(c(x))))))
C(b(a(b(x)))) → A1(a(a(c(c(x)))))
C(b(a(b(x)))) → A1(a(c(c(x))))
A1(A(x)) → A1(B(x))
C(b(a(b(x)))) → C(c(x))
A1(c(b(a(B(x))))) → A1(A(x))
C(b(a(b(x)))) → C(x)
C(b(a(B(x)))) → A1(A(x))
C(b(a(b(x)))) → A1(c(c(x)))
C(b(a(B(x)))) → A1(a(A(x)))

The TRS R consists of the following rules:

a(x) → b(x)
c(b(a(b(x)))) → a(a(a(a(c(c(x))))))
c(b(a(B(x)))) → a(a(A(x)))
c(b(a(B(x)))) → a(A(x))
c(b(a(B(x)))) → a(a(a(A(x))))
a(c(b(a(B(x))))) → a(A(x))
a(A(x)) → a(B(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 9 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
QDP
                              ↳ Narrowing
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(a(b(x)))) → C(c(x))
C(b(a(b(x)))) → C(x)

The TRS R consists of the following rules:

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

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

C(b(a(b(b(a(B(x0))))))) → C(a(A(x0)))
C(b(a(b(b(a(B(x0))))))) → C(a(a(A(x0))))
C(b(a(b(b(a(B(x0))))))) → C(a(a(a(A(x0)))))
C(b(a(b(b(a(b(x0))))))) → C(a(a(a(a(c(c(x0)))))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
QDP
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(a(b(b(a(B(x0))))))) → C(a(A(x0)))
C(b(a(b(b(a(b(x0))))))) → C(a(a(a(a(c(c(x0)))))))
C(b(a(b(b(a(B(x0))))))) → C(a(a(A(x0))))
C(b(a(b(x)))) → C(x)
C(b(a(b(b(a(B(x0))))))) → C(a(a(a(A(x0)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                      ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

a(x) → b(x)
c(b(a(b(x)))) → a(a(a(a(c(c(x))))))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

a(x) → b(x)
c(b(a(b(x)))) → a(a(a(a(c(c(x))))))

Q is empty.

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

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

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

a(x) → b(x)
c(b(a(b(x)))) → a(a(a(a(c(c(x))))))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

a(x) → b(x)
c(b(a(b(x)))) → a(a(a(a(c(c(x))))))

Q is empty.