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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → b(c(b(a(x1))))
b(x1) → a(x1)
c(c(x1)) → 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
          ↳ QDPToSRSProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

B1(A(x)) → C(B(x))
B1(a(x)) → B1(x)
B1(a(x)) → A1(b(c(b(x))))
B1(a(x)) → C(b(x))
B1(a(x)) → B1(c(b(x)))
B1(A(x)) → A1(B(x))
B1(A(x)) → B1(c(B(x)))
B1(A(x)) → A1(b(c(B(x))))
B1(x) → A1(x)
B1(A(x)) → B2(x)

The TRS R consists of the following rules:

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

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

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

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

B1(A(x)) → C(B(x))
B1(a(x)) → B1(x)
B1(a(x)) → A1(b(c(b(x))))
B1(a(x)) → C(b(x))
B1(a(x)) → B1(c(b(x)))
B1(A(x)) → A1(B(x))
B1(A(x)) → B1(c(B(x)))
B1(A(x)) → A1(b(c(B(x))))
B1(x) → A1(x)
B1(A(x)) → B2(x)

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(b(c(b(x))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(B(x))
b(A(x)) → a(b(c(B(x))))
B(x) → A(x)
b(A(x)) → A(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 7 less nodes.

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

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

B1(a(x)) → B1(x)
B1(a(x)) → B1(c(b(x)))
B1(A(x)) → B1(c(B(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(x)) → B1(c(B(x))) at position [0] we obtained the following new rules:

B1(A(x0)) → B1(c(A(x0)))



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

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

B1(A(x0)) → B1(c(A(x0)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(b(x)))

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(b(c(b(x))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(B(x))
b(A(x)) → a(b(c(B(x))))
B(x) → A(x)
b(A(x)) → A(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
  ↳ QTRS Reverse

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

B1(a(x)) → B1(x)
B1(a(x)) → B1(c(b(x)))

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.