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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



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

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

C(c(x1)) → A(c(x1))
A(b(b(x0))) → C(b(c(a(x0))))
A(b(x0)) → C(x0)
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

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

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

C(c(x1)) → A(c(x1))
A(b(x0)) → C(x0)
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

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

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

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



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

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

C(c(x0)) → A(b(x0))
A(b(x0)) → C(x0)
C(c(c(x0))) → A(a(c(x0)))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(x1) → b(x1)
c(c(x1)) → a(c(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
      ↳ Narrowing
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ 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(a(x1)))
c(x1) → b(x1)
c(c(x1)) → a(c(x1))
C(c(x0)) → A(b(x0))
A(b(x0)) → C(x0)
C(c(c(x0))) → A(a(c(x0)))
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(a(x1)))
c(x1) → b(x1)
c(c(x1)) → a(c(x1))
C(c(x0)) → A(b(x0))
A(b(x0)) → C(x0)
C(c(c(x0))) → A(a(c(x0)))
A(b(x1)) → A(x1)

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ 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(c(b(x)))
c(x) → b(x)
c(c(x)) → c(a(x))
c(C(x)) → b(A(x))
b(A(x)) → C(x)
c(c(C(x))) → c(a(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:

B(a(x)) → A1(c(b(x)))
C1(x) → B(x)
C1(c(x)) → A1(x)
C1(c(C(x))) → C1(a(A(x)))
B(a(x)) → B(x)
C1(C(x)) → B(A(x))
B(a(x)) → C1(b(x))
C1(c(C(x))) → A1(A(x))
C1(c(x)) → C1(a(x))

The TRS R consists of the following rules:

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

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

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

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

B(a(x)) → A1(c(b(x)))
C1(x) → B(x)
C1(c(x)) → A1(x)
C1(c(C(x))) → C1(a(A(x)))
B(a(x)) → B(x)
C1(C(x)) → B(A(x))
B(a(x)) → C1(b(x))
C1(c(C(x))) → A1(A(x))
C1(c(x)) → C1(a(x))

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(x)))
c(x) → b(x)
c(c(x)) → c(a(x))
c(C(x)) → b(A(x))
b(A(x)) → C(x)
c(c(C(x))) → c(a(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 4 less nodes.

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

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

C1(x) → B(x)
C1(c(C(x))) → C1(a(A(x)))
B(a(x)) → B(x)
B(a(x)) → C1(b(x))
C1(c(x)) → C1(a(x))

The TRS R consists of the following rules:

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

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

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

Q is empty.