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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → b(a(c(x1)))
b(x1) → x1
c(c(x1)) → b(a(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

A(b(c(x0))) → A(b(a(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → c(a(b(x)))
b(x) → x
c(c(x)) → a(b(x))
c(C(x)) → A(x)
c(b(A(x))) → a(b(A(x)))
b(A(x)) → C(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 3 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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