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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

A(b(c(b(c(x0))))) → A(c(c(b(a(c(a(x0)))))))
A(b(c(x0))) → A(c(b(x0)))
A(b(c(x0))) → A(c(x0))
A(b(c(y0))) → A(a(y0))



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

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

A(b(c(x0))) → A(c(b(x0)))
A(b(c(b(c(x0))))) → A(c(c(b(a(c(a(x0)))))))
A(b(c(y0))) → A(a(y0))
A(b(c(x0))) → A(c(x0))
A(b(c(x1))) → A(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(c(x0))) → A(c(b(x0))) at position [0] we obtained the following new rules:

A(b(c(y0))) → A(b(y0))



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

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

A(b(c(b(c(x0))))) → A(c(c(b(a(c(a(x0)))))))
A(b(c(y0))) → A(b(y0))
A(b(c(x1))) → A(x1)
A(b(c(x0))) → A(c(x0))
A(b(c(y0))) → A(a(y0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(c(x0))) → A(c(x0)) at position [0] we obtained the following new rules:

A(b(c(x0))) → A(x0)



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

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

A(b(c(b(c(x0))))) → A(c(c(b(a(c(a(x0)))))))
A(b(c(y0))) → A(b(y0))
A(b(c(y0))) → A(a(y0))
A(b(c(x1))) → A(x1)

The TRS R consists of the following rules:

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.