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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.