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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

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

The TRS R consists of the following rules:

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

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