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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

C(c(x1)) → B(x1)
B(x1) → C(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


C(c(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.

B(x1) → C(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (4)x_1   
POL(c(x1)) = 2 + (4)x_1   
POL(B(x1)) = (4)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
          ↳ QDP

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

B(x1) → C(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

A(b(x1)) → A(a(x1))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

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

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