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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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


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

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

POL(C(x1)) = 0   
POL(c(x1)) = 0   
POL(B(x1)) = 1/4   
POL(a(x1)) = x_1   
POL(A(x1)) = (9/4)x_1   
POL(b(x1)) = 1/4 + x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(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 2 SCCs with 1 less node.

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

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

B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))

The TRS R consists of the following rules:

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

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

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

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

C(a(x1)) → C(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(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(a(x1)) → C(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (4)x_1   
POL(a(x1)) = 1 + (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.