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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(b(b(a(x1)))) → a(c(a(b(x1))))
a(c(x1)) → c(c(a(x1)))
c(c(c(x1))) → b(c(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 1 SCC with 4 less nodes.

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

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

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

The TRS R consists of the following rules:

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


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

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

POL(c(x1)) = x_1   
POL(a(x1)) = 1/2 + (4)x_1   
POL(A(x1)) = (1/4)x_1   
POL(b(x1)) = x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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