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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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