Termination w.r.t. Q proof of /tmp/tmpuPisrL/3.srs

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (4)x_1
POL(c(x₁)) = 1/4 + (4)x_1
POL(B(x₁)) = (4)x_1
POL(a(x₁)) = 1/4 + (4)x_1
POL(A(x₁)) = (4)x_1
POL(b(x₁)) = 1/4 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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