Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

B2(A(x1)) → C2(x1)
A1(a(a(a(a(x1))))) → A2(A(A(x1)))
B2(B(B(c(x1)))) → A2(x1)
B2(B(B(B(x1)))) → B1(x1)
B2(a(a(a(a(x1))))) → A2(A(A(x1)))
A2(A(A(A(x1)))) → A1(a(a(x1)))
C1(C(C(a(x1)))) → C2(B(x1))
C1(B(x1)) → A1(x1)
C1(b(b(b(b(x1))))) → B2(x1)
B2(B(B(B(x1)))) → B1(b(x1))
C1(b(b(b(b(x1))))) → B2(B(x1))
C2(c(c(c(c(x1))))) → C1(C(C(x1)))
A2(A(A(b(x1)))) → A1(a(C(x1)))
C1(C(C(C(x1)))) → C2(x1)
B2(B(B(c(x1)))) → B1(b(b(A(x1))))
B1(b(b(b(b(x1))))) → B2(B(x1))
B2(B(B(B(x1)))) → B1(b(b(b(x1))))
A2(c(c(c(c(x1))))) → C1(x1)
C1(C(C(a(x1)))) → C2(c(c(B(x1))))
C2(c(c(c(c(x1))))) → C1(x1)
C1(b(b(b(b(x1))))) → A1(B(B(B(x1))))
A2(A(A(b(x1)))) → C1(x1)
A1(b(x1)) → C1(x1)
A2(c(c(c(c(x1))))) → C1(C(C(x1)))
A2(A(A(A(x1)))) → A1(x1)
B2(a(a(a(a(x1))))) → A2(x1)
B2(a(a(a(a(x1))))) → A2(A(x1))
C1(C(C(C(x1)))) → C2(c(c(c(x1))))
C1(C(C(C(x1)))) → C2(c(c(x1)))
C1(C(C(a(x1)))) → C2(c(c(c(B(x1)))))
B1(c(x1)) → A2(x1)
A2(A(A(b(x1)))) → A1(a(a(C(x1))))
A2(A(A(A(x1)))) → A1(a(x1))
A2(A(A(b(x1)))) → A1(C(x1))
A2(C(x1)) → B1(x1)
B1(b(b(b(b(x1))))) → B2(x1)
C2(a(x1)) → B2(x1)
C2(c(c(c(c(x1))))) → C1(C(x1))
A2(c(c(c(c(x1))))) → B1(C(C(C(x1))))
B2(a(a(a(a(x1))))) → C2(A(A(A(x1))))
C1(C(C(a(x1)))) → B2(x1)
A1(a(a(a(a(x1))))) → A2(x1)
B2(B(B(c(x1)))) → B1(b(A(x1)))
B2(B(B(c(x1)))) → B1(A(x1))
C1(b(b(b(b(x1))))) → B2(B(B(x1)))
B2(B(B(B(x1)))) → B1(b(b(x1)))
A2(c(c(c(c(x1))))) → C1(C(x1))
A2(A(A(b(x1)))) → A1(a(a(a(C(x1)))))
C1(C(C(a(x1)))) → C2(c(B(x1)))
B1(b(b(b(b(x1))))) → B2(B(B(x1)))
A1(a(a(a(a(x1))))) → A2(A(x1))
B2(B(B(c(x1)))) → B1(b(b(b(A(x1)))))
A2(A(A(A(x1)))) → A1(a(a(a(x1))))
C1(C(C(C(x1)))) → C2(c(x1))

The TRS R consists of the following rules:

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

B2(A(x1)) → C2(x1)
A1(a(a(a(a(x1))))) → A2(A(A(x1)))
B2(B(B(c(x1)))) → A2(x1)
B2(B(B(B(x1)))) → B1(x1)
B2(a(a(a(a(x1))))) → A2(A(A(x1)))
A2(A(A(A(x1)))) → A1(a(a(x1)))
C1(C(C(a(x1)))) → C2(B(x1))
C1(B(x1)) → A1(x1)
C1(b(b(b(b(x1))))) → B2(x1)
B2(B(B(B(x1)))) → B1(b(x1))
C1(b(b(b(b(x1))))) → B2(B(x1))
C2(c(c(c(c(x1))))) → C1(C(C(x1)))
A2(A(A(b(x1)))) → A1(a(C(x1)))
C1(C(C(C(x1)))) → C2(x1)
B2(B(B(c(x1)))) → B1(b(b(A(x1))))
B1(b(b(b(b(x1))))) → B2(B(x1))
B2(B(B(B(x1)))) → B1(b(b(b(x1))))
A2(c(c(c(c(x1))))) → C1(x1)
C1(C(C(a(x1)))) → C2(c(c(B(x1))))
C2(c(c(c(c(x1))))) → C1(x1)
C1(b(b(b(b(x1))))) → A1(B(B(B(x1))))
A2(A(A(b(x1)))) → C1(x1)
A1(b(x1)) → C1(x1)
A2(c(c(c(c(x1))))) → C1(C(C(x1)))
A2(A(A(A(x1)))) → A1(x1)
B2(a(a(a(a(x1))))) → A2(x1)
B2(a(a(a(a(x1))))) → A2(A(x1))
C1(C(C(C(x1)))) → C2(c(c(c(x1))))
C1(C(C(C(x1)))) → C2(c(c(x1)))
C1(C(C(a(x1)))) → C2(c(c(c(B(x1)))))
B1(c(x1)) → A2(x1)
A2(A(A(b(x1)))) → A1(a(a(C(x1))))
A2(A(A(A(x1)))) → A1(a(x1))
A2(A(A(b(x1)))) → A1(C(x1))
A2(C(x1)) → B1(x1)
B1(b(b(b(b(x1))))) → B2(x1)
C2(a(x1)) → B2(x1)
C2(c(c(c(c(x1))))) → C1(C(x1))
A2(c(c(c(c(x1))))) → B1(C(C(C(x1))))
B2(a(a(a(a(x1))))) → C2(A(A(A(x1))))
C1(C(C(a(x1)))) → B2(x1)
A1(a(a(a(a(x1))))) → A2(x1)
B2(B(B(c(x1)))) → B1(b(A(x1)))
B2(B(B(c(x1)))) → B1(A(x1))
C1(b(b(b(b(x1))))) → B2(B(B(x1)))
B2(B(B(B(x1)))) → B1(b(b(x1)))
A2(c(c(c(c(x1))))) → C1(C(x1))
A2(A(A(b(x1)))) → A1(a(a(a(C(x1)))))
C1(C(C(a(x1)))) → C2(c(B(x1)))
B1(b(b(b(b(x1))))) → B2(B(B(x1)))
A1(a(a(a(a(x1))))) → A2(A(x1))
B2(B(B(c(x1)))) → B1(b(b(b(A(x1)))))
A2(A(A(A(x1)))) → A1(a(a(a(x1))))
C1(C(C(C(x1)))) → C2(c(x1))

The TRS R consists of the following rules:

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


B2(A(x1)) → C2(x1)
B2(B(B(c(x1)))) → A2(x1)
B2(B(B(B(x1)))) → B1(x1)
B2(a(a(a(a(x1))))) → A2(A(A(x1)))
A2(A(A(A(x1)))) → A1(a(a(x1)))
C1(C(C(a(x1)))) → C2(B(x1))
C1(B(x1)) → A1(x1)
C1(b(b(b(b(x1))))) → B2(x1)
B2(B(B(B(x1)))) → B1(b(x1))
C1(b(b(b(b(x1))))) → B2(B(x1))
C2(c(c(c(c(x1))))) → C1(C(C(x1)))
A2(A(A(b(x1)))) → A1(a(C(x1)))
C1(C(C(C(x1)))) → C2(x1)
B2(B(B(c(x1)))) → B1(b(b(A(x1))))
B1(b(b(b(b(x1))))) → B2(B(x1))
B2(B(B(B(x1)))) → B1(b(b(b(x1))))
A2(c(c(c(c(x1))))) → C1(x1)
C1(C(C(a(x1)))) → C2(c(c(B(x1))))
C2(c(c(c(c(x1))))) → C1(x1)
C1(b(b(b(b(x1))))) → A1(B(B(B(x1))))
A2(A(A(b(x1)))) → C1(x1)
A2(c(c(c(c(x1))))) → C1(C(C(x1)))
A2(A(A(A(x1)))) → A1(x1)
B2(a(a(a(a(x1))))) → A2(x1)
B2(a(a(a(a(x1))))) → A2(A(x1))
C1(C(C(C(x1)))) → C2(c(c(c(x1))))
C1(C(C(C(x1)))) → C2(c(c(x1)))
B1(c(x1)) → A2(x1)
A2(A(A(b(x1)))) → A1(a(a(C(x1))))
A2(A(A(A(x1)))) → A1(a(x1))
A2(A(A(b(x1)))) → A1(C(x1))
A2(C(x1)) → B1(x1)
B1(b(b(b(b(x1))))) → B2(x1)
C2(a(x1)) → B2(x1)
C2(c(c(c(c(x1))))) → C1(C(x1))
C1(C(C(a(x1)))) → B2(x1)
A1(a(a(a(a(x1))))) → A2(x1)
B2(B(B(c(x1)))) → B1(b(A(x1)))
B2(B(B(c(x1)))) → B1(A(x1))
C1(b(b(b(b(x1))))) → B2(B(B(x1)))
B2(B(B(B(x1)))) → B1(b(b(x1)))
A2(c(c(c(c(x1))))) → C1(C(x1))
A2(A(A(b(x1)))) → A1(a(a(a(C(x1)))))
C1(C(C(a(x1)))) → C2(c(B(x1)))
B1(b(b(b(b(x1))))) → B2(B(B(x1)))
A1(a(a(a(a(x1))))) → A2(A(x1))
A2(A(A(A(x1)))) → A1(a(a(a(x1))))
C1(C(C(C(x1)))) → C2(c(x1))
The remaining pairs can at least be oriented weakly.

A1(a(a(a(a(x1))))) → A2(A(A(x1)))
A1(b(x1)) → C1(x1)
C1(C(C(a(x1)))) → C2(c(c(c(B(x1)))))
A2(c(c(c(c(x1))))) → B1(C(C(C(x1))))
B2(a(a(a(a(x1))))) → C2(A(A(A(x1))))
B2(B(B(c(x1)))) → B1(b(b(b(A(x1)))))
Used ordering: Polynomial interpretation [25,35]:

POL(B2(x1)) = 2 + x_1   
POL(C(x1)) = 3 + x_1   
POL(c(x1)) = 2 + x_1   
POL(A1(x1)) = x_1   
POL(B(x1)) = 3 + x_1   
POL(C2(x1)) = 1 + x_1   
POL(a(x1)) = 2 + x_1   
POL(A(x1)) = 3 + x_1   
POL(B1(x1)) = 1 + x_1   
POL(b(x1)) = 2 + x_1   
POL(A2(x1)) = 2 + x_1   
POL(C1(x1)) = 2 + x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

B2(a(a(a(a(x1))))) → C2(A(A(A(x1))))
A1(a(a(a(a(x1))))) → A2(A(A(x1)))
B2(B(B(c(x1)))) → B1(b(b(b(A(x1)))))
C1(C(C(a(x1)))) → C2(c(c(c(B(x1)))))
A2(c(c(c(c(x1))))) → B1(C(C(C(x1))))
A1(b(x1)) → C1(x1)

The TRS R consists of the following rules:

a(b(x1)) → C(x1)
b(c(x1)) → A(x1)
c(a(x1)) → B(x1)
A(C(x1)) → b(x1)
C(B(x1)) → a(x1)
B(A(x1)) → c(x1)
a(a(a(a(a(x1))))) → A(A(A(x1)))
A(A(A(A(x1)))) → a(a(a(a(x1))))
b(b(b(b(b(x1))))) → B(B(B(x1)))
B(B(B(B(x1)))) → b(b(b(b(x1))))
c(c(c(c(c(x1))))) → C(C(C(x1)))
C(C(C(C(x1)))) → c(c(c(c(x1))))
B(a(a(a(a(x1))))) → c(A(A(A(x1))))
A(A(A(b(x1)))) → a(a(a(a(C(x1)))))
C(b(b(b(b(x1))))) → a(B(B(B(x1))))
B(B(B(c(x1)))) → b(b(b(b(A(x1)))))
A(c(c(c(c(x1))))) → b(C(C(C(x1))))
C(C(C(a(x1)))) → c(c(c(c(B(x1)))))
a(A(x1)) → x1
A(a(x1)) → x1
b(B(x1)) → x1
B(b(x1)) → x1
c(C(x1)) → x1
C(c(x1)) → 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 6 less nodes.