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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


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

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

POL(C(x1)) = (2)x_1   
POL(B2(x1)) = x_1   
POL(c(x1)) = x_1   
POL(A1(x1)) = x_1   
POL(B(x1)) = x_1   
POL(a(x1)) = x_1   
POL(C2(x1)) = (2)x_1   
POL(B1(x1)) = x_1   
POL(A(x1)) = 2 + (4)x_1   
POL(b(x1)) = x_1   
POL(A2(x1)) = 2 + (4)x_1   
POL(C1(x1)) = x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

a(b(c(x1))) → c(b(a(x1)))
C(B(A(x1))) → A(B(C(x1)))
b(a(C(x1))) → C(a(b(x1)))
c(A(B(x1))) → B(A(c(x1)))
A(c(b(x1))) → b(c(A(x1)))
B(C(a(x1))) → a(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 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

POL(B2(x1)) = 3 + (4)x_1   
POL(C(x1)) = 2 + (4)x_1   
POL(c(x1)) = x_1   
POL(A1(x1)) = 3 + (4)x_1   
POL(B(x1)) = x_1   
POL(a(x1)) = x_1   
POL(A(x1)) = x_1   
POL(B1(x1)) = 3 + x_1   
POL(b(x1)) = x_1   
POL(A2(x1)) = 3 + (4)x_1   
POL(C1(x1)) = 3 + (4)x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

a(b(c(x1))) → c(b(a(x1)))
C(B(A(x1))) → A(B(C(x1)))
b(a(C(x1))) → C(a(b(x1)))
c(A(B(x1))) → B(A(c(x1)))
A(c(b(x1))) → b(c(A(x1)))
B(C(a(x1))) → a(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 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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


A2(c(b(x1))) → C1(A(x1))
A1(b(c(x1))) → A1(x1)
A2(c(b(x1))) → A2(x1)
The remaining pairs can at least be oriented weakly.

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

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

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



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

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

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

The TRS R consists of the following rules:

a(b(c(x1))) → c(b(a(x1)))
C(B(A(x1))) → A(B(C(x1)))
b(a(C(x1))) → C(a(b(x1)))
c(A(B(x1))) → B(A(c(x1)))
A(c(b(x1))) → b(c(A(x1)))
B(C(a(x1))) → a(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 1 SCC with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

A1(b(c(x1))) → C1(b(a(x1)))
Used ordering: Polynomial interpretation [25,35]:

POL(B2(x1)) = 1 + (2)x_1   
POL(C(x1)) = x_1   
POL(c(x1)) = x_1   
POL(A1(x1)) = 4 + x_1   
POL(B(x1)) = (2)x_1   
POL(a(x1)) = 2 + x_1   
POL(A(x1)) = (4)x_1   
POL(b(x1)) = 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:

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



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

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

A1(b(c(x1))) → C1(b(a(x1)))

The TRS R consists of the following rules:

a(b(c(x1))) → c(b(a(x1)))
C(B(A(x1))) → A(B(C(x1)))
b(a(C(x1))) → C(a(b(x1)))
c(A(B(x1))) → B(A(c(x1)))
A(c(b(x1))) → b(c(A(x1)))
B(C(a(x1))) → a(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 1 less node.