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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 2 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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:

C1(a(c(x))) → B1(a(x))
C1(a(C(x))) → C1(b(A(x)))
C1(a(c(x))) → A1(x)
C1(a(c(x))) → A1(c(b(a(x))))
A1(x) → B1(x)
C1(a(c(x))) → C1(b(a(x)))
B1(b(x)) → C1(x)
C1(a(C(x))) → B1(A(x))
C1(a(C(x))) → A2(x)
C1(a(C(x))) → A1(c(b(A(x))))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C1(a(c(x))) → B1(a(x))
C1(a(C(x))) → C1(b(A(x)))
C1(a(c(x))) → A1(x)
C1(a(c(x))) → A1(c(b(a(x))))
A1(x) → B1(x)
C1(a(c(x))) → C1(b(a(x)))
B1(b(x)) → C1(x)
C1(a(C(x))) → B1(A(x))
C1(a(C(x))) → A2(x)
C1(a(C(x))) → A1(c(b(A(x))))

The TRS R consists of the following rules:

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

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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C1(a(c(x))) → B1(a(x))
C1(a(C(x))) → C1(b(A(x)))
C1(a(c(x))) → A1(x)
A1(x) → B1(x)
C1(a(c(x))) → A1(c(b(a(x))))
C1(a(c(x))) → C1(b(a(x)))
B1(b(x)) → C1(x)
C1(a(C(x))) → B1(A(x))
C1(a(C(x))) → A1(c(b(A(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

C1(a(c(x))) → B1(a(x))
C1(a(C(x))) → C1(b(A(x)))
C1(a(c(x))) → A1(x)
C1(a(c(x))) → C1(b(a(x)))
C1(a(C(x))) → B1(A(x))


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(A1(x1)) = 1 + x1   
POL(B(x1)) = x1   
POL(B1(x1)) = 1 + x1   
POL(C(x1)) = 1 + x1   
POL(C1(x1)) = 2 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 2 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ QDPOrderProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C1(a(c(x))) → A1(c(b(a(x))))
A1(x) → B1(x)
B1(b(x)) → C1(x)
C1(a(C(x))) → A1(c(b(A(x))))

The TRS R consists of the following rules:

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

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(C(x))) → A1(c(b(A(x))))
The remaining pairs can at least be oriented weakly.

C1(a(c(x))) → A1(c(b(a(x))))
A1(x) → B1(x)
B1(b(x)) → C1(x)
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( C(x1) ) = 1


POL( c(x1) ) = max{0, -1}


POL( A1(x1) ) = x1


POL( B(x1) ) = 1


POL( a(x1) ) = x1


POL( A(x1) ) = max{0, x1 - 1}


POL( B1(x1) ) = x1


POL( b(x1) ) = x1


POL( C1(x1) ) = x1



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(x) → B1(x)
C1(a(c(x))) → A1(c(b(a(x))))
B1(b(x)) → C1(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                    ↳ QTRS
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
                    ↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

a(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(c(b(a(x))))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse

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

a(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(c(b(a(x))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

a(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(c(b(a(x))))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS

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

a(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(c(b(a(x))))

Q is empty.