Termination w.r.t. Q proof of /tmp/tmpQ74vr-/3.srs

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:

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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ 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.

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

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))))

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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))))

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

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

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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

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

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

B(a(a(a(x1)))) → A(a(a(b(x1))))

The remaining pairs can at least be oriented weakly.

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))))

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 42.
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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

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

C(b(a(a(x1)))) → A(a(b(c(x1))))
A(b(c(x1))) → C(b(a(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.
By narrowing [15] the rule C(b(a(a(x1)))) → A(a(b(c(x1)))) at position [0] we obtained the following new rules:

C(b(a(a(x0)))) → A(c(b(a(x0))))
C(b(a(a(c(b(b(x0))))))) → A(a(b(b(b(c(c(x0)))))))
C(b(a(a(b(a(a(x0))))))) → A(a(b(a(a(b(c(x0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

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

C(b(a(a(x0)))) → A(c(b(a(x0))))
C(b(a(a(b(a(a(x0))))))) → A(a(b(a(a(b(c(x0)))))))
A(b(c(x1))) → C(b(a(x1)))
C(b(a(a(c(b(b(x0))))))) → A(a(b(b(b(c(c(x0)))))))

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.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

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))))
C(b(a(a(x0)))) → A(c(b(a(x0))))
C(b(a(a(b(a(a(x0))))))) → A(a(b(a(a(b(c(x0)))))))
A(b(c(x1))) → C(b(a(x1)))
C(b(a(a(c(b(b(x0))))))) → A(a(b(b(b(c(c(x0)))))))

Q is empty.

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

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))))
C(b(a(a(x0)))) → A(c(b(a(x0))))
C(b(a(a(b(a(a(x0))))))) → A(a(b(a(a(b(c(x0)))))))
A(b(c(x1))) → C(b(a(x1)))
C(b(a(a(c(b(b(x0))))))) → A(a(b(b(b(c(c(x0)))))))

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(a(b(c(x)))) → c(b(a(a(x))))
a(a(a(b(x)))) → b(a(a(a(x))))
c(b(a(x))) → a(b(c(x)))
b(b(c(c(x)))) → c(c(b(b(x))))
a(a(b(C(x)))) → a(b(c(A(x))))
a(a(b(a(a(b(C(x))))))) → c(b(a(a(b(a(A(x)))))))
c(b(A(x))) → a(b(C(x)))
b(b(c(a(a(b(C(x))))))) → c(c(b(b(b(a(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 16 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ RuleRemovalProof

                                            ↳ QDP

                                              ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C¹(b(a(x))) → A¹(b(c(x))) at position [0] we obtained the following new rules:

C¹(b(a(b(A(x0))))) → A¹(b(a(b(C(x0)))))
C¹(b(a(b(a(x0))))) → A¹(b(a(b(c(x0)))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ RuleRemovalProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

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

C¹(b(a(b(A(x0))))) → A¹(b(a(b(C(x0)))))
A¹(a(b(a(a(b(C(x))))))) → C¹(b(a(a(b(a(A(x)))))))
A¹(a(b(c(x)))) → C¹(b(a(a(x))))
B(b(c(c(x)))) → C¹(c(b(b(x))))
C¹(b(a(b(a(x0))))) → A¹(b(a(b(c(x0)))))
A¹(a(a(b(x)))) → B(a(a(a(x))))

The TRS R consists of the following rules:

a(a(b(c(x)))) → c(b(a(a(x))))
a(a(a(b(x)))) → b(a(a(a(x))))
c(b(a(x))) → a(b(c(x)))
b(b(c(c(x)))) → c(c(b(b(x))))
a(a(b(C(x)))) → a(b(c(A(x))))
a(a(b(a(a(b(C(x))))))) → c(b(a(a(b(a(A(x)))))))
c(b(A(x))) → a(b(C(x)))
b(b(c(a(a(b(C(x))))))) → c(c(b(b(b(a(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 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ RuleRemovalProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ QDPOrderProof

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

B(b(c(c(x)))) → C¹(c(b(b(x))))
A¹(a(b(c(x)))) → C¹(b(a(a(x))))
A¹(a(a(b(x)))) → B(a(a(a(x))))
C¹(b(a(b(a(x0))))) → A¹(b(a(b(c(x0)))))

The TRS R consists of the following rules:

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

A¹(a(b(c(x)))) → C¹(b(a(a(x))))
A¹(a(a(b(x)))) → B(a(a(a(x))))
C¹(b(a(b(a(x0))))) → A¹(b(a(b(c(x0)))))

The remaining pairs can at least be oriented weakly.

B(b(c(c(x)))) → C¹(c(b(b(x))))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 1/4 + (2)x_1
POL(c(x₁)) = (4)x_1
POL(A¹(x₁)) = 2 + (1/4)x_1
POL(B(x₁)) = 9/4
POL(a(x₁)) = 9/4
POL(A(x₁)) = 4 + (1/4)x_1
POL(b(x₁)) = (1/4)x_1
POL(C¹(x₁)) = 9/4

The value of delta used in the strict ordering is 7/64.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ RuleRemovalProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ QDPOrderProof

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

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

B(b(c(c(x)))) → C¹(c(b(b(x))))

The TRS R consists of the following rules:

a(a(b(c(x)))) → c(b(a(a(x))))
a(a(a(b(x)))) → b(a(a(a(x))))
c(b(a(x))) → a(b(c(x)))
b(b(c(c(x)))) → c(c(b(b(x))))
a(a(b(C(x)))) → a(b(c(A(x))))
a(a(b(a(a(b(C(x))))))) → c(b(a(a(b(a(A(x)))))))
c(b(A(x))) → a(b(C(x)))
b(b(c(a(a(b(C(x))))))) → c(c(b(b(b(a(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 0 SCCs with 1 less node.