Termination w.r.t. Q proof of /tmp/tmpPRaksH/09.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:

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(s(x1)) → A(b(a(b(x1)))) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(a(a(x1)))) → B(a(b(a(x1)))) at position [0] we obtained the following new rules:

A(b(a(a(b(a(a(x0))))))) → B(a(b(b(a(b(a(x0)))))))
A(b(a(a(b(b(x0)))))) → B(a(c(s(x0))))
A(b(a(a(a(x0))))) → B(b(a(b(a(x0)))))
A(b(a(a(b(s(x0)))))) → B(a(a(b(s(a(x0))))))
A(b(a(a(s(x0))))) → B(a(b(s(a(x0)))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(s(x1)) → s(a(x1))
b(a(b(s(x1)))) → a(b(s(a(x1))))
b(a(b(b(x1)))) → c(s(x1))
c(s(x1)) → a(b(a(b(x1))))
a(b(a(a(x1)))) → b(a(b(a(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 3 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                        ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                        ↳ QTRS

                                      ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(s(x)) → s(a(x))
b(a(b(s(x)))) → a(b(s(a(x))))
b(a(b(b(x)))) → c(s(x))
c(s(x)) → a(b(a(b(x))))
a(b(a(a(x)))) → b(a(b(a(x))))
C(s(a(a(x)))) → A(b(b(a(b(a(x))))))
C(s(b(x))) → A(c(s(x)))
A(b(a(a(a(x))))) → B(b(a(b(a(x)))))
A(b(a(a(b(b(x)))))) → B(a(c(s(x))))
A(b(a(a(b(a(a(x))))))) → B(a(b(b(a(b(a(x)))))))
C(s(a(b(b(x))))) → A(b(a(c(s(x)))))
B(a(b(b(x)))) → C(s(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(a(b(A(x))))) → B¹(a(b(B(x))))
S(c(x)) → B¹(a(x))
A¹(a(a(b(A(x))))) → A¹(b(B(x)))
S(a(x)) → A¹(s(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(B(x)))
S(b(a(b(x)))) → A¹(x)
B¹(b(a(a(b(A(x)))))) → S(c(a(B(x))))
A¹(a(s(C(x)))) → A¹(b(a(b(b(A(x))))))
A¹(a(a(b(A(x))))) → B¹(B(x))
B¹(b(a(B(x)))) → S(C(x))
A¹(a(a(b(A(x))))) → A¹(b(a(b(B(x)))))
A¹(a(b(a(x)))) → A¹(b(a(b(x))))
S(b(a(b(x)))) → S(b(a(x)))
A¹(a(b(a(x)))) → B¹(x)
S(c(x)) → B¹(a(b(a(x))))
A¹(a(s(C(x)))) → B¹(b(A(x)))
S(a(x)) → S(x)
B¹(b(a(s(C(x))))) → B¹(A(x))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(a(b(b(a(B(x)))))))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(b(a(B(x)))))
A¹(a(s(C(x)))) → A¹(b(b(A(x))))
A¹(a(b(a(x)))) → A¹(b(x))
A¹(a(s(C(x)))) → B¹(a(b(b(A(x)))))
B¹(b(a(a(b(A(x)))))) → A¹(B(x))
S(b(a(b(x)))) → A¹(s(b(a(x))))
B¹(b(a(s(C(x))))) → A¹(b(A(x)))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(b(b(a(B(x))))))
A¹(a(b(a(a(b(A(x))))))) → A¹(B(x))
B¹(s(C(x))) → S(c(A(x)))
S(c(x)) → A¹(b(a(x)))
B¹(b(a(b(x)))) → S(c(x))
A¹(a(b(a(a(b(A(x))))))) → B¹(b(a(B(x))))
A¹(a(s(C(x)))) → B¹(A(x))
A¹(a(b(a(x)))) → B¹(a(b(x)))
S(b(a(b(x)))) → B¹(a(x))
S(c(x)) → A¹(x)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

A¹(a(a(b(A(x))))) → B¹(a(b(B(x))))
S(c(x)) → B¹(a(x))
A¹(a(a(b(A(x))))) → A¹(b(B(x)))
S(a(x)) → A¹(s(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(B(x)))
S(b(a(b(x)))) → A¹(x)
B¹(b(a(a(b(A(x)))))) → S(c(a(B(x))))
A¹(a(s(C(x)))) → A¹(b(a(b(b(A(x))))))
A¹(a(a(b(A(x))))) → B¹(B(x))
B¹(b(a(B(x)))) → S(C(x))
A¹(a(a(b(A(x))))) → A¹(b(a(b(B(x)))))
A¹(a(b(a(x)))) → A¹(b(a(b(x))))
S(b(a(b(x)))) → S(b(a(x)))
A¹(a(b(a(x)))) → B¹(x)
S(c(x)) → B¹(a(b(a(x))))
A¹(a(s(C(x)))) → B¹(b(A(x)))
S(a(x)) → S(x)
B¹(b(a(s(C(x))))) → B¹(A(x))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(a(b(b(a(B(x)))))))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(b(a(B(x)))))
A¹(a(s(C(x)))) → A¹(b(b(A(x))))
A¹(a(b(a(x)))) → A¹(b(x))
A¹(a(s(C(x)))) → B¹(a(b(b(A(x)))))
B¹(b(a(a(b(A(x)))))) → A¹(B(x))
S(b(a(b(x)))) → A¹(s(b(a(x))))
B¹(b(a(s(C(x))))) → A¹(b(A(x)))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(b(b(a(B(x))))))
A¹(a(b(a(a(b(A(x))))))) → A¹(B(x))
B¹(s(C(x))) → S(c(A(x)))
S(c(x)) → A¹(b(a(x)))
B¹(b(a(b(x)))) → S(c(x))
A¹(a(b(a(a(b(A(x))))))) → B¹(b(a(B(x))))
A¹(a(s(C(x)))) → B¹(A(x))
A¹(a(b(a(x)))) → B¹(a(b(x)))
S(b(a(b(x)))) → B¹(a(x))
S(c(x)) → A¹(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 20 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(a(b(a(a(b(A(x))))))) → A¹(b(b(a(B(x)))))
S(c(x)) → B¹(a(x))
A¹(a(b(a(x)))) → A¹(b(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(b(b(a(B(x))))))
B¹(b(a(a(b(A(x)))))) → S(c(a(B(x))))
B¹(s(C(x))) → S(c(A(x)))
S(c(x)) → A¹(b(a(x)))
B¹(b(a(b(x)))) → S(c(x))
A¹(a(b(a(x)))) → A¹(b(a(b(x))))
A¹(a(b(a(x)))) → B¹(x)
S(c(x)) → B¹(a(b(a(x))))
A¹(a(b(a(x)))) → B¹(a(b(x)))
S(c(x)) → A¹(x)
A¹(a(b(a(a(b(A(x))))))) → A¹(b(a(b(b(a(B(x)))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(a(b(a(a(b(A(x))))))) → A¹(b(b(a(B(x)))))
S(c(x)) → B¹(a(x))
A¹(a(b(a(x)))) → A¹(b(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(b(b(a(B(x))))))
B¹(b(a(a(b(A(x)))))) → S(c(a(B(x))))
B¹(s(C(x))) → S(c(A(x)))
S(c(x)) → A¹(b(a(x)))
B¹(b(a(b(x)))) → S(c(x))
A¹(a(b(a(x)))) → A¹(b(a(b(x))))
A¹(a(b(a(x)))) → B¹(x)
S(c(x)) → B¹(a(b(a(x))))
A¹(a(b(a(x)))) → B¹(a(b(x)))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(a(b(b(a(B(x)))))))
S(c(x)) → A¹(x)

The TRS R consists of the following rules:

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

A¹(a(b(a(a(b(A(x))))))) → A¹(b(b(a(B(x)))))
S(c(x)) → B¹(a(x))
A¹(a(b(a(x)))) → A¹(b(x))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(b(b(a(B(x))))))
S(c(x)) → A¹(b(a(x)))
A¹(a(b(a(x)))) → B¹(x)
A¹(a(b(a(x)))) → B¹(a(b(x)))
S(c(x)) → A¹(x)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

B¹(s(C(x))) → S(c(A(x)))
A¹(a(b(a(x)))) → A¹(b(a(b(x))))
B¹(b(a(b(x)))) → S(c(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
S(c(x)) → B¹(a(b(a(x))))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(a(b(b(a(B(x)))))))
B¹(b(a(a(b(A(x)))))) → S(c(a(B(x))))

The TRS R consists of the following rules:

b(b(a(B(x)))) → s(C(x))
a(a(b(a(a(b(A(x))))))) → a(b(a(b(b(a(B(x)))))))
a(a(b(a(x)))) → a(b(a(b(x))))
a(a(s(C(x)))) → a(b(a(b(b(A(x))))))
a(a(a(b(A(x))))) → a(b(a(b(B(x)))))
b(b(a(a(b(A(x)))))) → s(c(a(B(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))
b(b(a(s(C(x))))) → s(c(a(b(A(x)))))
b(s(C(x))) → s(c(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 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

                                                            ↳ QDP

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(a(b(a(x)))) → A¹(b(a(b(x))))

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( C(x₁) ) = max{0, x₁ - 1}

POL( c(x₁) ) = x₁

POL( A¹(x₁) ) = x₁

POL( B(x₁) ) = x₁

POL( a(x₁) ) = 1

POL( A(x₁) ) = x₁ + 1

POL( s(x₁) ) = max{0, -1}

POL( b(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

                                                                ↳ QDP

                                                                  ↳ PisEmptyProof

                                                            ↳ QDP

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

B¹(s(C(x))) → S(c(A(x)))
B¹(b(a(b(x)))) → S(c(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
S(c(x)) → B¹(a(b(a(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule S(c(x)) → B¹(a(b(a(x)))) at position [0] we obtained the following new rules:

S(c(a(b(a(a(b(A(x0)))))))) → B¹(a(b(a(b(a(b(b(a(B(x0))))))))))
S(c(a(s(C(x0))))) → B¹(a(b(a(b(a(b(b(A(x0)))))))))
S(c(a(a(b(A(x0)))))) → B¹(a(b(a(b(a(b(B(x0))))))))
S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

B¹(s(C(x))) → S(c(A(x)))
B¹(b(a(b(x)))) → S(c(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
S(c(a(b(a(a(b(A(x0)))))))) → B¹(a(b(a(b(a(b(b(a(B(x0))))))))))
S(c(a(s(C(x0))))) → B¹(a(b(a(b(a(b(b(A(x0)))))))))
S(c(a(a(b(A(x0)))))) → B¹(a(b(a(b(a(b(B(x0))))))))
S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

The TRS R consists of the following rules:

b(b(a(B(x)))) → s(C(x))
a(a(b(a(a(b(A(x))))))) → a(b(a(b(b(a(B(x)))))))
a(a(b(a(x)))) → a(b(a(b(x))))
a(a(s(C(x)))) → a(b(a(b(b(A(x))))))
a(a(a(b(A(x))))) → a(b(a(b(B(x)))))
b(b(a(a(b(A(x)))))) → s(c(a(B(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))
b(b(a(s(C(x))))) → s(c(a(b(A(x)))))
b(s(C(x))) → s(c(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 4 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

B¹(b(a(b(x)))) → S(c(x))
S(c(a(b(a(a(b(A(x0)))))))) → B¹(a(b(a(b(a(b(b(a(B(x0))))))))))
S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule S(c(a(b(a(a(b(A(x0)))))))) → B¹(a(b(a(b(a(b(b(a(B(x0)))))))))) at position [0] we obtained the following new rules:

S(c(a(b(a(a(b(A(x0)))))))) → B¹(a(b(a(b(a(s(C(x0))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ DependencyGraphProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

B¹(b(a(b(x)))) → S(c(x))
S(c(a(b(a(a(b(A(x0)))))))) → B¹(a(b(a(b(a(s(C(x0))))))))
S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

The TRS R consists of the following rules:

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ DependencyGraphProof

                                                                            ↳ QDP

                                                                              ↳ QDPOrderProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

B¹(b(a(b(x)))) → S(c(x))
S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

The TRS R consists of the following rules:

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

B¹(b(a(b(x)))) → S(c(x))

The remaining pairs can at least be oriented weakly.

S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( C(x₁) ) = x₁

POL( c(x₁) ) = max{0, -1}

POL( B(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = max{0, -1}

POL( B¹(x₁) ) = x₁ + 1

POL( A(x₁) ) = max{0, -1}

POL( s(x₁) ) = 1

POL( b(x₁) ) = 1

POL( S(x₁) ) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ DependencyGraphProof

                                                                            ↳ QDP

                                                                              ↳ QDPOrderProof

                                                                                ↳ QDP

                                                                                  ↳ DependencyGraphProof

                                                ↳ UsableRulesProof

                                              ↳ QDP

  ↳ QTRS Reverse

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

S(c(a(b(a(x0))))) → B¹(a(b(a(b(a(b(x0)))))))

The TRS R consists of the following rules:

b(b(a(B(x)))) → s(C(x))
a(a(b(a(a(b(A(x))))))) → a(b(a(b(b(a(B(x)))))))
a(a(b(a(x)))) → a(b(a(b(x))))
a(a(s(C(x)))) → a(b(a(b(b(A(x))))))
a(a(a(b(A(x))))) → a(b(a(b(B(x)))))
b(b(a(a(b(A(x)))))) → s(c(a(B(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))
b(b(a(s(C(x))))) → s(c(a(b(A(x)))))
b(s(C(x))) → s(c(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.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(a(b(a(a(b(A(x))))))) → A¹(b(b(a(B(x)))))
S(c(x)) → B¹(a(x))
A¹(a(b(a(x)))) → A¹(b(x))
B¹(b(a(s(C(x))))) → S(c(a(b(A(x)))))
A¹(a(b(a(a(b(A(x))))))) → B¹(a(b(b(a(B(x))))))
B¹(b(a(a(b(A(x)))))) → S(c(a(B(x))))
B¹(s(C(x))) → S(c(A(x)))
S(c(x)) → A¹(b(a(x)))
B¹(b(a(b(x)))) → S(c(x))
A¹(a(b(a(x)))) → A¹(b(a(b(x))))
A¹(a(b(a(x)))) → B¹(x)
S(c(x)) → B¹(a(b(a(x))))
A¹(a(b(a(x)))) → B¹(a(b(x)))
A¹(a(b(a(a(b(A(x))))))) → A¹(b(a(b(b(a(B(x)))))))
S(c(x)) → A¹(x)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                ↳ UsableRulesProof

  ↳ QTRS Reverse

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

S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                ↳ UsableRulesProof

  ↳ QTRS Reverse

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

S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)

The TRS R consists of the following rules:

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

S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ PisEmptyProof

                                                ↳ UsableRulesProof

  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPToSRSProof

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                      ↳ QTRS Reverse

                                      ↳ QTRS Reverse

                                      ↳ DependencyPairsProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                ↳ UsableRulesProof

                                                  ↳ QDP

  ↳ QTRS Reverse

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

S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.