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)))

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:

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))

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
                                              ↳ 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))

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:

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.