Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ Narrowing
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(b(y0)) → A(c(b(y0)))
A(b(x0)) → A(a(c(a(x0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
QDP
              ↳ Narrowing
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(b(x0)) → A(c(a(x0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
QDP
                  ↳ QDPToSRSProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
QTRS
                      ↳ QTRS Reverse
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                          ↳ DependencyPairsProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

B1(a(x)) → B1(x)
B1(A(x)) → A1(a(B(x)))
B1(a(x)) → A1(a(b(x)))
B1(a(x)) → A1(b(x))
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(x)) → B2(x)
B1(A(x)) → C(a(A(x)))
B1(A(x)) → C(a(a(B(x))))
B1(A(x)) → C(A(x))
B1(A(x)) → B1(c(a(a(B(x)))))
B1(A(x)) → A1(c(a(A(x))))
B1(A(x)) → A1(A(x))
B1(A(x)) → A1(B(x))
B1(x) → A1(x)
B1(a(x)) → C(a(a(b(x))))
B1(A(x)) → A1(c(A(x)))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
QDP
                              ↳ DependencyGraphProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

B1(a(x)) → B1(x)
B1(A(x)) → A1(a(B(x)))
B1(a(x)) → A1(a(b(x)))
B1(a(x)) → A1(b(x))
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(x)) → B2(x)
B1(A(x)) → C(a(A(x)))
B1(A(x)) → C(a(a(B(x))))
B1(A(x)) → C(A(x))
B1(A(x)) → B1(c(a(a(B(x)))))
B1(A(x)) → A1(c(a(A(x))))
B1(A(x)) → A1(A(x))
B1(A(x)) → A1(B(x))
B1(x) → A1(x)
B1(a(x)) → C(a(a(b(x))))
B1(A(x)) → A1(c(A(x)))

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(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 13 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
QDP
                                  ↳ Narrowing
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B1(A(x0)) → B1(c(a(a(A(x0)))))
B1(A(y0)) → B1(c(a(B(y0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
QDP
                                      ↳ Narrowing
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

B1(A(y0)) → B1(c(a(B(y0))))
B1(A(x0)) → B1(c(a(a(A(x0)))))
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))

The TRS R consists of the following rules:

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

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

B1(A(y0)) → B1(c(a(A(y0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ Narrowing
QDP
                                          ↳ Narrowing
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

B1(A(y0)) → B1(c(a(B(y0))))
B1(a(x)) → B1(x)
B1(A(y0)) → B1(c(a(A(y0))))
B1(a(x)) → B1(c(a(a(b(x)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(y0)) → B1(c(a(B(y0)))) at position [0] we obtained the following new rules:

B1(A(y0)) → B1(c(B(y0)))
B1(A(x0)) → B1(c(a(A(x0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
QDP
                                              ↳ Narrowing
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

B1(a(x)) → B1(x)
B1(A(y0)) → B1(c(a(A(y0))))
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(y0)) → B1(c(B(y0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(y0)) → B1(c(a(A(y0)))) at position [0] we obtained the following new rules:

B1(A(y0)) → B1(c(A(y0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
QDP
                                                  ↳ DependencyGraphProof
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

B1(A(y0)) → B1(c(A(y0)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(y0)) → B1(c(B(y0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
QDP
                                                      ↳ Narrowing
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(y0)) → B1(c(B(y0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(y0)) → B1(c(B(y0))) at position [0] we obtained the following new rules:

B1(A(x0)) → B1(c(A(x0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
QDP
                                                          ↳ DependencyGraphProof
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
QDP
                                  ↳ NonTerminationProof
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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


s = B1(c(a(a(b(A(x')))))) evaluates to t =B1(c(a(a(b(A(x'))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

B1(c(a(a(b(A(x'))))))B1(c(a(b(A(x')))))
with rule a(x) → x at position [0,0] and matcher [x / a(b(A(x')))]

B1(c(a(b(A(x')))))B1(c(b(A(x'))))
with rule a(x) → x at position [0,0] and matcher [x / b(A(x'))]

B1(c(b(A(x'))))B1(c(a(c(a(A(x'))))))
with rule b(A(x'')) → a(c(a(A(x'')))) at position [0,0] and matcher [x'' / x']

B1(c(a(c(a(A(x'))))))B1(c(c(a(A(x')))))
with rule a(x'') → x'' at position [0,0] and matcher [x'' / c(a(A(x')))]

B1(c(c(a(A(x')))))B1(a(A(x')))
with rule c(c(x'')) → x'' at position [0] and matcher [x'' / a(A(x'))]

B1(a(A(x')))B1(c(a(a(b(A(x'))))))
with rule B1(a(x)) → B1(c(a(a(b(x)))))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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


s = A(a(c(b(a(a(c(b(x1')))))))) evaluates to t =A(a(c(b(a(a(c(b(x1'))))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

A(a(c(b(a(a(c(b(x1'))))))))A(a(c(b(a(c(b(x1')))))))
with rule a(x1) → x1 at position [0,0,0,0] and matcher [x1 / a(c(b(x1')))]

A(a(c(b(a(c(b(x1')))))))A(a(c(b(c(b(x1'))))))
with rule a(x1) → x1 at position [0,0,0,0] and matcher [x1 / c(b(x1'))]

A(a(c(b(c(b(x1'))))))A(a(c(a(c(b(x1'))))))
with rule b(x1) → a(x1) at position [0,0,0] and matcher [x1 / c(b(x1'))]

A(a(c(a(c(b(x1'))))))A(a(c(c(b(x1')))))
with rule a(x1) → x1 at position [0,0,0] and matcher [x1 / c(b(x1'))]

A(a(c(c(b(x1')))))A(a(b(x1')))
with rule c(c(x1)) → x1 at position [0,0] and matcher [x1 / b(x1')]

A(a(b(x1')))A(b(a(a(c(b(x1'))))))
with rule a(b(x1'')) → b(a(a(c(b(x1''))))) at position [0] and matcher [x1'' / x1']

A(b(a(a(c(b(x1'))))))A(a(c(b(a(a(c(b(x1'))))))))
with rule A(b(x1)) → A(a(c(b(x1))))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.