Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(a(c(b(x1))))
b(b(x1)) → a(a(x1))
c(a(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
          ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(b(x1)) → A(x1)
A(a(x1)) → A(c(b(x1)))
A(a(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.

A(a(x1)) → B(a(c(b(x1))))
B(b(x1)) → A(a(x1))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = x1 + 1


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


POL( b(x1) ) = x1 + 1


POL( B(x1) ) = x1 + 1


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(a(b(x0))) → B(a(c(a(a(x0)))))



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

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(a(c(b(x1))))
b(b(x1)) → a(a(x1))
c(a(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
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B1(b(x)) → A1(x)
B1(a(A(x))) → A1(a(c(a(B(x)))))
B1(b(x)) → A1(a(x))
The remaining pairs can at least be oriented weakly.

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

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


POL( A1(x1) ) = x1


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


POL( a(x1) ) = x1 + 1


POL( B1(x1) ) = x1 + 1


POL( A(x1) ) = 0


POL( b(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

A1(a(x)) → A1(b(x))
A1(a(x)) → B1(x)

The TRS R consists of the following rules:

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

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

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

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

A1(a(x)) → A1(b(x))

The TRS R consists of the following rules:

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

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

A1(a(a(A(x0)))) → A1(a(a(c(a(B(x0))))))
A1(a(B(x0))) → A1(a(A(x0)))
A1(a(b(x0))) → A1(a(a(x0)))



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

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

A1(a(a(A(x0)))) → A1(a(a(c(a(B(x0))))))
A1(a(b(x0))) → A1(a(a(x0)))
A1(a(B(x0))) → A1(a(A(x0)))

The TRS R consists of the following rules:

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

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

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

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

A1(a(a(A(x0)))) → A1(a(a(c(a(B(x0))))))
A1(a(b(x0))) → A1(a(a(x0)))

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.