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

Q is empty.


QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The set Q consists of the following terms:

a(a(b(x0)))
a(c(x0))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(x0)))
a(c(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(x0)))
a(c(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
QDP
          ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all (P,Q,R)-chains.
By forward instantiating [14] the rule A(c(x1)) → A(x1) we obtained the following new rules:

A(c(c(y_0))) → A(c(y_0))
A(c(a(b(y_0)))) → A(a(b(y_0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
QDP
              ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A(c(c(y_0))) → A(c(y_0))
A(a(b(x1))) → A(c(x1))
A(c(x1)) → A(a(x1))
A(c(a(b(y_0)))) → A(a(b(y_0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(x0)))
a(c(x0))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
QTRS
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(a(b(x1))) → b(b(c(a(c(x1)))))
a(c(x1)) → a(a(x1))
A(c(c(y_0))) → A(c(y_0))
A(a(b(x1))) → A(c(x1))
A(c(x1)) → A(a(x1))
A(c(a(b(y_0)))) → A(a(b(y_0)))

The set Q consists of the following terms:

a(a(b(x0)))
a(c(x0))


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

a(a(b(x1))) → b(b(c(a(c(x1)))))
a(c(x1)) → a(a(x1))
A(c(c(y_0))) → A(c(y_0))
A(a(b(x1))) → A(c(x1))
A(c(x1)) → A(a(x1))
A(c(a(b(y_0)))) → A(a(b(y_0)))

The set Q is {a(a(b(x0))), a(c(x0))}.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ 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:

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ 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(b(x))) → b(b(c(a(c(x)))))
a(c(x)) → a(a(x))
A(c(c(x))) → A(c(x))
A(a(b(x))) → A(c(x))
A(c(x)) → A(a(x))
A(c(a(b(x)))) → 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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ 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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
QDP
                              ↳ Narrowing
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(a(a(x))) → B(x)
B(a(a(x))) → B(b(x))

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
QDP
                                  ↳ Narrowing
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B(a(a(a(A(x0))))) → B(a(A(x0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
                                ↳ QDP
                                  ↳ Narrowing
QDP
                                      ↳ DependencyGraphProof
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ ForwardInstantiation
            ↳ 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(b(x))) → b(b(c(a(c(x)))))
a(c(x)) → a(a(x))
A(c(c(x))) → A(c(x))
A(a(b(x))) → A(c(x))
A(c(x)) → A(a(x))
A(c(a(b(x)))) → A(a(b(x)))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.