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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
QTRS
              ↳ DependencyPairsProof
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x) → x
a(a(x)) → b(x)
b(x) → a(x)
c(b(x)) → a(b(c(c(x))))
c(B(x)) → A(x)
c(B(x)) → a(B(x))
B(x) → A(x)
a(A(x)) → 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 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                      ↳ QDP
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
B1(x) → A1(x)

The TRS R consists of the following rules:

a(x) → x
a(a(x)) → b(x)
b(x) → a(x)
c(b(x)) → a(b(c(c(x))))
c(B(x)) → A(x)
c(B(x)) → a(B(x))
B(x) → A(x)
a(A(x)) → B(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ UsableRulesReductionPairsProof
                        ↳ UsableRulesProof
                      ↳ QDP
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
B1(x) → A1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

A1(a(x)) → B1(x)
B1(x) → A1(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A1(x1)) = x1   
POL(B1(x1)) = 1 + 2·x1   
POL(a(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
QDP
                                ↳ PisEmptyProof
                        ↳ UsableRulesProof
                      ↳ QDP
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
P is empty.
R is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
QDP
                      ↳ QDP
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
B1(x) → A1(x)

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

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

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

C(b(x)) → C(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(x) → x
a(a(x)) → b(x)
b(x) → a(x)
c(b(x)) → a(b(c(c(x))))
c(B(x)) → A(x)
c(B(x)) → a(B(x))
B(x) → A(x)
a(A(x)) → 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
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
              ↳ QTRS Reverse
QTRS
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
              ↳ QTRS Reverse
              ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.