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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

C(c(x1)) → B(x1)
B(x1) → C(x1)

The TRS R consists of the following rules:

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

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
            ↳ UsableRulesProof
          ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(c(x1)) → B(x1)
B(x1) → C(x1)

R is empty.
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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
QDP
                ↳ UsableRulesReductionPairsProof
          ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(c(x1)) → B(x1)
B(x1) → C(x1)

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:

C(c(x1)) → B(x1)
B(x1) → C(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(B(x1)) = 1 + 2·x1   
POL(C(x1)) = x1   
POL(c(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
  ↳ 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.

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

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

A(b(x1)) → A(a(x1))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → c(b(a(a(x1))))
b(x1) → c(x1)
c(c(x1)) → b(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
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ 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(b(x1)) → c(b(a(a(x1))))
b(x1) → c(x1)
c(c(x1)) → b(x1)
A(b(b(x0))) → A(c(b(a(a(x0)))))
A(b(x1)) → A(x1)

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ 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
b(a(x)) → a(a(b(c(x))))
b(x) → c(x)
c(c(x)) → b(x)
b(b(A(x))) → a(a(b(c(A(x)))))
b(A(x)) → 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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ 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:

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(a(b(c(x))))
b(x) → c(x)
c(c(x)) → b(x)
b(b(A(x))) → a(a(b(c(A(x)))))
b(A(x)) → 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 5 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.