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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

B(b(x1)) → A(x1)
A(a(x1)) → B(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(B(x1)) = 2·x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

a(a(x1)) → a(b(x1))
a(c(x1)) → b(x1)
b(b(x1)) → c(b(a(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 1 less node.

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

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

B(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 0
B: 0
a: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

B.0(b.1(a.0(x0))) → B.0(a.0(b.0(x0)))
B.0(b.0(c.0(x0))) → B.0(b.0(x0))
B.0(b.1(a.1(x0))) → B.1(a.0(b.1(x0)))
B.0(b.1(a.1(x0))) → B.0(a.0(b.1(x0)))
B.0(b.1(a.0(x0))) → B.1(a.0(b.0(x0)))
B.0(b.0(c.1(x0))) → B.0(b.1(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
a.0(c.0(x1)) → b.0(x1)
b.0(b.1(x1)) → c.0(b.1(a.1(x1)))
b.1(x0) → b.0(x0)
a.1(a.0(x1)) → a.0(b.0(x1))
a.1(a.1(x1)) → a.0(b.1(x1))
a.1(x0) → a.0(x0)
b.0(b.0(x1)) → c.0(b.1(a.0(x1)))
a.0(c.1(x1)) → b.1(x1)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
QDP
                        ↳ DependencyGraphProof
                    ↳ SemLabProof2
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B.0(b.1(a.0(x0))) → B.0(a.0(b.0(x0)))
B.0(b.0(c.0(x0))) → B.0(b.0(x0))
B.0(b.1(a.1(x0))) → B.1(a.0(b.1(x0)))
B.0(b.1(a.1(x0))) → B.0(a.0(b.1(x0)))
B.0(b.1(a.0(x0))) → B.1(a.0(b.0(x0)))
B.0(b.0(c.1(x0))) → B.0(b.1(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
a.0(c.0(x1)) → b.0(x1)
b.0(b.1(x1)) → c.0(b.1(a.1(x1)))
b.1(x0) → b.0(x0)
a.1(a.0(x1)) → a.0(b.0(x1))
a.1(a.1(x1)) → a.0(b.1(x1))
a.1(x0) → a.0(x0)
b.0(b.0(x1)) → c.0(b.1(a.0(x1)))
a.0(c.1(x1)) → b.1(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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                      ↳ QDP
                        ↳ DependencyGraphProof
QDP
                            ↳ UsableRulesReductionPairsProof
                    ↳ SemLabProof2
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B.0(b.1(a.0(x0))) → B.0(a.0(b.0(x0)))
B.0(b.0(c.0(x0))) → B.0(b.0(x0))
B.0(b.1(a.1(x0))) → B.0(a.0(b.1(x0)))
B.0(b.0(c.1(x0))) → B.0(b.1(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
a.0(c.0(x1)) → b.0(x1)
b.0(b.1(x1)) → c.0(b.1(a.1(x1)))
b.1(x0) → b.0(x0)
a.1(a.0(x1)) → a.0(b.0(x1))
a.1(a.1(x1)) → a.0(b.1(x1))
a.1(x0) → a.0(x0)
b.0(b.0(x1)) → c.0(b.1(a.0(x1)))
a.0(c.1(x1)) → b.1(x1)

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:

B.0(b.0(c.1(x0))) → B.0(b.1(x0))
The following rules are removed from R:

a.0(c.1(x1)) → b.1(x1)
Used ordering: POLO with Polynomial interpretation [25]:

POL(B.0(x1)) = x1   
POL(a.0(x1)) = x1   
POL(a.1(x1)) = x1   
POL(b.0(x1)) = x1   
POL(b.1(x1)) = x1   
POL(c.0(x1)) = x1   
POL(c.1(x1)) = 1 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
QDP
                                ↳ DependencyGraphProof
                    ↳ SemLabProof2
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B.0(b.0(c.0(x0))) → B.0(b.0(x0))
B.0(b.1(a.0(x0))) → B.0(a.0(b.0(x0)))
B.0(b.1(a.1(x0))) → B.0(a.0(b.1(x0)))

The TRS R consists of the following rules:

b.1(x0) → b.0(x0)
b.0(b.1(x1)) → c.0(b.1(a.1(x1)))
b.0(b.0(x1)) → c.0(b.1(a.0(x1)))
a.0(c.0(x1)) → b.0(x1)
a.1(a.0(x1)) → a.0(b.0(x1))
a.1(a.1(x1)) → a.0(b.1(x1))
a.1(x0) → a.0(x0)

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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
                              ↳ QDP
                                ↳ DependencyGraphProof
QDP
                    ↳ SemLabProof2
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B.0(b.0(c.0(x0))) → B.0(b.0(x0))

The TRS R consists of the following rules:

b.1(x0) → b.0(x0)
b.0(b.1(x1)) → c.0(b.1(a.1(x1)))
b.0(b.0(x1)) → c.0(b.1(a.0(x1)))
a.0(c.0(x1)) → b.0(x1)
a.1(a.0(x1)) → a.0(b.0(x1))
a.1(a.1(x1)) → a.0(b.1(x1))
a.1(x0) → a.0(x0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
QDP
                        ↳ QDPToSRSProof
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(b(c(x0))) → B(b(x0))

The TRS R consists of the following rules:

a(a(x1)) → a(b(x1))
a(c(x1)) → b(x1)
b(b(x1)) → 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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
QTRS
                            ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
QTRS
                                ↳ DependencyPairsProof
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
QDP
                                    ↳ RuleRemovalProof
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

B1(b(x)) → B1(c(x))
C(a(x)) → B1(x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(B(x1)) = 2·x1   
POL(B1(x1)) = 2·x1   
POL(C(x1)) = 2 + x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
QDP
                                        ↳ DependencyGraphProof
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(x)) → b(a(x))
c(a(x)) → b(x)
b(b(x)) → a(b(c(x)))
c(b(B(x))) → b(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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ DependencyGraphProof
QDP
                                            ↳ Narrowing
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(a(a(x0))) → B1(b(a(x0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Narrowing
QDP
                                                ↳ Narrowing
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B1(b(a(x0))) → A(b(b(x0)))
B1(b(b(B(x0)))) → A(b(b(B(x0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Narrowing
                                              ↳ QDP
                                                ↳ Narrowing
QDP
                                                    ↳ Narrowing
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B1(b(a(x0))) → A(b(b(x0)))
A(a(a(x0))) → B1(b(a(x0)))
B1(b(b(B(x0)))) → A(b(b(B(x0))))

The TRS R consists of the following rules:

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

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

B1(b(b(B(y0)))) → A(a(b(c(B(y0)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Narrowing
                                              ↳ QDP
                                                ↳ Narrowing
                                                  ↳ QDP
                                                    ↳ Narrowing
QDP
                                                        ↳ DependencyGraphProof
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(x)) → b(a(x))
c(a(x)) → b(x)
b(b(x)) → a(b(c(x)))
c(b(B(x))) → b(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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Narrowing
                                              ↳ QDP
                                                ↳ Narrowing
                                                  ↳ QDP
                                                    ↳ Narrowing
                                                      ↳ QDP
                                                        ↳ DependencyGraphProof
QDP
                                ↳ QTRS Reverse
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ SemLabProof
                    ↳ SemLabProof2
                      ↳ QDP
                        ↳ QDPToSRSProof
                          ↳ QTRS
                            ↳ QTRS Reverse
                              ↳ QTRS
                                ↳ QTRS Reverse
                                ↳ DependencyPairsProof
                                ↳ QTRS Reverse
QTRS
              ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.


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

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

A(a(x1)) → A(b(x1))

The TRS R consists of the following rules:

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

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

A(a(b(x0))) → A(c(b(a(x0))))



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

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

A(a(b(x0))) → A(c(b(a(x0))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
We have reversed the following QTRS:
The set of rules R is

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

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

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

Q is empty.

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

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

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

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

Q is empty.