Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

A(b(b(b(x0)))) → A(b(x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ SemLabProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 1 + x0
a: x0
A: 0
b: 1 + x0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ SemLabProof
                    ↳ QDP
                      ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A.1(x1)) = x1   
POL(a.1(x1)) = x1   
POL(b.0(x1)) = x1   
POL(b.1(x1)) = x1   
POL(c.1(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ SemLabProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ AND
                          ↳ QDP
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A.0(x1)) = x1   
POL(a.0(x1)) = x1   
POL(b.0(x1)) = x1   
POL(b.1(x1)) = x1   
POL(c.0(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.