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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(x1) → b(b(c(b(c(x1)))))
c(b(b(x1))) → a(x1)
c(c(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: 0
c: 1 + x0
a: 1 + 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.0(b.1(b.0(x0))) → C.0(b.1(a.0(x0)))
A.1(c.0(x0)) → C.1(b.0(x0))
A.1(b.0(b.1(x0))) → C.1(b.0(a.1(x0)))
A.0(x1) → C.0(x1)
C.1(b.0(b.1(x1))) → A.1(x1)
A.1(x1) → C.1(x1)
C.0(b.1(b.0(x1))) → A.0(x1)
A.0(c.1(x0)) → C.0(b.1(x0))

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

c.0(c.1(x1)) → x1
c.1(b.0(b.1(x1))) → a.1(x1)
c.0(b.1(b.0(x1))) → a.0(x1)
a.1(x1) → b.1(b.0(c.1(b.0(c.1(x1)))))
a.0(x1) → b.0(b.1(c.0(b.1(c.0(x1)))))
c.1(c.0(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 2 SCCs.

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

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

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

The TRS R consists of the following rules:

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

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

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

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



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

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

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

The TRS R consists of the following rules:

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

C.1(b.0(b.1(x1))) → A.1(x1)


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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



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

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

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

The TRS R consists of the following rules:

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

C.0(b.1(b.0(x1))) → A.0(x1)


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

a.0(x1) → b.0(b.1(c.0(b.1(c.0(x1)))))
c.0(b.1(b.0(x1))) → 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 0 SCCs with 2 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.