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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ SemLabProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

c.0(b.1(b.0(x1))) → a.1(c.0(x1))
a.0(x1) → b.1(b.0(c.1(b.0(x1))))
c.0(c.1(x1)) → x1
c.1(b.0(b.1(x1))) → a.0(c.1(x1))
a.1(x1) → b.0(b.1(c.0(b.1(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ SemLabProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
              ↳ QDP
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

No dependency pairs are removed.

The following rules are removed from R:

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

POL(A.1(x1)) = x1   
POL(C.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)) = x1   



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

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

C.0(b.1(b.0(x1))) → A.1(c.0(x1))

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

No dependency pairs are removed.

The following rules are removed from R:

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

POL(A.0(x1)) = x1   
POL(C.1(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)) = x1   



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

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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(x1) → C.1(b.0(x1))

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.