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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

A(b(b(x0))) → C(b(b(a(x0))))



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

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The following dependency pairs can be deleted:

A.0(b.1(b.0(x0))) → C.1(b.0(b.1(a.0(x0))))
A.0(b.1(x1)) → C.1(x1)
The following rules are removed from R:

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

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



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

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

C.1(b.0(x1)) → A.0(x1)

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

The following dependency pairs can be deleted:

A.1(b.0(x1)) → C.0(x1)
A.1(b.0(b.1(x0))) → C.0(b.1(b.0(a.1(x0))))
The following rules are removed from R:

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

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



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

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

C.0(b.1(x1)) → A.1(x1)

The TRS R consists of the following rules:

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.