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

Q is empty.

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

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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Instantiation
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(c(y_1)) → B(c(y_1))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Instantiation
        ↳ QDP
          ↳ SemLabProof
  ↳ QTRS Reverse

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

B(c(c(x1))) → B(x1)
B(c(c(x1))) → A(c(b(x1)))
A(c(y_1)) → B(c(y_1))

The TRS R consists of the following rules:

a(x1) → b(x1)
b(b(x1)) → x1
b(c(c(x1))) → c(c(a(c(b(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
B: 0
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:

B.0(c.1(c.0(x1))) → A.0(c.1(b.0(x1)))
B.1(c.0(c.1(x1))) → B.1(x1)
B.1(c.0(c.1(x1))) → A.1(c.0(b.1(x1)))
A.0(c.1(y_1)) → B.0(c.1(y_1))
B.0(c.1(c.0(x1))) → B.0(x1)
A.1(c.0(y_1)) → B.1(c.0(y_1))

The TRS R consists of the following rules:

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

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

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

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

B.0(c.1(c.0(x1))) → A.0(c.1(b.0(x1)))
B.1(c.0(c.1(x1))) → B.1(x1)
B.1(c.0(c.1(x1))) → A.1(c.0(b.1(x1)))
A.0(c.1(y_1)) → B.0(c.1(y_1))
B.0(c.1(c.0(x1))) → B.0(x1)
A.1(c.0(y_1)) → B.1(c.0(y_1))

The TRS R consists of the following rules:

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

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

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

B.1(c.0(c.1(x1))) → B.1(x1)
B.1(c.0(c.1(x1))) → A.1(c.0(b.1(x1)))
A.1(c.0(y_1)) → B.1(c.0(y_1))

The TRS R consists of the following rules:

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

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

POL(A.1(x1)) = x1   
POL(B.1(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Instantiation
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                  ↳ QDP
  ↳ QTRS Reverse

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

A.1(c.0(y_1)) → B.1(c.0(y_1))

The TRS R consists of the following rules:

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

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

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

B.0(c.1(c.0(x1))) → A.0(c.1(b.0(x1)))
A.0(c.1(y_1)) → B.0(c.1(y_1))
B.0(c.1(c.0(x1))) → B.0(x1)

The TRS R consists of the following rules:

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

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

POL(A.0(x1)) = x1   
POL(B.0(x1)) = x1   
POL(a.0(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
      ↳ Instantiation
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                      ↳ QDP
                        ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

A.0(c.1(y_1)) → B.0(c.1(y_1))

The TRS R consists of the following rules:

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

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

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

Q is empty.