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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + 2·x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(c(x1)) → C(b(b(x1)))
B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.

A(b(x1)) → B(b(a(x1)))
A(b(x1)) → B(a(x1))
C(a(x1)) → A(c(c(x1)))
A(b(x1)) → A(x1)
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = 1


POL( C(x1) ) = 1


POL( c(x1) ) = x1 + 1


POL( b(x1) ) = x1


POL( B(x1) ) = x1 + 1


POL( a(x1) ) = max{0, -1}



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(x1)) → b(b(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(c(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 3 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
  ↳ QTRS Reverse

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

A(b(x1)) → A(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
  ↳ QTRS Reverse

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

A(b(x1)) → A(x1)

R is empty.
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(b(x1)) → A(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(b(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
                        ↳ QDP
                          ↳ PisEmptyProof
  ↳ QTRS Reverse

Q DP problem:
P is empty.
R is empty.
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(b(x1)) → b(b(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(c(x1)))

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

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

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

Q is empty.