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:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

F(a, f(b, x)) → F(b, f(a, x))
F(c, f(a, x)) → F(c, x)
F(c, f(a, x)) → F(a, f(c, x))
F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(b, x)
F(b, f(c, x)) → F(c, f(b, x))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof

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

F(a, f(b, x)) → F(b, f(a, x))
F(c, f(a, x)) → F(c, x)
F(c, f(a, x)) → F(a, f(c, x))
F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(b, x)
F(b, f(c, x)) → F(c, f(b, x))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
First, we A-transformed [17] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

a1(b(x)) → b1(a(x))
c1(a(x)) → c1(x)
c1(a(x)) → a1(c(x))
a1(b(x)) → a1(x)
b1(c(x)) → b1(x)
b1(c(x)) → c1(b(x))

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

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

Q is empty.

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.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(a(x1)) = x1   
POL(a1(x1)) = x1   
POL(b(x1)) = x1   
POL(b1(x1)) = x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RuleRemovalProof

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

a1(b(x)) → a1(x)
c1(a(x)) → c1(x)
b1(c(x)) → b1(x)
a1(b(x)) → b1(a(x))
b1(c(x)) → c1(b(x))
c1(a(x)) → a1(c(x))

The TRS R consists of the following rules:

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

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:

a1(b(x)) → a1(x)
c1(a(x)) → a1(c(x))

Strictly oriented rules of the TRS R:

a(b(x)) → b(a(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(a(x1)) = 2·x1   
POL(a1(x1)) = 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(b1(x1)) = 2 + 2·x1   
POL(c(x1)) = x1   
POL(c1(x1)) = 1 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof

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

c1(a(x)) → c1(x)
b1(c(x)) → b1(x)
a1(b(x)) → b1(a(x))
b1(c(x)) → c1(b(x))

The TRS R consists of the following rules:

c(a(x)) → a(c(x))
b(c(x)) → c(b(x))

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 with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP

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

c1(a(x)) → c1(x)

The TRS R consists of the following rules:

c(a(x)) → a(c(x))
b(c(x)) → c(b(x))

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
                  ↳ QDP

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

c1(a(x)) → c1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof

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

b1(c(x)) → b1(x)

The TRS R consists of the following rules:

c(a(x)) → a(c(x))
b(c(x)) → c(b(x))

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof

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

b1(c(x)) → b1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: