Termination w.r.t. Q proof of /tmp/tmpLF4Hje/z24.trs

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

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

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

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

R is empty.
The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

                ↳ QDP

                  ↳ QDPToSRSProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

c(a(x)) → a(c(x))
a(b(x)) → b(a(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:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

C(a(x)) → A(c(x))

Strictly oriented rules of the TRS R:

b(c(x)) → c(b(x))

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ RuleRemovalProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

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

A(b(x)) → B(a(x))
B(c(x)) → C(b(x))

The TRS R consists of the following rules:

c(a(x)) → a(c(x))
a(b(x)) → b(a(x))

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 2 less nodes.