Termination w.r.t. Q proof of /tmp/tmpXIHcmt/4.32.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:

a(b(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(b(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

Q is empty.

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

a(b(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

Used ordering:
Polynomial interpretation [25]:

POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁
POL(u(x₁)) = 1 + 2·x₁
POL(v(x₁)) = 2 + x₁
POL(w(x₁)) = 1 + 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = 1

POL( C(x₁) ) = x₁ + 1

POL( c(x₁) ) = x₁

POL( b(x₁) ) = max{0, -1}

POL( B(x₁) ) = 1

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

B(c(x)) → B(x)
B(c(x)) → B(b(x))

The TRS R consists of the following rules:

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

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(x)) → B(x)
B(c(x)) → B(b(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( c(x₁) ) = x₁ + 1

POL( b(x₁) ) = x₁

POL( B(x₁) ) = x₁ + 1

POL( a(x₁) ) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

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

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.

A(b(x)) → A(a(x))
A(b(x)) → A(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = x₁ + 1

POL( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁ + 1

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

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

b(a(x)) → a(a(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

b(a(x)) → a(a(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x

Q is empty.

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

a(b(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

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

b(a(x)) → a(a(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

b(a(x)) → a(a(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(c(a(x)))
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x

Q is empty.