Termination w.r.t. Q proof of /tmp/tmplbi5yS/size-12-alpha-3-num-565.srs

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x1)) → A(b(x1)) at position [0] we obtained the following new rules:

B(c(c(x0))) → A(a(b(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

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

B(c(c(x0))) → A(a(b(x0)))
A(c(x1)) → B(c(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

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

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

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

a(b(x1)) → x1
a(c(x1)) → c(c(b(c(x1))))
b(c(x1)) → a(b(x1))
B(c(c(x0))) → A(a(b(x0)))
A(c(x1)) → B(c(x1))
B(c(x1)) → B(x1)

Q is empty.

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

a(b(x1)) → x1
a(c(x1)) → c(c(b(c(x1))))
b(c(x1)) → a(b(x1))
B(c(c(x0))) → A(a(b(x0)))
A(c(x1)) → B(c(x1))
B(c(x1)) → B(x1)

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                  ↳ QTRS Reverse

                  ↳ QTRS Reverse

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

b(a(x)) → x
c(a(x)) → c(b(c(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → c(B(x))
c(B(x)) → 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(c(x))
C(a(x)) → C(x)
C(A(x)) → C(B(x))
C(b(x)) → B¹(a(x))
C(a(x)) → B¹(c(c(x)))
C(a(x)) → C(b(c(c(x))))
C(c(B(x))) → B¹(a(A(x)))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                  ↳ 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)
C(A(x)) → C(B(x))
C(b(x)) → B¹(a(x))
C(a(x)) → B¹(c(c(x)))
C(a(x)) → C(b(c(c(x))))
C(c(B(x))) → B¹(a(A(x)))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → c(b(c(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → c(B(x))
c(B(x)) → 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 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ 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)
C(a(x)) → C(b(c(c(x))))

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

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

POL( A(x₁) ) = x₁ + 1

POL( C(x₁) ) = x₁ + 1

POL( c(x₁) ) = x₁ + 1

POL( b(x₁) ) = max{0, x₁ - 1}

POL( B(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ 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(b(c(c(x))))

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (1/4)x_1
POL(c(x₁)) = (4)x_1
POL(B(x₁)) = 3/4 + (2)x_1
POL(a(x₁)) = 4 + (4)x_1
POL(A(x₁)) = 4 + (13/4)x_1
POL(b(x₁)) = 1/2 + (1/4)x_1

The value of delta used in the strict ordering is 7/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ PisEmptyProof

                  ↳ QTRS Reverse

                  ↳ QTRS Reverse

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

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                  ↳ QTRS Reverse

                    ↳ QTRS

                  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ QDPToSRSProof

            ↳ QTRS

              ↳ QTRS Reverse

                ↳ QTRS

                  ↳ DependencyPairsProof

                  ↳ QTRS Reverse

                  ↳ QTRS Reverse

                    ↳ QTRS

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

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

Q is empty.