Termination w.r.t. Q proof of /tmp/tmplrnG3I/size-12-alpha-3-num-561.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(x1))
b(c(x1)) → a(b(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(x1))
b(c(x1)) → a(b(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:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

B(c(x1)) → B(a(b(x1)))
B(c(x1)) → A(b(a(b(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(x1))
b(c(x1)) → a(b(a(b(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

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

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

The TRS R consists of the following rules:

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

      ↳ DependencyGraphProof

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                      ↳ QTRS Reverse

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                      ↳ DependencyPairsProof

                      ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

The TRS R consists of the following rules:

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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPOrderProof

                      ↳ QTRS Reverse

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

C(a(x)) → C(c(x))
C(a(x)) → C(x)

The TRS R consists of the following rules:

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

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

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                      ↳ QTRS Reverse

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

C(a(x)) → C(c(x))

The TRS R consists of the following rules:

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

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

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

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ PisEmptyProof

                      ↳ QTRS Reverse

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                      ↳ QTRS Reverse

                        ↳ QTRS

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

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

Q is empty.