Termination w.r.t. Q proof of /tmp/tmp6JUlWb/z119.srs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

Q is empty.

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

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

A(c(x1)) → D(d(d(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
C(f(x1)) → C(x1)
A(b(x1)) → D(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → B(x1)
C(f(x1)) → D(c(x1))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

A(c(x1)) → D(d(d(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
C(f(x1)) → C(x1)
A(b(x1)) → D(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → B(x1)
C(f(x1)) → D(c(x1))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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:

A(b(x1)) → D(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = 1 + x₁
POL(C(x₁)) = x₁
POL(D(x₁)) = x₁
POL(F(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 1 + x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(f(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
F(f(x1)) → A(x1)
A(c(x1)) → D(d(d(x1)))
A(b(x1)) → B(d(x1))
D(d(x1)) → F(x1)
C(f(x1)) → C(x1)
B(d(x1)) → B(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
C(f(x1)) → D(c(x1))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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(d(x1)) → C(b(x1))
A(c(x1)) → D(d(x1))
A(c(x1)) → D(d(d(x1)))
C(f(x1)) → C(x1)
B(d(x1)) → B(x1)
A(c(x1)) → D(x1)
C(f(x1)) → D(c(x1))

The remaining pairs can at least be oriented weakly.

C(f(x1)) → D(d(c(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → A(c(b(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = x_1
POL(f(x₁)) = 2 + x_1
POL(c(x₁)) = x_1
POL(B(x₁)) = (4)x_1
POL(D(x₁)) = 1 + x_1
POL(a(x₁)) = 4 + x_1
POL(A(x₁)) = 4 + x_1
POL(d(x₁)) = 1 + x_1
POL(b(x₁)) = (4)x_1
POL(F(x₁)) = 2 + x_1

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

a(c(x1)) → d(d(d(x1)))
d(d(x1)) → f(x1)
b(d(x1)) → a(c(b(x1)))
a(b(x1)) → b(d(x1))
f(f(x1)) → a(x1)
c(f(x1)) → d(d(c(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

C(f(x1)) → D(d(c(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → A(c(b(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(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 3 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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

B(d(d(x0))) → A(c(a(c(b(x0)))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                                ↳ QTRS

                              ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(b(x)) → b(d(x))
a(c(x)) → d(d(d(x)))
b(d(x)) → a(c(b(x)))
c(f(x)) → d(d(c(x)))
d(d(x)) → f(x)
f(f(x)) → a(x)
B(d(d(x))) → A(c(a(c(b(x)))))
A(b(x)) → B(d(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:

D(d(B(x))) → C(a(c(A(x))))
C(a(x)) → D(d(d(x)))
B¹(a(x)) → B¹(x)
F(c(x)) → C(d(d(x)))
C(a(x)) → D(d(x))
B¹(A(x)) → D(B(x))
D(d(B(x))) → C(A(x))
B¹(a(x)) → D(b(x))
D(b(x)) → B¹(c(a(x)))
D(b(x)) → C(a(x))
D(d(x)) → F(x)
F(c(x)) → D(x)
D(d(B(x))) → B¹(c(a(c(A(x)))))
F(c(x)) → D(d(x))
C(a(x)) → D(x)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

D(d(B(x))) → C(a(c(A(x))))
C(a(x)) → D(d(d(x)))
B¹(a(x)) → B¹(x)
F(c(x)) → C(d(d(x)))
C(a(x)) → D(d(x))
B¹(A(x)) → D(B(x))
D(d(B(x))) → C(A(x))
B¹(a(x)) → D(b(x))
D(b(x)) → B¹(c(a(x)))
D(b(x)) → C(a(x))
D(d(x)) → F(x)
F(c(x)) → D(x)
D(d(B(x))) → B¹(c(a(c(A(x)))))
F(c(x)) → D(d(x))
C(a(x)) → D(x)

The TRS R consists of the following rules:

b(a(x)) → d(b(x))
c(a(x)) → d(d(d(x)))
d(b(x)) → b(c(a(x)))
f(c(x)) → c(d(d(x)))
d(d(x)) → f(x)
f(f(x)) → a(x)
d(d(B(x))) → b(c(a(c(A(x)))))
b(A(x)) → d(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 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

D(d(B(x))) → C(a(c(A(x))))
C(a(x)) → D(d(d(x)))
B¹(a(x)) → B¹(x)
F(c(x)) → C(d(d(x)))
C(a(x)) → D(d(x))
D(b(x)) → B¹(c(a(x)))
B¹(a(x)) → D(b(x))
D(b(x)) → C(a(x))
D(d(x)) → F(x)
F(c(x)) → D(x)
D(d(B(x))) → B¹(c(a(c(A(x)))))
F(c(x)) → D(d(x))
C(a(x)) → D(x)

The TRS R consists of the following rules:

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

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = 2 + x₁
POL(B¹(x₁)) = 2 + x₁
POL(C(x₁)) = x₁
POL(D(x₁)) = x₁
POL(F(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 2 + x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(f(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

B¹(a(x)) → D(b(x))
D(b(x)) → B¹(c(a(x)))
C(a(x)) → D(d(d(x)))
B¹(a(x)) → B¹(x)
F(c(x)) → C(d(d(x)))
D(d(x)) → F(x)
F(c(x)) → D(x)
C(a(x)) → D(d(x))
D(d(B(x))) → B¹(c(a(c(A(x)))))
F(c(x)) → D(d(x))
C(a(x)) → D(x)

The TRS R consists of the following rules:

b(a(x)) → d(b(x))
c(a(x)) → d(d(d(x)))
d(b(x)) → b(c(a(x)))
f(c(x)) → c(d(d(x)))
d(d(x)) → f(x)
f(f(x)) → a(x)
d(d(B(x))) → b(c(a(c(A(x)))))
b(A(x)) → d(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.

F(c(x)) → C(d(d(x)))
F(c(x)) → D(x)
F(c(x)) → D(d(x))

The remaining pairs can at least be oriented weakly.

B¹(a(x)) → D(b(x))
D(b(x)) → B¹(c(a(x)))
C(a(x)) → D(d(d(x)))
B¹(a(x)) → B¹(x)
D(d(x)) → F(x)
C(a(x)) → D(d(x))
D(d(B(x))) → B¹(c(a(c(A(x)))))
C(a(x)) → D(x)

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

POL( C(x₁) ) = x₁ + 1

POL( f(x₁) ) = x₁

POL( c(x₁) ) = x₁ + 1

POL( D(x₁) ) = x₁ + 1

POL( B(x₁) ) = 0

POL( a(x₁) ) = x₁

POL( B¹(x₁) ) = 1

POL( A(x₁) ) = x₁

POL( b(x₁) ) = max{0, -1}

POL( d(x₁) ) = x₁

POL( F(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

D(b(x)) → B¹(c(a(x)))
B¹(a(x)) → D(b(x))
C(a(x)) → D(d(d(x)))
B¹(a(x)) → B¹(x)
D(d(x)) → F(x)
C(a(x)) → D(d(x))
D(d(B(x))) → B¹(c(a(c(A(x)))))
C(a(x)) → D(x)

The TRS R consists of the following rules:

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

  ↳ QTRS Reverse

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

B¹(a(x)) → D(b(x))
D(b(x)) → B¹(c(a(x)))
B¹(a(x)) → B¹(x)
D(d(B(x))) → B¹(c(a(c(A(x)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule D(b(x)) → B¹(c(a(x))) at position [0] we obtained the following new rules:

D(b(x0)) → B¹(d(d(d(x0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ Narrowing

  ↳ QTRS Reverse

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

B¹(a(x)) → D(b(x))
B¹(a(x)) → B¹(x)
D(d(B(x))) → B¹(c(a(c(A(x)))))
D(b(x0)) → B¹(d(d(d(x0))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule D(d(B(x))) → B¹(c(a(c(A(x))))) at position [0] we obtained the following new rules:

D(d(B(y0))) → B¹(d(d(d(c(A(y0))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

  ↳ QTRS Reverse

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

D(d(B(y0))) → B¹(d(d(d(c(A(y0))))))
B¹(a(x)) → D(b(x))
B¹(a(x)) → B¹(x)
D(b(x0)) → B¹(d(d(d(x0))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.