Termination w.r.t. Q proof of /tmp/tmpxfilH2/torpa4.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(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(x1))))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

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

C(a(c(x1))) → A(b(c(x1)))
C(d(x1)) → A(a(x1))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
C(a(c(x1))) → C(a(b(c(x1))))
A(d(x1)) → C(x1)
G(x1) → C(a(x1))
A(b(c(a(x1)))) → A(b(x1))
C(d(x1)) → A(x1)
B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
A(f(f(x1))) → G(x1)
G(x1) → A(x1)
A(b(c(a(x1)))) → B(x1)
C(a(c(x1))) → B(c(x1))
A(b(c(a(x1)))) → B(a(b(x1)))
A(b(c(a(x1)))) → C(b(a(b(x1))))
C(a(c(x1))) → B(c(a(b(c(x1)))))
B(g(x1)) → B(x1)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

C(a(c(x1))) → A(b(c(x1)))
C(d(x1)) → A(a(x1))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
C(a(c(x1))) → C(a(b(c(x1))))
A(d(x1)) → C(x1)
G(x1) → C(a(x1))
A(b(c(a(x1)))) → A(b(x1))
C(d(x1)) → A(x1)
B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
A(f(f(x1))) → G(x1)
G(x1) → A(x1)
A(b(c(a(x1)))) → B(x1)
C(a(c(x1))) → B(c(x1))
A(b(c(a(x1)))) → B(a(b(x1)))
A(b(c(a(x1)))) → C(b(a(b(x1))))
C(a(c(x1))) → B(c(a(b(c(x1)))))
B(g(x1)) → B(x1)

The TRS R consists of the following rules:

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

C(a(c(x1))) → A(b(c(x1)))
A(b(c(a(x1)))) → A(b(x1))
C(d(x1)) → A(x1)
G(x1) → A(x1)
A(b(c(a(x1)))) → B(x1)
C(a(c(x1))) → B(c(x1))
A(b(c(a(x1)))) → B(a(b(x1)))
A(b(c(a(x1)))) → C(b(a(b(x1))))
B(g(x1)) → B(x1)

The remaining pairs can at least be oriented weakly.

C(d(x1)) → A(a(x1))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
C(a(c(x1))) → C(a(b(c(x1))))
A(d(x1)) → C(x1)
G(x1) → C(a(x1))
B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
A(f(f(x1))) → G(x1)
C(a(c(x1))) → B(c(a(b(c(x1)))))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 8 + x₁
POL(B(x₁)) = 4 + x₁
POL(C(x₁)) = 10 + x₁
POL(G(x₁)) = 14 + x₁
POL(a(x₁)) = 4 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 2 + x₁
POL(f(x₁)) = 3 + x₁
POL(g(x₁)) = 10 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
A(f(f(x1))) → G(x1)
C(a(c(x1))) → C(a(b(c(x1))))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
C(d(x1)) → A(a(x1))
A(d(x1)) → C(x1)
C(a(c(x1))) → B(c(a(b(c(x1)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

A(f(f(x1))) → G(x1)
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
A(d(x1)) → C(x1)

The remaining pairs can at least be oriented weakly.

B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
C(a(c(x1))) → C(a(b(c(x1))))
C(d(x1)) → A(a(x1))
C(a(c(x1))) → B(c(a(b(c(x1)))))
G(x1) → C(a(x1))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 6 + x₁
POL(B(x₁)) = x₁
POL(C(x₁)) = 8 + x₁
POL(G(x₁)) = 13 + x₁
POL(a(x₁)) = 5 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 8 + x₁
POL(d(x₁)) = 3 + x₁
POL(f(x₁)) = 4 + x₁
POL(g(x₁)) = 13 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
C(d(x1)) → A(a(x1))
C(a(c(x1))) → C(a(b(c(x1))))
G(x1) → C(a(x1))
C(a(c(x1))) → B(c(a(b(c(x1)))))

The TRS R consists of the following rules:

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

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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                  ↳ QDP

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

A(b(c(a(x1)))) → A(c(b(a(b(x1)))))

The TRS R consists of the following rules:

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

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

A(b(c(a(c(a(x0)))))) → A(c(b(b(a(c(b(a(b(x0)))))))))
A(b(c(a(g(x0))))) → A(c(b(a(g(b(x0))))))
A(b(c(a(y0)))) → A(f(f(b(a(b(y0))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                  ↳ QDP

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

A(b(c(a(c(a(x0)))))) → A(c(b(b(a(c(b(a(b(x0)))))))))
A(b(c(a(g(x0))))) → A(c(b(a(g(b(x0))))))
A(b(c(a(y0)))) → A(f(f(b(a(b(y0))))))

The TRS R consists of the following rules:

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                  ↳ QDP

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

A(b(c(a(c(a(x0)))))) → A(c(b(b(a(c(b(a(b(x0)))))))))
A(b(c(a(g(x0))))) → A(c(b(a(g(b(x0))))))

The TRS R consists of the following rules:

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                  ↳ QDP

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                  ↳ QDP

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                      ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                  ↳ QDP

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                      ↳ QTRS

                                    ↳ DependencyPairsProof

                  ↳ QDP

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

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

A¹(c(b(a(x)))) → A¹(b(x))
A¹(c(b(a(x)))) → C(a(b(x)))
G(x) → A¹(c(x))
G(a(c(b(A(x))))) → A¹(b(c(A(x))))
G(x) → D(d(d(d(x))))
A¹(c(b(a(x)))) → A¹(b(c(a(b(x)))))
D(c(x)) → A¹(a(x))
F(f(a(x))) → G(x)
G(x) → D(d(x))
A¹(c(a(c(b(A(x)))))) → A¹(b(b(c(A(x)))))
G(a(c(b(A(x))))) → C(A(x))
C(a(c(x))) → C(b(x))
C(x) → F(f(x))
G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
D(c(x)) → A¹(x)
C(x) → F(x)
G(x) → D(d(d(x)))
G(x) → C(x)
C(a(c(x))) → A¹(c(b(x)))
A¹(c(a(c(b(A(x)))))) → C(A(x))
G(x) → D(x)
A¹(c(a(c(b(A(x)))))) → C(a(b(b(c(A(x))))))
A¹(c(a(c(b(A(x)))))) → A¹(b(c(a(b(b(c(A(x))))))))
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
D(a(x)) → C(x)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                  ↳ QDP

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

A¹(c(b(a(x)))) → A¹(b(x))
A¹(c(b(a(x)))) → C(a(b(x)))
G(x) → A¹(c(x))
G(a(c(b(A(x))))) → A¹(b(c(A(x))))
G(x) → D(d(d(d(x))))
A¹(c(b(a(x)))) → A¹(b(c(a(b(x)))))
D(c(x)) → A¹(a(x))
F(f(a(x))) → G(x)
G(x) → D(d(x))
A¹(c(a(c(b(A(x)))))) → A¹(b(b(c(A(x)))))
G(a(c(b(A(x))))) → C(A(x))
C(a(c(x))) → C(b(x))
C(x) → F(f(x))
G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
D(c(x)) → A¹(x)
C(x) → F(x)
G(x) → D(d(d(x)))
G(x) → C(x)
C(a(c(x))) → A¹(c(b(x)))
A¹(c(a(c(b(A(x)))))) → C(A(x))
G(x) → D(x)
A¹(c(a(c(b(A(x)))))) → C(a(b(b(c(A(x))))))
A¹(c(a(c(b(A(x)))))) → A¹(b(c(a(b(b(c(A(x))))))))
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
D(a(x)) → C(x)

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
a(c(a(c(b(A(x)))))) → b(a(b(c(a(b(b(c(A(x)))))))))
g(a(c(b(A(x))))) → b(g(a(b(c(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 1 SCC with 5 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                  ↳ QDP

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

G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
A¹(c(b(a(x)))) → C(a(b(x)))
G(x) → A¹(c(x))
G(x) → D(d(d(d(x))))
D(c(x)) → A¹(x)
C(x) → F(x)
D(c(x)) → A¹(a(x))
G(x) → D(d(d(x)))
F(f(a(x))) → G(x)
G(x) → C(x)
G(x) → D(d(x))
C(a(c(x))) → A¹(c(b(x)))
A¹(c(a(c(b(A(x)))))) → C(A(x))
G(x) → D(x)
A¹(c(a(c(b(A(x)))))) → C(a(b(b(c(A(x))))))
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
G(a(c(b(A(x))))) → C(A(x))
C(x) → F(f(x))
C(a(c(x))) → C(b(x))
D(a(x)) → C(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule D(c(x)) → A¹(a(x)) at position [0] we obtained the following new rules:

D(c(c(a(c(b(A(x0))))))) → A¹(b(a(b(c(a(b(b(c(A(x0))))))))))
D(c(c(b(a(x0))))) → A¹(b(a(b(c(a(b(x0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                  ↳ QDP

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

G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
G(x) → A¹(c(x))
A¹(c(b(a(x)))) → C(a(b(x)))
D(c(c(b(a(x0))))) → A¹(b(a(b(c(a(b(x0)))))))
G(x) → D(d(d(d(x))))
D(c(x)) → A¹(x)
C(x) → F(x)
F(f(a(x))) → G(x)
G(x) → D(d(d(x)))
G(x) → C(x)
D(c(c(a(c(b(A(x0))))))) → A¹(b(a(b(c(a(b(b(c(A(x0))))))))))
C(a(c(x))) → A¹(c(b(x)))
G(x) → D(d(x))
A¹(c(a(c(b(A(x)))))) → C(A(x))
A¹(c(a(c(b(A(x)))))) → C(a(b(b(c(A(x))))))
G(x) → D(x)
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
G(a(c(b(A(x))))) → C(A(x))
C(a(c(x))) → C(b(x))
C(x) → F(f(x))
D(a(x)) → C(x)

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
a(c(a(c(b(A(x)))))) → b(a(b(c(a(b(b(c(A(x)))))))))
g(a(c(b(A(x))))) → b(g(a(b(c(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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                  ↳ QDP

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

G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
G(x) → A¹(c(x))
A¹(c(b(a(x)))) → C(a(b(x)))
G(x) → D(d(d(d(x))))
D(c(x)) → A¹(x)
C(x) → F(x)
G(x) → D(d(d(x)))
F(f(a(x))) → G(x)
G(x) → C(x)
G(x) → D(d(x))
C(a(c(x))) → A¹(c(b(x)))
A¹(c(a(c(b(A(x)))))) → C(A(x))
G(x) → D(x)
A¹(c(a(c(b(A(x)))))) → C(a(b(b(c(A(x))))))
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
G(a(c(b(A(x))))) → C(A(x))
C(x) → F(f(x))
C(a(c(x))) → C(b(x))
D(a(x)) → C(x)

The TRS R consists of the following rules:

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

A¹(c(b(a(x)))) → C(a(b(x)))
D(c(x)) → A¹(x)
C(x) → F(x)
G(x) → D(d(d(x)))
F(f(a(x))) → G(x)
G(x) → C(x)
G(x) → D(d(x))
C(a(c(x))) → A¹(c(b(x)))
A¹(c(a(c(b(A(x)))))) → C(A(x))
G(x) → D(x)
A¹(c(a(c(b(A(x)))))) → C(a(b(b(c(A(x))))))
G(a(c(b(A(x))))) → C(A(x))
C(a(c(x))) → C(b(x))
D(a(x)) → C(x)

The remaining pairs can at least be oriented weakly.

G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
G(x) → A¹(c(x))
G(x) → D(d(d(d(x))))
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
C(x) → F(f(x))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 2·x₁
POL(A¹(x₁)) = 2 + x₁
POL(C(x₁)) = 5 + x₁
POL(D(x₁)) = 2 + x₁
POL(F(x₁)) = 2 + x₁
POL(G(x₁)) = 8 + x₁
POL(a(x₁)) = 4 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 2 + x₁
POL(f(x₁)) = 3 + x₁
POL(g(x₁)) = 10 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                  ↳ QDP

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

G(b(x)) → G(x)
C(a(c(x))) → C(b(a(c(b(x)))))
G(x) → A¹(c(x))
G(x) → D(d(d(d(x))))
G(a(c(b(A(x))))) → G(a(b(c(A(x)))))
C(x) → F(f(x))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
a(c(a(c(b(A(x)))))) → b(a(b(c(a(b(b(c(A(x)))))))))
g(a(c(b(A(x))))) → b(g(a(b(c(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 1 SCC with 5 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                  ↳ QDP

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

G(b(x)) → G(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

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

G(b(x)) → G(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

G(b(x)) → G(x)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(G(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPToSRSProof

                              ↳ QTRS

                                ↳ QTRS Reverse

                                  ↳ QTRS

                                    ↳ QTRS Reverse

                                    ↳ QTRS Reverse

                                    ↳ DependencyPairsProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ UsableRulesReductionPairsProof

                                                                  ↳ QDP

                                                                    ↳ PisEmptyProof

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

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

B(g(x1)) → G(b(x1))
C(a(c(x1))) → C(a(b(c(x1))))
C(a(c(x1))) → B(c(a(b(c(x1)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

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

C(a(c(y0))) → B(f(f(a(b(c(y0))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
C(a(c(x0))) → B(c(a(b(f(f(x0))))))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

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

C(a(c(y0))) → B(f(f(a(b(c(y0))))))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
B(g(x1)) → G(b(x1))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(x1))) → C(a(b(c(x1))))
C(a(c(x0))) → B(c(a(b(f(f(x0))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

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

C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
B(g(x1)) → G(b(x1))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(x1))) → C(a(b(c(x1))))
C(a(c(x0))) → B(c(a(b(f(f(x0))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

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

C(a(c(x0))) → C(a(b(f(f(x0)))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(d(x0)))) → C(a(b(a(a(x0)))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

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

C(a(c(x0))) → C(a(b(f(f(x0)))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
C(a(c(x0))) → B(c(a(b(f(f(x0))))))
C(a(c(d(x0)))) → C(a(b(a(a(x0)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

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

C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
C(a(c(x0))) → B(c(a(b(f(f(x0))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
C(a(c(d(x0)))) → C(a(b(a(a(x0)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(x0))) → B(c(a(b(f(f(x0)))))) at position [0] we obtained the following new rules:

C(a(c(y0))) → B(f(f(a(b(f(f(y0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

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

C(a(c(y0))) → B(f(f(a(b(f(f(y0)))))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
C(a(c(d(x0)))) → C(a(b(a(a(x0)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

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

C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
C(a(c(d(x0)))) → B(c(a(b(a(a(x0))))))
C(a(c(d(x0)))) → C(a(b(a(a(x0)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
G(x1) → C(a(x1))

Used ordering: Polynomial interpretation [25]:

POL(B(x₁)) = x₁
POL(C(x₁)) = 8 + x₁
POL(G(x₁)) = 13 + x₁
POL(a(x₁)) = 5 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 8 + x₁
POL(d(x₁)) = 3 + x₁
POL(f(x₁)) = 4 + x₁
POL(g(x₁)) = 13 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

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

B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

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

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(x1))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
G(x1) → C(a(x1))

Q is empty.

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

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(x1))))
B(g(x1)) → G(b(x1))
C(a(c(a(x0)))) → B(c(b(a(c(b(a(b(x0))))))))
C(a(c(a(x0)))) → C(b(a(c(b(a(b(x0)))))))
C(a(c(a(c(x0))))) → B(c(a(b(b(c(a(b(c(x0)))))))))
C(a(c(a(c(x0))))) → C(a(b(b(c(a(b(c(x0))))))))
G(x1) → C(a(x1))

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

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

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

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

Q is empty.

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

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

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

a(b(c(a(x)))) → b(a(c(b(a(b(x))))))
a(d(x)) → c(x)
a(f(f(x))) → g(x)
b(g(x)) → g(b(x))
c(x) → f(f(x))
c(a(c(x))) → b(c(a(b(c(x)))))
c(d(x)) → a(a(x))
g(x) → c(a(x))
g(x) → d(d(d(d(x))))
B(g(x)) → G(b(x))
C(a(c(a(x)))) → B(c(b(a(c(b(a(b(x))))))))
C(a(c(a(x)))) → C(b(a(c(b(a(b(x)))))))
C(a(c(a(c(x))))) → B(c(a(b(b(c(a(b(c(x)))))))))
C(a(c(a(c(x))))) → C(a(b(b(c(a(b(c(x))))))))
G(x) → C(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                          ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

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

a(b(c(a(x)))) → b(a(c(b(a(b(x))))))
a(d(x)) → c(x)
a(f(f(x))) → g(x)
b(g(x)) → g(b(x))
c(x) → f(f(x))
c(a(c(x))) → b(c(a(b(c(x)))))
c(d(x)) → a(a(x))
g(x) → c(a(x))
g(x) → d(d(d(d(x))))
B(g(x)) → G(b(x))
C(a(c(a(x)))) → B(c(b(a(c(b(a(b(x))))))))
C(a(c(a(x)))) → C(b(a(c(b(a(b(x)))))))
C(a(c(a(c(x))))) → B(c(a(b(b(c(a(b(c(x)))))))))
C(a(c(a(c(x))))) → C(a(b(b(c(a(b(c(x))))))))
G(x) → C(a(x))

Q is empty.

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

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

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

a(b(c(a(x)))) → b(a(c(b(a(b(x))))))
a(d(x)) → c(x)
a(f(f(x))) → g(x)
b(g(x)) → g(b(x))
c(x) → f(f(x))
c(a(c(x))) → b(c(a(b(c(x)))))
c(d(x)) → a(a(x))
g(x) → c(a(x))
g(x) → d(d(d(d(x))))
B(g(x)) → G(b(x))
C(a(c(a(x)))) → B(c(b(a(c(b(a(b(x))))))))
C(a(c(a(x)))) → C(b(a(c(b(a(b(x)))))))
C(a(c(a(c(x))))) → B(c(a(b(b(c(a(b(c(x)))))))))
C(a(c(a(c(x))))) → C(a(b(b(c(a(b(c(x))))))))
G(x) → C(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                          ↳ QTRS

                                                        ↳ DependencyPairsProof

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

a(b(c(a(x)))) → b(a(c(b(a(b(x))))))
a(d(x)) → c(x)
a(f(f(x))) → g(x)
b(g(x)) → g(b(x))
c(x) → f(f(x))
c(a(c(x))) → b(c(a(b(c(x)))))
c(d(x)) → a(a(x))
g(x) → c(a(x))
g(x) → d(d(d(d(x))))
B(g(x)) → G(b(x))
C(a(c(a(x)))) → B(c(b(a(c(b(a(b(x))))))))
C(a(c(a(x)))) → C(b(a(c(b(a(b(x)))))))
C(a(c(a(c(x))))) → B(c(a(b(b(c(a(b(c(x)))))))))
C(a(c(a(c(x))))) → C(a(b(b(c(a(b(c(x))))))))
G(x) → C(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:

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
A(c(a(C(x)))) → A(b(c(a(b(c(B(x)))))))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
A(c(a(C(x)))) → A(b(c(B(x))))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
G²(x) → A(C(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
C¹(a(c(a(C(x))))) → A(c(B(x)))
A(c(b(a(x)))) → A(b(c(a(b(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → A(c(b(b(a(C(x))))))
A(c(a(C(x)))) → A(b(c(a(b(C(x))))))
A(c(a(C(x)))) → A(b(C(x)))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
G¹(B(x)) → G²(x)
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))
A(c(b(a(x)))) → A(b(x))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
A(c(a(C(x)))) → A(b(c(a(b(c(B(x)))))))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
A(c(a(C(x)))) → A(b(c(B(x))))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
G²(x) → A(C(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
C¹(a(c(a(C(x))))) → A(c(B(x)))
A(c(b(a(x)))) → A(b(c(a(b(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → A(c(b(b(a(C(x))))))
A(c(a(C(x)))) → A(b(c(a(b(C(x))))))
A(c(a(C(x)))) → A(b(C(x)))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
G¹(B(x)) → G²(x)
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))
A(c(b(a(x)))) → A(b(x))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(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 1 SCC with 8 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
C¹(a(c(a(C(x))))) → A(c(B(x)))
G¹(x) → D(x)
D(a(x)) → C¹(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → A(c(b(b(a(C(x))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C¹(a(c(a(C(x))))) → A(c(B(x))) at position [0] we obtained the following new rules:

C¹(a(c(a(C(y0))))) → A(f(f(B(y0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
C¹(a(c(a(C(y0))))) → A(f(f(B(y0))))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
G¹(x) → D(d(d(d(x))))
C¹(x) → F(x)
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → A(c(b(b(a(C(x))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(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 1 SCC with 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
G¹(x) → D(x)
D(a(x)) → C¹(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → A(c(b(b(a(C(x))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C¹(a(c(a(C(x))))) → A(c(b(b(a(C(x)))))) at position [0] we obtained the following new rules:

C¹(a(c(a(C(y0))))) → A(f(f(b(b(a(C(y0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
C¹(a(c(a(C(y0))))) → A(f(f(b(b(a(C(y0)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
G¹(x) → D(d(d(d(x))))
C¹(x) → F(x)
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(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 1 SCC with 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
D(c(x)) → A(a(x))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
G¹(x) → D(x)
D(a(x)) → C¹(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

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

D(c(c(b(a(x0))))) → A(b(a(b(c(a(b(x0)))))))
D(c(c(a(C(x0))))) → A(b(a(b(c(a(b(c(B(x0)))))))))
D(c(c(a(C(x0))))) → A(b(a(b(c(a(b(C(x0))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

                                                                                  ↳ QDP

                                                                                    ↳ DependencyGraphProof

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
D(c(c(b(a(x0))))) → A(b(a(b(c(a(b(x0)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
D(c(c(a(C(x0))))) → A(b(a(b(c(a(b(c(B(x0)))))))))
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
G¹(x) → D(d(d(d(x))))
C¹(x) → F(x)
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
D(c(c(a(C(x0))))) → A(b(a(b(c(a(b(C(x0))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(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 1 SCC with 3 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

                                                                                  ↳ QDP

                                                                                    ↳ DependencyGraphProof

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

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

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
G¹(b(x)) → G¹(x)
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
F(f(a(x))) → G¹(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(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.

C¹(a(c(x))) → A(c(b(x)))
C¹(a(c(a(C(x))))) → C¹(b(b(a(c(B(x))))))
C¹(a(c(a(C(x))))) → C¹(b(b(a(C(x)))))
G¹(x) → D(d(d(x)))
G¹(x) → D(d(x))
C¹(a(c(a(C(x))))) → C¹(B(x))
C¹(a(c(a(C(x))))) → A(c(b(b(a(c(B(x)))))))
A(c(a(C(x)))) → C¹(a(b(c(B(x)))))
D(a(x)) → C¹(x)
G¹(x) → D(x)
A(c(a(C(x)))) → C¹(a(b(C(x))))
G¹(x) → A(c(x))
G¹(x) → C¹(x)
A(c(b(a(x)))) → C¹(a(b(x)))
A(c(a(C(x)))) → C¹(B(x))
C¹(x) → F(x)
G¹(x) → D(d(d(d(x))))
D(c(x)) → A(x)
C¹(a(c(x))) → C¹(b(x))

The remaining pairs can at least be oriented weakly.

G¹(b(x)) → G¹(x)
C¹(x) → F(f(x))
F(f(a(x))) → G¹(x)
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 1 + 2·x₁
POL(B(x₁)) = 2
POL(C(x₁)) = 8
POL(C¹(x₁)) = 6 + 2·x₁
POL(D(x₁)) = 2·x₁
POL(F(x₁)) = 2·x₁
POL(G(x₁)) = 12
POL(G¹(x₁)) = 14 + 2·x₁
POL(a(x₁)) = 4 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 2 + x₁
POL(f(x₁)) = 3 + x₁
POL(g(x₁)) = 10 + x₁

The following usable rules [17] were oriented:

G(x) → a(C(x))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
g(B(x)) → b(G(x))
g(x) → a(c(x))
d(c(x)) → a(a(x))
g(b(x)) → b(g(x))
c(a(c(x))) → c(b(a(c(b(x)))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(x) → f(f(x))
d(a(x)) → c(x)
g(x) → d(d(d(d(x))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
f(f(a(x))) → g(x)
a(c(b(a(x)))) → b(a(b(c(a(b(x))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

                                                                                  ↳ QDP

                                                                                    ↳ DependencyGraphProof

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

                                                                                          ↳ QDP

                                                                                            ↳ DependencyGraphProof

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

G¹(b(x)) → G¹(x)
C¹(a(c(x))) → C¹(b(a(c(b(x)))))
F(f(a(x))) → G¹(x)
C¹(x) → F(f(x))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(C(x))))))))
C¹(a(c(a(C(x))))) → C¹(b(a(c(b(b(a(c(B(x)))))))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(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 1 SCC with 5 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

                                                                                  ↳ QDP

                                                                                    ↳ DependencyGraphProof

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

                                                                                          ↳ QDP

                                                                                            ↳ DependencyGraphProof

                                                                                              ↳ QDP

                                                                                                ↳ UsableRulesProof

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

G¹(b(x)) → G¹(x)

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
d(a(x)) → c(x)
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
d(c(x)) → a(a(x))
g(x) → a(c(x))
g(x) → d(d(d(d(x))))
g(B(x)) → b(G(x))
a(c(a(C(x)))) → b(a(b(c(a(b(c(B(x))))))))
a(c(a(C(x)))) → b(a(b(c(a(b(C(x)))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(c(B(x)))))))))
c(a(c(a(C(x))))) → c(b(a(c(b(b(a(C(x))))))))
G(x) → a(C(x))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

                                                                                  ↳ QDP

                                                                                    ↳ DependencyGraphProof

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

                                                                                          ↳ QDP

                                                                                            ↳ DependencyGraphProof

                                                                                              ↳ QDP

                                                                                                ↳ UsableRulesProof

                                                                                                  ↳ QDP

                                                                                                    ↳ UsableRulesReductionPairsProof

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

G¹(b(x)) → G¹(x)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(G¹(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPToSRSProof

                                                  ↳ QTRS

                                                    ↳ QTRS Reverse

                                                      ↳ QTRS

                                                        ↳ QTRS Reverse

                                                        ↳ QTRS Reverse

                                                        ↳ DependencyPairsProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Narrowing

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ Narrowing

                                                                                  ↳ QDP

                                                                                    ↳ DependencyGraphProof

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

                                                                                          ↳ QDP

                                                                                            ↳ DependencyGraphProof

                                                                                              ↳ QDP

                                                                                                ↳ UsableRulesProof

                                                                                                  ↳ QDP

                                                                                                    ↳ UsableRulesReductionPairsProof

                                                                                                      ↳ QDP

                                                                                                        ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.