Termination w.r.t. Q proof of /tmp/tmp3m1vsL/12.srs

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

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

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

Q is empty.

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

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(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:

L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))
A(l(x1)) → L(a(x1))
L(r(a(a(x1)))) → C(r(x1))
C(a(r(x1))) → A(x1)
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(c(x1)) → A(x1)
L(r(a(a(x1)))) → L(c(c(c(r(x1)))))
L(r(a(a(x1)))) → C(c(r(x1)))
A(l(x1)) → A(x1)
A(c(x1)) → C(a(x1))
L(r(a(a(x1)))) → C(c(c(r(x1))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))
A(l(x1)) → L(a(x1))
L(r(a(a(x1)))) → C(r(x1))
C(a(r(x1))) → A(x1)
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(c(x1)) → A(x1)
L(r(a(a(x1)))) → L(c(c(c(r(x1)))))
L(r(a(a(x1)))) → C(c(r(x1)))
A(l(x1)) → A(x1)
A(c(x1)) → C(a(x1))
L(r(a(a(x1)))) → C(c(c(r(x1))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(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 4 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))
A(l(x1)) → L(a(x1))
C(a(r(x1))) → A(x1)
A(c(x1)) → A(x1)
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(l(x1)) → A(x1)
A(c(x1)) → C(a(x1))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A(l(x1)) → A(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(C(x₁)) = x₁
POL(L(x₁)) = 2 + 2·x₁
POL(a(x₁)) = x₁
POL(c(x₁)) = x₁
POL(l(x₁)) = 2 + 2·x₁
POL(r(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

A(l(x1)) → L(a(x1))
L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))
C(a(r(x1))) → A(x1)
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(c(x1)) → A(x1)
A(c(x1)) → C(a(x1))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

C(a(r(x1))) → A(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(C(x₁)) = x₁
POL(L(x₁)) = x₁
POL(a(x₁)) = x₁
POL(c(x₁)) = x₁
POL(l(x₁)) = x₁
POL(r(x₁)) = 1 + x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))
A(l(x1)) → L(a(x1))
A(c(x1)) → A(x1)
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(c(x1)) → C(a(x1))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))
A(l(x1)) → L(a(x1))
A(c(x1)) → A(x1)
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

L(r(a(a(x1)))) → A(l(c(c(c(r(x1))))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 2 + 2·x₁
POL(L(x₁)) = x₁
POL(a(x₁)) = 2 + 2·x₁
POL(c(x₁)) = x₁
POL(l(x₁)) = x₁
POL(r(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

A(l(x1)) → L(a(x1))
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(c(x1)) → A(x1)

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(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(c(x1)) → A(x1)

The remaining pairs can at least be oriented weakly.

A(l(x1)) → L(a(x1))
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))

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

POL( A(x₁) ) = x₁

POL( r(x₁) ) = max{0, x₁ - 1}

POL( c(x₁) ) = x₁ + 1

POL( l(x₁) ) = max{0, -1}

POL( a(x₁) ) = x₁

POL( L(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))
a(l(x1)) → l(a(x1))
c(a(r(x1))) → r(a(x1))
a(c(x1)) → c(a(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

  ↳ QTRS Reverse

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

A(l(x1)) → L(a(x1))
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

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

A(l(l(x0))) → L(l(a(x0)))
A(l(c(x0))) → L(c(a(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

  ↳ QTRS Reverse

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

A(l(c(x0))) → L(c(a(x0)))
L(r(a(a(x1)))) → A(a(l(c(c(c(r(x1)))))))
A(l(l(x0))) → L(l(a(x0)))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

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

L(r(a(a(y0)))) → A(l(a(c(c(c(r(y0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

  ↳ QTRS Reverse

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

A(l(c(x0))) → L(c(a(x0)))
A(l(l(x0))) → L(l(a(x0)))
L(r(a(a(y0)))) → A(l(a(c(c(c(r(y0)))))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(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(l(l(x0))) → L(l(a(x0)))

The remaining pairs can at least be oriented weakly.

A(l(c(x0))) → L(c(a(x0)))
L(r(a(a(y0)))) → A(l(a(c(c(c(r(y0)))))))

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

POL( A(x₁) ) = 1

POL( r(x₁) ) = 1

POL( c(x₁) ) = 1

POL( l(x₁) ) = 0

POL( a(x₁) ) = x₁

POL( L(x₁) ) = x₁

The following usable rules [17] were oriented:

l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))
a(l(x1)) → l(a(x1))
c(a(r(x1))) → r(a(x1))
a(c(x1)) → c(a(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

  ↳ QTRS Reverse

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

A(l(c(x0))) → L(c(a(x0)))
L(r(a(a(y0)))) → A(l(a(c(c(c(r(y0)))))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))
A(l(c(x0))) → L(c(a(x0)))
L(r(a(a(y0)))) → A(l(a(c(c(c(r(y0)))))))

Q is empty.

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

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))
A(l(c(x0))) → L(c(a(x0)))
L(r(a(a(y0)))) → A(l(a(c(c(c(r(y0)))))))

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

Q is empty.

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

a(l(x)) → l(a(x))
a(c(x)) → c(a(x))
c(a(r(x))) → r(a(x))
l(r(a(a(x)))) → a(a(l(c(c(c(r(x)))))))
A(l(c(x))) → L(c(a(x)))
L(r(a(a(x)))) → A(l(a(c(c(c(r(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                    ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(l(x)) → l(a(x))
a(c(x)) → c(a(x))
c(a(r(x))) → r(a(x))
l(r(a(a(x)))) → a(a(l(c(c(c(r(x)))))))
A(l(c(x))) → L(c(a(x)))
L(r(a(a(x)))) → A(l(a(c(c(c(r(x)))))))

Q is empty.

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

a(l(x)) → l(a(x))
a(c(x)) → c(a(x))
c(a(r(x))) → r(a(x))
l(r(a(a(x)))) → a(a(l(c(c(c(r(x)))))))
A(l(c(x))) → L(c(a(x)))
L(r(a(a(x)))) → A(l(a(c(c(c(r(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                    ↳ QTRS

                                                  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(l(x)) → l(a(x))
a(c(x)) → c(a(x))
c(a(r(x))) → r(a(x))
l(r(a(a(x)))) → a(a(l(c(c(c(r(x)))))))
A(l(c(x))) → L(c(a(x)))
L(r(a(a(x)))) → A(l(a(c(c(c(r(x)))))))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C(a(x)) → A¹(c(x))
A¹(a(r(L(x)))) → R(c(c(c(a(l(A(x)))))))
C(a(x)) → C(x)
L¹(a(x)) → A¹(l(x))
A¹(a(r(L(x)))) → C(c(c(a(l(A(x))))))
A¹(a(r(l(x)))) → C(l(a(a(x))))
A¹(a(r(l(x)))) → C(c(c(l(a(a(x))))))
A¹(a(r(L(x)))) → C(c(a(l(A(x)))))
L¹(a(x)) → L¹(x)
A¹(a(r(L(x)))) → L¹(A(x))
A¹(a(r(L(x)))) → C(a(l(A(x))))
C(l(A(x))) → A¹(c(L(x)))
A¹(a(r(l(x)))) → A¹(x)
R(a(c(x))) → R(x)
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
A¹(a(r(l(x)))) → A¹(a(x))
R(a(c(x))) → A¹(r(x))
A¹(a(r(L(x)))) → A¹(l(A(x)))
A¹(a(r(l(x)))) → L¹(a(a(x)))
C(l(A(x))) → C(L(x))
A¹(a(r(l(x)))) → C(c(l(a(a(x)))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

C(a(x)) → A¹(c(x))
A¹(a(r(L(x)))) → R(c(c(c(a(l(A(x)))))))
C(a(x)) → C(x)
L¹(a(x)) → A¹(l(x))
A¹(a(r(L(x)))) → C(c(c(a(l(A(x))))))
A¹(a(r(l(x)))) → C(l(a(a(x))))
A¹(a(r(l(x)))) → C(c(c(l(a(a(x))))))
A¹(a(r(L(x)))) → C(c(a(l(A(x)))))
L¹(a(x)) → L¹(x)
A¹(a(r(L(x)))) → L¹(A(x))
A¹(a(r(L(x)))) → C(a(l(A(x))))
C(l(A(x))) → A¹(c(L(x)))
A¹(a(r(l(x)))) → A¹(x)
R(a(c(x))) → R(x)
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
A¹(a(r(l(x)))) → A¹(a(x))
R(a(c(x))) → A¹(r(x))
A¹(a(r(L(x)))) → A¹(l(A(x)))
A¹(a(r(l(x)))) → L¹(a(a(x)))
C(l(A(x))) → C(L(x))
A¹(a(r(l(x)))) → C(c(l(a(a(x)))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(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 4 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

C(a(x)) → A¹(c(x))
A¹(a(r(L(x)))) → R(c(c(c(a(l(A(x)))))))
C(a(x)) → C(x)
L¹(a(x)) → A¹(l(x))
A¹(a(r(L(x)))) → C(c(c(a(l(A(x))))))
A¹(a(r(l(x)))) → C(l(a(a(x))))
A¹(a(r(l(x)))) → C(c(c(l(a(a(x))))))
A¹(a(r(L(x)))) → C(c(a(l(A(x)))))
L¹(a(x)) → L¹(x)
A¹(a(r(L(x)))) → C(a(l(A(x))))
A¹(a(r(l(x)))) → A¹(x)
R(a(c(x))) → R(x)
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
A¹(a(r(l(x)))) → A¹(a(x))
R(a(c(x))) → A¹(r(x))
A¹(a(r(l(x)))) → L¹(a(a(x)))
A¹(a(r(l(x)))) → C(c(l(a(a(x)))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

A¹(a(r(L(x)))) → C(c(c(a(l(A(x))))))
A¹(a(r(L(x)))) → C(c(a(l(A(x)))))
A¹(a(r(L(x)))) → C(a(l(A(x))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + x₁
POL(A¹(x₁)) = x₁
POL(C(x₁)) = x₁
POL(L(x₁)) = 2 + 2·x₁
POL(L¹(x₁)) = 2·x₁
POL(R(x₁)) = 2·x₁
POL(a(x₁)) = x₁
POL(c(x₁)) = x₁
POL(l(x₁)) = 2·x₁
POL(r(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

                                                            ↳ QDP

                                                              ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

C(a(x)) → A¹(c(x))
A¹(a(r(L(x)))) → R(c(c(c(a(l(A(x)))))))
C(a(x)) → C(x)
L¹(a(x)) → A¹(l(x))
A¹(a(r(l(x)))) → C(l(a(a(x))))
A¹(a(r(l(x)))) → C(c(c(l(a(a(x))))))
L¹(a(x)) → L¹(x)
A¹(a(r(l(x)))) → A¹(x)
R(a(c(x))) → R(x)
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
A¹(a(r(l(x)))) → A¹(a(x))
R(a(c(x))) → A¹(r(x))
A¹(a(r(l(x)))) → L¹(a(a(x)))
A¹(a(r(l(x)))) → C(c(l(a(a(x)))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

C(a(x)) → C(x)
L¹(a(x)) → L¹(x)
A¹(a(r(l(x)))) → A¹(x)
R(a(c(x))) → R(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + 2·x₁
POL(A¹(x₁)) = 1 + 2·x₁
POL(C(x₁)) = x₁
POL(L(x₁)) = x₁
POL(L¹(x₁)) = x₁
POL(R(x₁)) = x₁
POL(a(x₁)) = 1 + 2·x₁
POL(c(x₁)) = x₁
POL(l(x₁)) = x₁
POL(r(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

                                                            ↳ QDP

                                                              ↳ RuleRemovalProof

                                                                ↳ QDP

                                                                  ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

C(a(x)) → A¹(c(x))
A¹(a(r(L(x)))) → R(c(c(c(a(l(A(x)))))))
A¹(a(r(l(x)))) → A¹(a(x))
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
L¹(a(x)) → A¹(l(x))
A¹(a(r(l(x)))) → C(l(a(a(x))))
R(a(c(x))) → A¹(r(x))
A¹(a(r(l(x)))) → C(c(c(l(a(a(x))))))
A¹(a(r(l(x)))) → L¹(a(a(x)))
A¹(a(r(l(x)))) → C(c(l(a(a(x)))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

C(a(x)) → A¹(c(x))
A¹(a(r(l(x)))) → A¹(a(x))
L¹(a(x)) → A¹(l(x))
A¹(a(r(l(x)))) → C(l(a(a(x))))
A¹(a(r(l(x)))) → C(c(c(l(a(a(x))))))
A¹(a(r(l(x)))) → L¹(a(a(x)))
A¹(a(r(l(x)))) → C(c(l(a(a(x)))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + 2·x₁
POL(A¹(x₁)) = 2·x₁
POL(C(x₁)) = x₁
POL(L(x₁)) = x₁
POL(L¹(x₁)) = 2 + x₁
POL(R(x₁)) = 2·x₁
POL(a(x₁)) = 1 + 2·x₁
POL(c(x₁)) = x₁
POL(l(x₁)) = x₁
POL(r(x₁)) = 1 + 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

                                                            ↳ QDP

                                                              ↳ RuleRemovalProof

                                                                ↳ QDP

                                                                  ↳ RuleRemovalProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

  ↳ QTRS Reverse

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

A¹(a(r(L(x)))) → R(c(c(c(a(l(A(x)))))))
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
R(a(c(x))) → A¹(r(x))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

A¹(a(r(L(y0)))) → R(c(c(a(c(l(A(y0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

                                                            ↳ QDP

                                                              ↳ RuleRemovalProof

                                                                ↳ QDP

                                                                  ↳ RuleRemovalProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ Narrowing

  ↳ QTRS Reverse

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

A¹(a(r(L(y0)))) → R(c(c(a(c(l(A(y0)))))))
A¹(a(r(l(x)))) → R(c(c(c(l(a(a(x)))))))
R(a(c(x))) → A¹(r(x))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

A¹(a(r(l(a(r(l(x0))))))) → R(c(c(c(l(a(r(c(c(c(l(a(a(x0)))))))))))))
A¹(a(r(l(a(r(L(x0))))))) → R(c(c(c(l(a(r(c(c(c(a(l(A(x0)))))))))))))
A¹(a(r(l(r(l(x0)))))) → R(c(c(c(l(r(c(c(c(l(a(a(x0))))))))))))
A¹(a(r(l(r(L(x0)))))) → R(c(c(c(l(r(c(c(c(a(l(A(x0))))))))))))
A¹(a(r(l(y0)))) → R(c(c(c(a(l(a(y0)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

                                                            ↳ QDP

                                                              ↳ RuleRemovalProof

                                                                ↳ QDP

                                                                  ↳ RuleRemovalProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ Narrowing

                                                                            ↳ QDP

                                                                              ↳ Narrowing

  ↳ QTRS Reverse

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

A¹(a(r(l(r(l(x0)))))) → R(c(c(c(l(r(c(c(c(l(a(a(x0))))))))))))
A¹(a(r(L(y0)))) → R(c(c(a(c(l(A(y0)))))))
A¹(a(r(l(a(r(l(x0))))))) → R(c(c(c(l(a(r(c(c(c(l(a(a(x0)))))))))))))
A¹(a(r(l(a(r(L(x0))))))) → R(c(c(c(l(a(r(c(c(c(a(l(A(x0)))))))))))))
A¹(a(r(l(r(L(x0)))))) → R(c(c(c(l(r(c(c(c(a(l(A(x0))))))))))))
R(a(c(x))) → A¹(r(x))
A¹(a(r(l(y0)))) → R(c(c(c(a(l(a(y0)))))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

R(a(c(a(c(x0))))) → A¹(a(r(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPToSRSProof

                                            ↳ QTRS

                                              ↳ QTRS Reverse

                                                ↳ QTRS

                                                  ↳ QTRS Reverse

                                                  ↳ QTRS Reverse

                                                  ↳ DependencyPairsProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ RuleRemovalProof

                                                            ↳ QDP

                                                              ↳ RuleRemovalProof

                                                                ↳ QDP

                                                                  ↳ RuleRemovalProof

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ Narrowing

                                                                            ↳ QDP

                                                                              ↳ Narrowing

                                                                                ↳ QDP

  ↳ QTRS Reverse

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

A¹(a(r(l(r(l(x0)))))) → R(c(c(c(l(r(c(c(c(l(a(a(x0))))))))))))
A¹(a(r(L(y0)))) → R(c(c(a(c(l(A(y0)))))))
R(a(c(a(c(x0))))) → A¹(a(r(x0)))
A¹(a(r(l(a(r(l(x0))))))) → R(c(c(c(l(a(r(c(c(c(l(a(a(x0)))))))))))))
A¹(a(r(l(a(r(L(x0))))))) → R(c(c(c(l(a(r(c(c(c(a(l(A(x0)))))))))))))
A¹(a(r(l(r(L(x0)))))) → R(c(c(c(l(r(c(c(c(a(l(A(x0))))))))))))
A¹(a(r(l(y0)))) → R(c(c(c(a(l(a(y0)))))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))
c(l(A(x))) → a(c(L(x)))
a(a(r(L(x)))) → r(c(c(c(a(l(A(x)))))))

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

a(l(x1)) → l(a(x1))
a(c(x1)) → c(a(x1))
c(a(r(x1))) → r(a(x1))
l(r(a(a(x1)))) → a(a(l(c(c(c(r(x1)))))))

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

l(a(x)) → a(l(x))
c(a(x)) → a(c(x))
r(a(c(x))) → a(r(x))
a(a(r(l(x)))) → r(c(c(c(l(a(a(x)))))))

Q is empty.