Termination w.r.t. Q proof of /tmp/tmpy-xkW6/12.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(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

  ↳ DependencyPairsProof

  ↳ 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)))))))

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ 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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

A(l(x1)) → L(a(x1))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = (1/2)x_1
POL(l(x₁)) = 1/2 + (4)x_1
POL(a(x₁)) = (2)x_1
POL(L(x₁)) = 1/4 + (2)x_1
POL(A(x₁)) = x_1
POL(r(x₁)) = 3/4

The value of delta used in the strict ordering is 1/4.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
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

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ 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.

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

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ 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.