Termination w.r.t. Q proof of /tmp/tmpx6Yz

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(x1))) → a(l(c(c(r(x1)))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

L(r(a(x1))) → C(c(r(x1)))
A(l(x1)) → L(a(x1))
C(a(r(x1))) → A(x1)
A(c(x1)) → A(x1)
A(l(x1)) → A(x1)
L(r(a(x1))) → C(r(x1))
L(r(a(x1))) → A(l(c(c(r(x1)))))
A(c(x1)) → C(a(x1))
L(r(a(x1))) → L(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(x1))) → a(l(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A(l(x1)) → L(a(x1))
C(a(r(x1))) → A(x1)
A(c(x1)) → A(x1)
L(r(a(x1))) → A(l(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(x1))) → a(l(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 2 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

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

L(r(a(x1))) → A(l(c(c(r(x1)))))

The remaining pairs can at least be oriented weakly.

A(l(x1)) → L(a(x1))

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 1/32.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                  ↳ QDP

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

A(l(x1)) → L(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(x1))) → a(l(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

C(a(r(x1))) → A(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(x1))) → a(l(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.

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

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

POL(C(x₁)) = 1 + (2)x_1
POL(c(x₁)) = 4 + (4)x_1
POL(l(x₁)) = 0
POL(a(x₁)) = x_1
POL(A(x₁)) = 1/2 + (3)x_1
POL(r(x₁)) = (4)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

Q DP problem:
P is empty.
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(x1))) → a(l(c(c(r(x1)))))

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.