Termination w.r.t. Q proof of /tmp/tmpxQshKo/z100.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:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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:

R2(a(x1)) → A(a(a(r2(x1))))
B(l2(x1)) → R1(x1)
A(l1(x1)) → A(a(x1))
R2(b(x1)) → A(b(x1))
R2(a(x1)) → R2(x1)
A(a(l2(x1))) → A(x1)
B(l1(x1)) → R2(x1)
A(l1(x1)) → A(x1)
R2(a(x1)) → A(r2(x1))
R2(a(x1)) → A(a(r2(x1)))
R1(a(x1)) → A(a(a(r1(x1))))
R1(a(x1)) → A(r1(x1))
R1(a(x1)) → A(a(r1(x1)))
B(l2(x1)) → B(r1(x1))
B(l1(x1)) → B(r2(x1))
R1(a(x1)) → R1(x1)
A(l1(x1)) → A(a(a(x1)))
A(a(l2(x1))) → A(a(x1))

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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:

R2(a(x1)) → A(a(a(r2(x1))))
B(l2(x1)) → R1(x1)
A(l1(x1)) → A(a(x1))
R2(b(x1)) → A(b(x1))
R2(a(x1)) → R2(x1)
A(a(l2(x1))) → A(x1)
B(l1(x1)) → R2(x1)
A(l1(x1)) → A(x1)
R2(a(x1)) → A(r2(x1))
R2(a(x1)) → A(a(r2(x1)))
R1(a(x1)) → A(a(a(r1(x1))))
R1(a(x1)) → A(r1(x1))
R1(a(x1)) → A(a(r1(x1)))
B(l2(x1)) → B(r1(x1))
B(l1(x1)) → B(r2(x1))
R1(a(x1)) → R1(x1)
A(l1(x1)) → A(a(a(x1)))
A(a(l2(x1))) → A(a(x1))

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

A(l1(x1)) → A(a(x1))
A(a(l2(x1))) → A(x1)
A(l1(x1)) → A(a(a(x1)))
A(a(l2(x1))) → A(a(x1))
A(l1(x1)) → A(x1)

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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(a(l2(x1))) → A(x1)
A(a(l2(x1))) → A(a(x1))

The remaining pairs can at least be oriented weakly.

A(l1(x1)) → A(a(x1))
A(l1(x1)) → A(a(a(x1)))
A(l1(x1)) → A(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(l2(x₁)) = 1/4 + (5/2)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = (1/2)x_1
POL(l1(x₁)) = x_1

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
a(a(x1)) → x1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

A(l1(x1)) → A(a(x1))
A(l1(x1)) → A(a(a(x1)))
A(l1(x1)) → A(x1)

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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(l1(x1)) → A(a(x1))
A(l1(x1)) → A(a(a(x1)))
A(l1(x1)) → A(x1)

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

POL(l2(x₁)) = 1/4 + (3)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = (4)x_1
POL(l1(x₁)) = 1/4 + (11/4)x_1

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

a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
a(a(x1)) → x1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

R2(a(x1)) → R2(x1)

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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.

R2(a(x1)) → R2(x1)

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

POL(R2(x₁)) = (2)x_1
POL(a(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

R1(a(x1)) → R1(x1)

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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.

R1(a(x1)) → R1(x1)

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

POL(a(x₁)) = 1/4 + (7/2)x_1
POL(R1(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

B(l1(x1)) → B(r2(x1))
B(l2(x1)) → B(r1(x1))

The TRS R consists of the following rules:

r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1

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