Termination w.r.t. Q proof of /tmp/tmp4slWSp/while1.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:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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:

2¹(c(1(x1))) → R¹(1(x1))
2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
0¹(L(x1)) → 2¹(R(x1))
R¹(1(x1)) → 3¹(x1)
R¹(3(x1)) → 3¹(R(x1))
3¹(L(x1)) → 3¹(x1)
R¹(3(x1)) → R¹(x1)
R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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:

2¹(c(1(x1))) → R¹(1(x1))
2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
0¹(L(x1)) → 2¹(R(x1))
R¹(1(x1)) → 3¹(x1)
R¹(3(x1)) → 3¹(R(x1))
3¹(L(x1)) → 3¹(x1)
R¹(3(x1)) → R¹(x1)
R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

3¹(L(x1)) → 3¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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.

3¹(L(x1)) → 3¹(x1)

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

POL(3¹(x₁)) = (2)x_1
POL(L(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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

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

2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
R¹(3(x1)) → R¹(x1)
R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)

The remaining pairs can at least be oriented weakly.

2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
R¹(3(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(2¹(x₁)) = (2)x_1
POL(1(x₁)) = (2)x_1
POL(c(x₁)) = x_1
POL(0¹(x₁)) = (2)x_1
POL(3(x₁)) = (2)x_1
POL(2(x₁)) = 1/4 + (4)x_1
POL(L(x₁)) = x_1
POL(b(x₁)) = 0
POL(0(x₁)) = 1/4 + (4)x_1
POL(R¹(x₁)) = (1/2)x_1
POL(R(x₁)) = x_1

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
3(L(x1)) → L(3(x1))
R(b(x1)) → c(1(b(x1)))
0(L(x1)) → 2(R(x1))
2(c(1(x1))) → c(0(R(1(x1))))
3(c(x1)) → c(1(x1))
2(c(0(x1))) → c(0(0(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

2¹(L(x1)) → 2¹(x1)
2¹(c(1(x1))) → 0¹(R(1(x1)))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(0(x1))) → 0¹(0(x1))
R¹(3(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ QDPOrderProof

                    ↳ QDP

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

R¹(3(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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(3(x1)) → R¹(x1)

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

POL(3(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

                    ↳ QDP

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

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

2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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