Termination w.r.t. Q proof of TRS/various12.trs

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:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

*¹(I(x), y) → *¹(x, y)
+¹(I(x), I(y)) → O¹(+(+(x, y), I(0)))
+¹(O(x), O(y)) → O¹(+(x, y))
+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), I(y)) → +¹(x, y)
*¹(I(x), y) → +¹(O(*(x, y)), y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
*¹(O(x), y) → *¹(x, y)
+¹(I(x), I(y)) → +¹(x, y)
*¹(I(x), y) → O¹(*(x, y))
+¹(O(x), O(y)) → +¹(x, y)
*¹(O(x), y) → O¹(*(x, y))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

*¹(I(x), y) → *¹(x, y)
+¹(I(x), I(y)) → O¹(+(+(x, y), I(0)))
+¹(O(x), O(y)) → O¹(+(x, y))
+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), I(y)) → +¹(x, y)
*¹(I(x), y) → +¹(O(*(x, y)), y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
*¹(O(x), y) → *¹(x, y)
+¹(I(x), I(y)) → +¹(x, y)
*¹(I(x), y) → O¹(*(x, y))
+¹(O(x), O(y)) → +¹(x, y)
*¹(O(x), y) → O¹(*(x, y))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

+¹(O(x), O(y)) → O¹(+(x, y))
+¹(O(x), I(y)) → +¹(x, y)
+¹(I(x), O(y)) → +¹(x, y)
+¹(I(x), I(y)) → +¹(x, y)
*¹(O(x), y) → *¹(x, y)
*¹(I(x), y) → *¹(x, y)
+¹(I(x), I(y)) → O¹(+(+(x, y), I(0)))
*¹(I(x), y) → +¹(O(*(x, y)), y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(O(x), O(y)) → +¹(x, y)
*¹(I(x), y) → O¹(*(x, y))
*¹(O(x), y) → O¹(*(x, y))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), I(y)) → +¹(x, y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(I(x), I(y)) → +¹(x, y)
+¹(O(x), O(y)) → +¹(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

+¹(O(x), I(y)) → +¹(x, y)
+¹(I(x), I(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.

+¹(I(x), O(y)) → +¹(x, y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(O(x), O(y)) → +¹(x, y)

Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x2)
I(x1) = I(x1)
O(x1) = x1
+(x1, x2) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

+^1₁ > I₁ > 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

+¹(I(x), O(y)) → +¹(x, y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(O(x), O(y)) → +¹(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

              ↳ QDP

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

+¹(I(x), I(y)) → +¹(+(x, y), I(0))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ QDPOrderProof

              ↳ QDP

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

+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), O(y)) → +¹(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), O(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x2)
I(x1) = I
O(x1) = O(x1)

Lexicographic Path Order [19].
Precedence:

I > +^1₁
O₁ > +^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ PisEmptyProof

              ↳ QDP

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

*¹(I(x), y) → *¹(x, y)
*¹(O(x), y) → *¹(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

*¹(O(x), y) → *¹(x, y)

The remaining pairs can at least be oriented weakly.

*¹(I(x), y) → *¹(x, y)

Used ordering: Combined order from the following AFS and order.
*¹(x1, x2) = x1
I(x1) = x1
O(x1) = O(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

*¹(I(x), y) → *¹(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

*¹(I(x), y) → *¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*¹(x1, x2) = *¹(x1)
I(x1) = I(x1)

Lexicographic Path Order [19].
Precedence:

I₁ > *^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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.