Termination w.r.t. Q proof of TRS/SK904.02.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:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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:

I(*(x, y)) → I(x)
*¹(x, *(y, z)) → *¹(x, y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)
I(*(x, y)) → *¹(i(y), i(x))
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → K(*(*(i(x), y), x), *(*(i(x), z), x))
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
I(*(x, y)) → I(y)
*¹(x, *(y, z)) → *¹(*(x, y), z)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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:

I(*(x, y)) → I(x)
*¹(x, *(y, z)) → *¹(x, y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)
I(*(x, y)) → *¹(i(y), i(x))
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → K(*(*(i(x), y), x), *(*(i(x), z), x))
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
I(*(x, y)) → I(y)
*¹(x, *(y, z)) → *¹(*(x, y), z)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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:

I(*(x, y)) → I(x)
*¹(x, *(y, z)) → *¹(x, y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
I(*(x, y)) → *¹(i(y), i(x))
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → K(*(*(i(x), y), x), *(*(i(x), z), x))
I(*(x, y)) → I(y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
*¹(x, *(y, z)) → *¹(*(x, y), z)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

*¹(x, *(y, z)) → *¹(x, y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
*¹(x, *(y, z)) → *¹(*(x, y), z)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

I(*(x, y)) → I(x)
I(*(x, y)) → I(y)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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.

I(*(x, y)) → I(x)
I(*(x, y)) → I(y)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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.