Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/SK90SLASH4.58.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:

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, 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:

I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(h₁(x), j₂(y, z))
F₂(x, g₁(y)) -> I₂(x, y)
I₂(h₁(x), j₂(j₂(y, z), 0)) -> J₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
I₂(x, j₂(y, z)) -> J₂(g₁(y), i₂(x, z))
J₂(g₁(x), g₁(y)) -> J₂(x, y)
I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(x, j₂(y, z))
I₂(x, j₂(y, z)) -> I₂(x, z)
F₂(x, g₁(y)) -> F₂(h₁(x), i₂(x, y))

The TRS R consists of the following rules:

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, y))

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:

I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(h₁(x), j₂(y, z))
F₂(x, g₁(y)) -> I₂(x, y)
I₂(h₁(x), j₂(j₂(y, z), 0)) -> J₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
I₂(x, j₂(y, z)) -> J₂(g₁(y), i₂(x, z))
J₂(g₁(x), g₁(y)) -> J₂(x, y)
I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(x, j₂(y, z))
I₂(x, j₂(y, z)) -> I₂(x, z)
F₂(x, g₁(y)) -> F₂(h₁(x), i₂(x, y))

The TRS R consists of the following rules:

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

J₂(g₁(x), g₁(y)) -> J₂(x, y)

The TRS R consists of the following rules:

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, 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 strictly oriented and are deleted.

J₂(g₁(x), g₁(y)) -> J₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
J₂(x1, x2) = x2
g₁(x1) = g₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, 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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(h₁(x), j₂(y, z))
I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(x, j₂(y, z))
I₂(x, j₂(y, z)) -> I₂(x, z)

The TRS R consists of the following rules:

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, 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 strictly oriented and are deleted.

I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(h₁(x), j₂(y, z))
I₂(h₁(x), j₂(j₂(y, z), 0)) -> I₂(x, j₂(y, z))
I₂(x, j₂(y, z)) -> I₂(x, z)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
I₂(x1, x2) = I₁(x2)
h₁(x1) = h
j₂(x1, x2) = j₂(x1, x2)
0 = 0
g₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

h > I₁
j₂ > I₁

The following usable rules [14] were oriented:

j₂(g₁(x), g₁(y)) -> g₁(j₂(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, 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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₂(x, g₁(y)) -> F₂(h₁(x), i₂(x, y))

The TRS R consists of the following rules:

f₂(x, g₁(y)) -> f₂(h₁(x), i₂(x, y))
i₂(x, j₂(0, 0)) -> g₁(0)
i₂(x, j₂(y, z)) -> j₂(g₁(y), i₂(x, z))
i₂(h₁(x), j₂(j₂(y, z), 0)) -> j₂(i₂(h₁(x), j₂(y, z)), i₂(x, j₂(y, z)))
j₂(g₁(x), g₁(y)) -> g₁(j₂(x, y))

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